Mixing
Proof that $sumlimits _n=1^infty nsumlimits _j=2^infty frac (-1)^j-1j^2 left( 1-j^-1 right)^n-1 = -frac12$
Clash Royale CLAN TAG#URR8PPP
up vote
5
down vote
favorite
How can we prove the following?
$-1+frac112pi ^2-frac12sumlimits_n=2^infty Gamma left( n+1 right) sumlimits_k=0^n+2
,frac zeta left( k right) left( -
1 right) ^-2+k left( 2^2-k-2 right) Gamma left( -1+k
right) left( n+1-k right) ! =sumlimits_n=1^infty n
sumlimits_j=2^infty frac left( -1 right) ^j-1j^
2 left( 1-j^-1 right) ^n-1 =-frac12$
We also have
$chi(n)=-sumlimits_k=0^n+2zeta left( k
right) left( k-n-2 right) nchoose k-2 left( left( -1
right) ^1+k2^1-k+ left( -1 right) ^k right) $ by replacing the $Gamma$ factors with the corresponding binomial occeficient
Related question at References for $ chi(n)=nsumlimits_j=2^inftyfrac (-1)^j-1j^2left(1-j^-1right)^n-1$ in $zeta$ expansion?
Digits := 37; flist(proc (n) options operator, arrow; evalf(coolchi(n)) end proc, 0 .. 75);
Digits := 37
[
[0., -0.1775329665758867817637924166769874055,
-0.158151287891164991627192075621149797,
-0.100756475454096876860689273411921431,
-0.052729122560581908477394880842614692,
-0.022173447806981260740844120495648785,
-0.005724758139923986695354623118140584,
0.001711181525440404872086057781767772,
0.004163504754435619487164038876686477,
0.004224146796392587120306729731994248,
0.003361436685455899377754627000698078,
0.00231507684250186191279422185478205,
0.00140312527387483007618661260684936,
0.00072272177218128957979732200349640,
0.00026840905319980399461855630763639,
-0.00000386786514313326830151239928860,
-0.00014529652172375174501064342971419,
-0.00020079200685610596363763783920190,
-0.00020486397550091241285507339640177,
-0.00018177251376674866978557090651028,
-0.00014732720568299891079533324844990,
-0.00011103259250611536839455447354313,
-0.00007801852557445637063802766476294,
-0.00005057002459089850935720449502326,
-0.00002924737982070438622370684257219,
-0.0000136577182218588710540784871811,
-0.0000029575122625340335255970335431,
0.0000038395199899767419714034429197,
0.0000076881193734840299247444522077,
0.0000094245398419219327153653943207,
0.0000097337529690318816412614692587,
0.0000091480274459270312096627770217,
0.0000080619145517535381759258176097,
0.0000067542476407336638941225910227,
0.000005411593443449269790648097261,
0.000004150125437131325581547336324,
0.000003034497471576312110001308212,
0.000002093267538135562252003012925,
0.000001330969825279401198332450463,
-7
7.37209507955248062779620826 10 ,
-7
2.93263622398134958434524015 10 ,
-8
-2.3317072362848976102839973 10 ,
-7
-2.35650675570634484892471134 10 ,
-7
-3.65620346716670870284369471 10 ,
-7
-4.32803484845776951918534983 10 ,
-7
-4.53980382773092409894967670 10 ,
-7
-4.43028271969806636513667532 10 ,
-7
-4.11056601063391173974634568 10 ,
-7
-3.66679744724574133777868780 10 ,
-7
-3.16354929829636515323370167 10 ,
-7
-2.64737199886948816611138728 10 ,
-7
-2.15021028592691266141174466 10 ,
-7
-1.69250991341964200913477377 10 ,
-7
-1.28592821342563655928047463 10 ,
-8
-9.3562155186891501184884726 10 ,
-8
-6.4212079021714840683989162 10 ,
-8
-4.0282829593958497425360774 10 ,
-8
-2.1318155241615525408999560 10 ,
-9
-6.753256088468931634905523 10 ,
-9
4.020837075521115896921341 10 ,
-8
1.1611930078301706186481030 10 ,
-8
1.6596359914684719233773543 10 ,
-8
1.9500255425207775038798882 10 ,
-8
2.0789463805547333601557044 10 ,
-8
2.0865770984321044922048033 10 ,
-8
2.0067518073257429000271847 10 ,
-8
1.8673131375018625836228485 10 ,
-8
1.6906432963871255429917949 10 ,
-8
1.4942881292166557781340237 10 ,
-8
1.2916159703946332890495351 10 ,
-8
1.0924629207316840757383290 10 ,
-9
9.037454965393441382004053 10 ,
-9
7.300036840122634764357642 10 ,
-9
5.739399681529520904444054 10 ,
-9 -9
4.367571268906899802263293 10 , 3.185931532251771457815357 10
]
]
listsum((20))= -0.4999999932944069400650356605249833795
Note: when evaluating this sum, at appears at least 28 digits of precision are required for accurate evaluation. If you use less than this (Maple defaults to 10) then the result will diverge quite wildly to an inaccurate answer.
This table is the partial sums of the terms, so I think the equality =-1/2 is correct
[0., -.1775329665758867817637924166769874055, -.3356842544670517733909844922981372025, -.4364407299211486502516737657100586335, -.4891698524817305587290686465526733255, -.5113433002887118194699127670483221105, -.5170680584286358061652673901664626945, -.5153568769031954012931813323846949225, -.5111933721487597818060172935080084455, -.5069692253523671946857105637760141975, -.5036077886669112953079559367753161195, -.5012927118244094333951617149205340695, -.4998895865505346033189751023136847095, -.4991668647783533137391777803101883095, -.4988984557251535097445592240025519195, -.4989023235902966430128607364018405195, -.4990476201120203947578713798315547095, -.4992484121188765007215090176707566095, -.4994532760943774131343640910671583795, -.4996350486081441618041496619736686595, -.4997823758138271607149449952221185595, -.4998934084063332760833395496956616895, -.4999714269319077324539775773604246295, -.5000219969564986309633347818554478895, -.5000512443363193353495584886980200795, -.5000649020545411942206125671852011795, -.5000678595668037282541381642187442795, -.5000640200468137515121667607758245795, -.5000563319274402674822420163236168795, -.5000469073875983455495266509292961795, -.5000371736346293136678853894600374795, -.5000280256071833866366757266830157795, -.5000199636926316330984998008654060795, -.5000132094449908994346056782743833795, -.5000077978515474501648150301771223795, -.5000036477261103188392334828407983795, -.5000006132286387425271234815325863795, -.4999985199611006069648714785196613795, -.4999971889912753275636731460691983795, -.4999964517817673723156103664483723795, -.4999961585181449741806519319243573795, -.4999961818352173370296280347643303795, -.4999964174858929076641129272354643795, -.4999967831062396243349832116049353795, -.4999972159097244701119351301399183795, -.4999976698901072432043450251075883795, -.4999981129183792130109815387751203795, -.4999985239749802764021555134096883795, -.4999988906547250009762892912784683795, -.4999992070096548306128046146486353795, -.4999994717468547175616212257873633795, -.4999996867678833102528873669618293795, -.4999998560188746522170882804392063795, -.4999999846116959947807442084866693795, -.5000000781738511816722453933713953795, -.5000001423859302033870860773605573795, -.5000001826687597973455835027213313795, -.5000002039869150389611089117208913795, -.5000002107401711274300405466264143795, -.5000002067193340519089246497050733795, -.5000001951074039736072184632240433795, -.5000001785110440589224992294505003795, -.5000001590107886337147241906516183795, -.5000001382213248281673905890945743795, -.5000001173555538438463456670465413795, -.5000000972880357705889166667746943795, -.5000000786149043955702908305462093795, -.5000000617084714316990354006282603795, -.5000000467655901395324776192880233795, -.5000000338494304355861447287926723795, -.5000000229248012282693039714093823795, -.5000000138873462628758625894053293795, -.5000000065873094227532278250476873795, -.5000000008479097412237069206036333795, -.4999999964803384723168071183403403795, -.4999999932944069400650356605249833795]
sequences-and-series taylor-expansion binomial-coefficients riemann-zeta binomial-theorem
asked Jul 15 at 20:45
crow
492416
 |Â
show 8 more comments
up vote
5
down vote
favorite
How can we prove the following?
$-1+frac112pi ^2-frac12sumlimits_n=2^infty Gamma left( n+1 right) sumlimits_k=0^n+2
,frac zeta left( k right) left( -
1 right) ^-2+k left( 2^2-k-2 right) Gamma left( -1+k
right) left( n+1-k right) ! =sumlimits_n=1^infty n
sumlimits_j=2^infty frac left( -1 right) ^j-1j^
2 left( 1-j^-1 right) ^n-1 =-frac12$
We also have
$chi(n)=-sumlimits_k=0^n+2zeta left( k
right) left( k-n-2 right) nchoose k-2 left( left( -1
right) ^1+k2^1-k+ left( -1 right) ^k right) $ by replacing the $Gamma$ factors with the corresponding binomial occeficient
Related question at References for $ chi(n)=nsumlimits_j=2^inftyfrac (-1)^j-1j^2left(1-j^-1right)^n-1$ in $zeta$ expansion?
Digits := 37; flist(proc (n) options operator, arrow; evalf(coolchi(n)) end proc, 0 .. 75);
Digits := 37
[
[0., -0.1775329665758867817637924166769874055,
-0.158151287891164991627192075621149797,
-0.100756475454096876860689273411921431,
-0.052729122560581908477394880842614692,
-0.022173447806981260740844120495648785,
-0.005724758139923986695354623118140584,
0.001711181525440404872086057781767772,
0.004163504754435619487164038876686477,
0.004224146796392587120306729731994248,
0.003361436685455899377754627000698078,
0.00231507684250186191279422185478205,
0.00140312527387483007618661260684936,
0.00072272177218128957979732200349640,
0.00026840905319980399461855630763639,
-0.00000386786514313326830151239928860,
-0.00014529652172375174501064342971419,
-0.00020079200685610596363763783920190,
-0.00020486397550091241285507339640177,
-0.00018177251376674866978557090651028,
-0.00014732720568299891079533324844990,
-0.00011103259250611536839455447354313,
-0.00007801852557445637063802766476294,
-0.00005057002459089850935720449502326,
-0.00002924737982070438622370684257219,
-0.0000136577182218588710540784871811,
-0.0000029575122625340335255970335431,
0.0000038395199899767419714034429197,
0.0000076881193734840299247444522077,
0.0000094245398419219327153653943207,
0.0000097337529690318816412614692587,
0.0000091480274459270312096627770217,
0.0000080619145517535381759258176097,
0.0000067542476407336638941225910227,
0.000005411593443449269790648097261,
0.000004150125437131325581547336324,
0.000003034497471576312110001308212,
0.000002093267538135562252003012925,
0.000001330969825279401198332450463,
-7
7.37209507955248062779620826 10 ,
-7
2.93263622398134958434524015 10 ,
-8
-2.3317072362848976102839973 10 ,
-7
-2.35650675570634484892471134 10 ,
-7
-3.65620346716670870284369471 10 ,
-7
-4.32803484845776951918534983 10 ,
-7
-4.53980382773092409894967670 10 ,
-7
-4.43028271969806636513667532 10 ,
-7
-4.11056601063391173974634568 10 ,
-7
-3.66679744724574133777868780 10 ,
-7
-3.16354929829636515323370167 10 ,
-7
-2.64737199886948816611138728 10 ,
-7
-2.15021028592691266141174466 10 ,
-7
-1.69250991341964200913477377 10 ,
-7
-1.28592821342563655928047463 10 ,
-8
-9.3562155186891501184884726 10 ,
-8
-6.4212079021714840683989162 10 ,
-8
-4.0282829593958497425360774 10 ,
-8
-2.1318155241615525408999560 10 ,
-9
-6.753256088468931634905523 10 ,
-9
4.020837075521115896921341 10 ,
-8
1.1611930078301706186481030 10 ,
-8
1.6596359914684719233773543 10 ,
-8
1.9500255425207775038798882 10 ,
-8
2.0789463805547333601557044 10 ,
-8
2.0865770984321044922048033 10 ,
-8
2.0067518073257429000271847 10 ,
-8
1.8673131375018625836228485 10 ,
-8
1.6906432963871255429917949 10 ,
-8
1.4942881292166557781340237 10 ,
-8
1.2916159703946332890495351 10 ,
-8
1.0924629207316840757383290 10 ,
-9
9.037454965393441382004053 10 ,
-9
7.300036840122634764357642 10 ,
-9
5.739399681529520904444054 10 ,
-9 -9
4.367571268906899802263293 10 , 3.185931532251771457815357 10
]
]
listsum((20))= -0.4999999932944069400650356605249833795
Note: when evaluating this sum, at appears at least 28 digits of precision are required for accurate evaluation. If you use less than this (Maple defaults to 10) then the result will diverge quite wildly to an inaccurate answer.
This table is the partial sums of the terms, so I think the equality =-1/2 is correct
[0., -.1775329665758867817637924166769874055, -.3356842544670517733909844922981372025, -.4364407299211486502516737657100586335, -.4891698524817305587290686465526733255, -.5113433002887118194699127670483221105, -.5170680584286358061652673901664626945, -.5153568769031954012931813323846949225, -.5111933721487597818060172935080084455, -.5069692253523671946857105637760141975, -.5036077886669112953079559367753161195, -.5012927118244094333951617149205340695, -.4998895865505346033189751023136847095, -.4991668647783533137391777803101883095, -.4988984557251535097445592240025519195, -.4989023235902966430128607364018405195, -.4990476201120203947578713798315547095, -.4992484121188765007215090176707566095, -.4994532760943774131343640910671583795, -.4996350486081441618041496619736686595, -.4997823758138271607149449952221185595, -.4998934084063332760833395496956616895, -.4999714269319077324539775773604246295, -.5000219969564986309633347818554478895, -.5000512443363193353495584886980200795, -.5000649020545411942206125671852011795, -.5000678595668037282541381642187442795, -.5000640200468137515121667607758245795, -.5000563319274402674822420163236168795, -.5000469073875983455495266509292961795, -.5000371736346293136678853894600374795, -.5000280256071833866366757266830157795, -.5000199636926316330984998008654060795, -.5000132094449908994346056782743833795, -.5000077978515474501648150301771223795, -.5000036477261103188392334828407983795, -.5000006132286387425271234815325863795, -.4999985199611006069648714785196613795, -.4999971889912753275636731460691983795, -.4999964517817673723156103664483723795, -.4999961585181449741806519319243573795, -.4999961818352173370296280347643303795, -.4999964174858929076641129272354643795, -.4999967831062396243349832116049353795, -.4999972159097244701119351301399183795, -.4999976698901072432043450251075883795, -.4999981129183792130109815387751203795, -.4999985239749802764021555134096883795, -.4999988906547250009762892912784683795, -.4999992070096548306128046146486353795, -.4999994717468547175616212257873633795, -.4999996867678833102528873669618293795, -.4999998560188746522170882804392063795, -.4999999846116959947807442084866693795, -.5000000781738511816722453933713953795, -.5000001423859302033870860773605573795, -.5000001826687597973455835027213313795, -.5000002039869150389611089117208913795, -.5000002107401711274300405466264143795, -.5000002067193340519089246497050733795, -.5000001951074039736072184632240433795, -.5000001785110440589224992294505003795, -.5000001590107886337147241906516183795, -.5000001382213248281673905890945743795, -.5000001173555538438463456670465413795, -.5000000972880357705889166667746943795, -.5000000786149043955702908305462093795, -.5000000617084714316990354006282603795, -.5000000467655901395324776192880233795, -.5000000338494304355861447287926723795, -.5000000229248012282693039714093823795, -.5000000138873462628758625894053293795, -.5000000065873094227532278250476873795, -.5000000008479097412237069206036333795, -.4999999964803384723168071183403403795, -.4999999932944069400650356605249833795]
sequences-and-series taylor-expansion binomial-coefficients riemann-zeta binomial-theorem
asked Jul 15 at 20:45
crow
492416
1
A confirmation about the numeric evaluation being so slow would be nice though.
– Diger
Jul 15 at 22:03
1
To be honest: I don't really know! I guess this is quite delicate as for when you would replace $(-1)^j-1$ by $1^j-1=1$ then the entire result diverges. Also the last sum with alternating $1$s and $-1$s is not really defined, but only in the sense of analytic continuation. Therefore I guess your sum above would have to be replaced by some function $$ f(x)=sum _n=1^infty left( sum _j=2^infty frac x^j-1nj^ 2 left( 1-j^-1 right) ^n-1 right) $$ with $|x|<1$, because then the series is absolutely and uniformly convergent and can you interchange
– Diger
Jul 16 at 1:02
1
summation order. Then $f(x)$ is evaluated at $-1$ which amounts for, if your sum is the result of meromorphic function represented by this series, then the value of this function can be attached to it.
– Diger
Jul 16 at 1:06
1
Mathematica code:Sum[Sum[((-1)^(j - 1)*n)/j^2*(1 - j^-1)^(n - 1), n, 1, Infinity, Assumptions -> j > 1] // Simplify, j, 2, Infinity, Regularization -> "Dirichlet"]gives: $-1/2$
– Mariusz Iwaniuk
Jul 18 at 14:36
1
@crow Do you want a proof of the single identity in the title or of the double idendity below the title? Plus, what have you tried for proving them?
– Alex Francisco
Jul 18 at 15:58
 |Â
show 8 more comments
up vote
5
down vote
favorite
up vote
5
down vote
favorite
How can we prove the following?
$-1+frac112pi ^2-frac12sumlimits_n=2^infty Gamma left( n+1 right) sumlimits_k=0^n+2
,frac zeta left( k right) left( -
1 right) ^-2+k left( 2^2-k-2 right) Gamma left( -1+k
right) left( n+1-k right) ! =sumlimits_n=1^infty n
sumlimits_j=2^infty frac left( -1 right) ^j-1j^
2 left( 1-j^-1 right) ^n-1 =-frac12$
We also have
$chi(n)=-sumlimits_k=0^n+2zeta left( k
right) left( k-n-2 right) nchoose k-2 left( left( -1
right) ^1+k2^1-k+ left( -1 right) ^k right) $ by replacing the $Gamma$ factors with the corresponding binomial occeficient
Related question at References for $ chi(n)=nsumlimits_j=2^inftyfrac (-1)^j-1j^2left(1-j^-1right)^n-1$ in $zeta$ expansion?
Digits := 37; flist(proc (n) options operator, arrow; evalf(coolchi(n)) end proc, 0 .. 75);
Digits := 37
[
[0., -0.1775329665758867817637924166769874055,
-0.158151287891164991627192075621149797,
-0.100756475454096876860689273411921431,
-0.052729122560581908477394880842614692,
-0.022173447806981260740844120495648785,
-0.005724758139923986695354623118140584,
0.001711181525440404872086057781767772,
0.004163504754435619487164038876686477,
0.004224146796392587120306729731994248,
0.003361436685455899377754627000698078,
0.00231507684250186191279422185478205,
0.00140312527387483007618661260684936,
0.00072272177218128957979732200349640,
0.00026840905319980399461855630763639,
-0.00000386786514313326830151239928860,
-0.00014529652172375174501064342971419,
-0.00020079200685610596363763783920190,
-0.00020486397550091241285507339640177,
-0.00018177251376674866978557090651028,
-0.00014732720568299891079533324844990,
-0.00011103259250611536839455447354313,
-0.00007801852557445637063802766476294,
-0.00005057002459089850935720449502326,
-0.00002924737982070438622370684257219,
-0.0000136577182218588710540784871811,
-0.0000029575122625340335255970335431,
0.0000038395199899767419714034429197,
0.0000076881193734840299247444522077,
0.0000094245398419219327153653943207,
0.0000097337529690318816412614692587,
0.0000091480274459270312096627770217,
0.0000080619145517535381759258176097,
0.0000067542476407336638941225910227,
0.000005411593443449269790648097261,
0.000004150125437131325581547336324,
0.000003034497471576312110001308212,
0.000002093267538135562252003012925,
0.000001330969825279401198332450463,
-7
7.37209507955248062779620826 10 ,
-7
2.93263622398134958434524015 10 ,
-8
-2.3317072362848976102839973 10 ,
-7
-2.35650675570634484892471134 10 ,
-7
-3.65620346716670870284369471 10 ,
-7
-4.32803484845776951918534983 10 ,
-7
-4.53980382773092409894967670 10 ,
-7
-4.43028271969806636513667532 10 ,
-7
-4.11056601063391173974634568 10 ,
-7
-3.66679744724574133777868780 10 ,
-7
-3.16354929829636515323370167 10 ,
-7
-2.64737199886948816611138728 10 ,
-7
-2.15021028592691266141174466 10 ,
-7
-1.69250991341964200913477377 10 ,
-7
-1.28592821342563655928047463 10 ,
-8
-9.3562155186891501184884726 10 ,
-8
-6.4212079021714840683989162 10 ,
-8
-4.0282829593958497425360774 10 ,
-8
-2.1318155241615525408999560 10 ,
-9
-6.753256088468931634905523 10 ,
-9
4.020837075521115896921341 10 ,
-8
1.1611930078301706186481030 10 ,
-8
1.6596359914684719233773543 10 ,
-8
1.9500255425207775038798882 10 ,
-8
2.0789463805547333601557044 10 ,
-8
2.0865770984321044922048033 10 ,
-8
2.0067518073257429000271847 10 ,
-8
1.8673131375018625836228485 10 ,
-8
1.6906432963871255429917949 10 ,
-8
1.4942881292166557781340237 10 ,
-8
1.2916159703946332890495351 10 ,
-8
1.0924629207316840757383290 10 ,
-9
9.037454965393441382004053 10 ,
-9
7.300036840122634764357642 10 ,
-9
5.739399681529520904444054 10 ,
-9 -9
4.367571268906899802263293 10 , 3.185931532251771457815357 10
]
]
listsum((20))= -0.4999999932944069400650356605249833795
Note: when evaluating this sum, at appears at least 28 digits of precision are required for accurate evaluation. If you use less than this (Maple defaults to 10) then the result will diverge quite wildly to an inaccurate answer.
This table is the partial sums of the terms, so I think the equality =-1/2 is correct
[0., -.1775329665758867817637924166769874055, -.3356842544670517733909844922981372025, -.4364407299211486502516737657100586335, -.4891698524817305587290686465526733255, -.5113433002887118194699127670483221105, -.5170680584286358061652673901664626945, -.5153568769031954012931813323846949225, -.5111933721487597818060172935080084455, -.5069692253523671946857105637760141975, -.5036077886669112953079559367753161195, -.5012927118244094333951617149205340695, -.4998895865505346033189751023136847095, -.4991668647783533137391777803101883095, -.4988984557251535097445592240025519195, -.4989023235902966430128607364018405195, -.4990476201120203947578713798315547095, -.4992484121188765007215090176707566095, -.4994532760943774131343640910671583795, -.4996350486081441618041496619736686595, -.4997823758138271607149449952221185595, -.4998934084063332760833395496956616895, -.4999714269319077324539775773604246295, -.5000219969564986309633347818554478895, -.5000512443363193353495584886980200795, -.5000649020545411942206125671852011795, -.5000678595668037282541381642187442795, -.5000640200468137515121667607758245795, -.5000563319274402674822420163236168795, -.5000469073875983455495266509292961795, -.5000371736346293136678853894600374795, -.5000280256071833866366757266830157795, -.5000199636926316330984998008654060795, -.5000132094449908994346056782743833795, -.5000077978515474501648150301771223795, -.5000036477261103188392334828407983795, -.5000006132286387425271234815325863795, -.4999985199611006069648714785196613795, -.4999971889912753275636731460691983795, -.4999964517817673723156103664483723795, -.4999961585181449741806519319243573795, -.4999961818352173370296280347643303795, -.4999964174858929076641129272354643795, -.4999967831062396243349832116049353795, -.4999972159097244701119351301399183795, -.4999976698901072432043450251075883795, -.4999981129183792130109815387751203795, -.4999985239749802764021555134096883795, -.4999988906547250009762892912784683795, -.4999992070096548306128046146486353795, -.4999994717468547175616212257873633795, -.4999996867678833102528873669618293795, -.4999998560188746522170882804392063795, -.4999999846116959947807442084866693795, -.5000000781738511816722453933713953795, -.5000001423859302033870860773605573795, -.5000001826687597973455835027213313795, -.5000002039869150389611089117208913795, -.5000002107401711274300405466264143795, -.5000002067193340519089246497050733795, -.5000001951074039736072184632240433795, -.5000001785110440589224992294505003795, -.5000001590107886337147241906516183795, -.5000001382213248281673905890945743795, -.5000001173555538438463456670465413795, -.5000000972880357705889166667746943795, -.5000000786149043955702908305462093795, -.5000000617084714316990354006282603795, -.5000000467655901395324776192880233795, -.5000000338494304355861447287926723795, -.5000000229248012282693039714093823795, -.5000000138873462628758625894053293795, -.5000000065873094227532278250476873795, -.5000000008479097412237069206036333795, -.4999999964803384723168071183403403795, -.4999999932944069400650356605249833795]
sequences-and-series taylor-expansion binomial-coefficients riemann-zeta binomial-theorem
asked Jul 15 at 20:45
crow
492416
How can we prove the following?
$-1+frac112pi ^2-frac12sumlimits_n=2^infty Gamma left( n+1 right) sumlimits_k=0^n+2
,frac zeta left( k right) left( -
1 right) ^-2+k left( 2^2-k-2 right) Gamma left( -1+k
right) left( n+1-k right) ! =sumlimits_n=1^infty n
sumlimits_j=2^infty frac left( -1 right) ^j-1j^
2 left( 1-j^-1 right) ^n-1 =-frac12$
We also have
$chi(n)=-sumlimits_k=0^n+2zeta left( k
right) left( k-n-2 right) nchoose k-2 left( left( -1
right) ^1+k2^1-k+ left( -1 right) ^k right) $ by replacing the $Gamma$ factors with the corresponding binomial occeficient
Related question at References for $ chi(n)=nsumlimits_j=2^inftyfrac (-1)^j-1j^2left(1-j^-1right)^n-1$ in $zeta$ expansion?
Digits := 37; flist(proc (n) options operator, arrow; evalf(coolchi(n)) end proc, 0 .. 75);
Digits := 37
[
[0., -0.1775329665758867817637924166769874055,
-0.158151287891164991627192075621149797,
-0.100756475454096876860689273411921431,
-0.052729122560581908477394880842614692,
-0.022173447806981260740844120495648785,
-0.005724758139923986695354623118140584,
0.001711181525440404872086057781767772,
0.004163504754435619487164038876686477,
0.004224146796392587120306729731994248,
0.003361436685455899377754627000698078,
0.00231507684250186191279422185478205,
0.00140312527387483007618661260684936,
0.00072272177218128957979732200349640,
0.00026840905319980399461855630763639,
-0.00000386786514313326830151239928860,
-0.00014529652172375174501064342971419,
-0.00020079200685610596363763783920190,
-0.00020486397550091241285507339640177,
-0.00018177251376674866978557090651028,
-0.00014732720568299891079533324844990,
-0.00011103259250611536839455447354313,
-0.00007801852557445637063802766476294,
-0.00005057002459089850935720449502326,
-0.00002924737982070438622370684257219,
-0.0000136577182218588710540784871811,
-0.0000029575122625340335255970335431,
0.0000038395199899767419714034429197,
0.0000076881193734840299247444522077,
0.0000094245398419219327153653943207,
0.0000097337529690318816412614692587,
0.0000091480274459270312096627770217,
0.0000080619145517535381759258176097,
0.0000067542476407336638941225910227,
0.000005411593443449269790648097261,
0.000004150125437131325581547336324,
0.000003034497471576312110001308212,
0.000002093267538135562252003012925,
0.000001330969825279401198332450463,
-7
7.37209507955248062779620826 10 ,
-7
2.93263622398134958434524015 10 ,
-8
-2.3317072362848976102839973 10 ,
-7
-2.35650675570634484892471134 10 ,
-7
-3.65620346716670870284369471 10 ,
-7
-4.32803484845776951918534983 10 ,
-7
-4.53980382773092409894967670 10 ,
-7
-4.43028271969806636513667532 10 ,
-7
-4.11056601063391173974634568 10 ,
-7
-3.66679744724574133777868780 10 ,
-7
-3.16354929829636515323370167 10 ,
-7
-2.64737199886948816611138728 10 ,
-7
-2.15021028592691266141174466 10 ,
-7
-1.69250991341964200913477377 10 ,
-7
-1.28592821342563655928047463 10 ,
-8
-9.3562155186891501184884726 10 ,
-8
-6.4212079021714840683989162 10 ,
-8
-4.0282829593958497425360774 10 ,
-8
-2.1318155241615525408999560 10 ,
-9
-6.753256088468931634905523 10 ,
-9
4.020837075521115896921341 10 ,
-8
1.1611930078301706186481030 10 ,
-8
1.6596359914684719233773543 10 ,
-8
1.9500255425207775038798882 10 ,
-8
2.0789463805547333601557044 10 ,
-8
2.0865770984321044922048033 10 ,
-8
2.0067518073257429000271847 10 ,
-8
1.8673131375018625836228485 10 ,
-8
1.6906432963871255429917949 10 ,
-8
1.4942881292166557781340237 10 ,
-8
1.2916159703946332890495351 10 ,
-8
1.0924629207316840757383290 10 ,
-9
9.037454965393441382004053 10 ,
-9
7.300036840122634764357642 10 ,
-9
5.739399681529520904444054 10 ,
-9 -9
4.367571268906899802263293 10 , 3.185931532251771457815357 10
]
]
listsum((20))= -0.4999999932944069400650356605249833795
Note: when evaluating this sum, at appears at least 28 digits of precision are required for accurate evaluation. If you use less than this (Maple defaults to 10) then the result will diverge quite wildly to an inaccurate answer.
This table is the partial sums of the terms, so I think the equality =-1/2 is correct
[0., -.1775329665758867817637924166769874055, -.3356842544670517733909844922981372025, -.4364407299211486502516737657100586335, -.4891698524817305587290686465526733255, -.5113433002887118194699127670483221105, -.5170680584286358061652673901664626945, -.5153568769031954012931813323846949225, -.5111933721487597818060172935080084455, -.5069692253523671946857105637760141975, -.5036077886669112953079559367753161195, -.5012927118244094333951617149205340695, -.4998895865505346033189751023136847095, -.4991668647783533137391777803101883095, -.4988984557251535097445592240025519195, -.4989023235902966430128607364018405195, -.4990476201120203947578713798315547095, -.4992484121188765007215090176707566095, -.4994532760943774131343640910671583795, -.4996350486081441618041496619736686595, -.4997823758138271607149449952221185595, -.4998934084063332760833395496956616895, -.4999714269319077324539775773604246295, -.5000219969564986309633347818554478895, -.5000512443363193353495584886980200795, -.5000649020545411942206125671852011795, -.5000678595668037282541381642187442795, -.5000640200468137515121667607758245795, -.5000563319274402674822420163236168795, -.5000469073875983455495266509292961795, -.5000371736346293136678853894600374795, -.5000280256071833866366757266830157795, -.5000199636926316330984998008654060795, -.5000132094449908994346056782743833795, -.5000077978515474501648150301771223795, -.5000036477261103188392334828407983795, -.5000006132286387425271234815325863795, -.4999985199611006069648714785196613795, -.4999971889912753275636731460691983795, -.4999964517817673723156103664483723795, -.4999961585181449741806519319243573795, -.4999961818352173370296280347643303795, -.4999964174858929076641129272354643795, -.4999967831062396243349832116049353795, -.4999972159097244701119351301399183795, -.4999976698901072432043450251075883795, -.4999981129183792130109815387751203795, -.4999985239749802764021555134096883795, -.4999988906547250009762892912784683795, -.4999992070096548306128046146486353795, -.4999994717468547175616212257873633795, -.4999996867678833102528873669618293795, -.4999998560188746522170882804392063795, -.4999999846116959947807442084866693795, -.5000000781738511816722453933713953795, -.5000001423859302033870860773605573795, -.5000001826687597973455835027213313795, -.5000002039869150389611089117208913795, -.5000002107401711274300405466264143795, -.5000002067193340519089246497050733795, -.5000001951074039736072184632240433795, -.5000001785110440589224992294505003795, -.5000001590107886337147241906516183795, -.5000001382213248281673905890945743795, -.5000001173555538438463456670465413795, -.5000000972880357705889166667746943795, -.5000000786149043955702908305462093795, -.5000000617084714316990354006282603795, -.5000000467655901395324776192880233795, -.5000000338494304355861447287926723795, -.5000000229248012282693039714093823795, -.5000000138873462628758625894053293795, -.5000000065873094227532278250476873795, -.5000000008479097412237069206036333795, -.4999999964803384723168071183403403795, -.4999999932944069400650356605249833795]
sequences-and-series taylor-expansion binomial-coefficients riemann-zeta binomial-theorem
asked Jul 15 at 20:45
crow
492416
edited Jul 21 at 19:57
asked Jul 15 at 20:45
crow
492416
asked Jul 15 at 20:45
crow
492416
asked Jul 15 at 20:45
crow
492416
492416
1
A confirmation about the numeric evaluation being so slow would be nice though.
– Diger
Jul 15 at 22:03
1
To be honest: I don't really know! I guess this is quite delicate as for when you would replace $(-1)^j-1$ by $1^j-1=1$ then the entire result diverges. Also the last sum with alternating $1$s and $-1$s is not really defined, but only in the sense of analytic continuation. Therefore I guess your sum above would have to be replaced by some function $$ f(x)=sum _n=1^infty left( sum _j=2^infty frac x^j-1nj^ 2 left( 1-j^-1 right) ^n-1 right) $$ with $|x|<1$, because then the series is absolutely and uniformly convergent and can you interchange
– Diger
Jul 16 at 1:02
1
summation order. Then $f(x)$ is evaluated at $-1$ which amounts for, if your sum is the result of meromorphic function represented by this series, then the value of this function can be attached to it.
– Diger
Jul 16 at 1:06
1
Mathematica code:Sum[Sum[((-1)^(j - 1)*n)/j^2*(1 - j^-1)^(n - 1), n, 1, Infinity, Assumptions -> j > 1] // Simplify, j, 2, Infinity, Regularization -> "Dirichlet"]gives: $-1/2$
– Mariusz Iwaniuk
Jul 18 at 14:36
1
@crow Do you want a proof of the single identity in the title or of the double idendity below the title? Plus, what have you tried for proving them?
– Alex Francisco
Jul 18 at 15:58
 |Â
show 8 more comments
1
A confirmation about the numeric evaluation being so slow would be nice though.
– Diger
Jul 15 at 22:03
1
To be honest: I don't really know! I guess this is quite delicate as for when you would replace $(-1)^j-1$ by $1^j-1=1$ then the entire result diverges. Also the last sum with alternating $1$s and $-1$s is not really defined, but only in the sense of analytic continuation. Therefore I guess your sum above would have to be replaced by some function $$ f(x)=sum _n=1^infty left( sum _j=2^infty frac x^j-1nj^ 2 left( 1-j^-1 right) ^n-1 right) $$ with $|x|<1$, because then the series is absolutely and uniformly convergent and can you interchange
– Diger
Jul 16 at 1:02
1
summation order. Then $f(x)$ is evaluated at $-1$ which amounts for, if your sum is the result of meromorphic function represented by this series, then the value of this function can be attached to it.
– Diger
Jul 16 at 1:06
1
Mathematica code:Sum[Sum[((-1)^(j - 1)*n)/j^2*(1 - j^-1)^(n - 1), n, 1, Infinity, Assumptions -> j > 1] // Simplify, j, 2, Infinity, Regularization -> "Dirichlet"]gives: $-1/2$
– Mariusz Iwaniuk
Jul 18 at 14:36
1
@crow Do you want a proof of the single identity in the title or of the double idendity below the title? Plus, what have you tried for proving them?
– Alex Francisco
Jul 18 at 15:58
1
1
A confirmation about the numeric evaluation being so slow would be nice though.
– Diger
Jul 15 at 22:03
A confirmation about the numeric evaluation being so slow would be nice though.
– Diger
Jul 15 at 22:03
1
1
To be honest: I don't really know! I guess this is quite delicate as for when you would replace $(-1)^j-1$ by $1^j-1=1$ then the entire result diverges. Also the last sum with alternating $1$s and $-1$s is not really defined, but only in the sense of analytic continuation. Therefore I guess your sum above would have to be replaced by some function $$ f(x)=sum _n=1^infty left( sum _j=2^infty frac x^j-1nj^ 2 left( 1-j^-1 right) ^n-1 right) $$ with $|x|<1$, because then the series is absolutely and uniformly convergent and can you interchange
– Diger
Jul 16 at 1:02
To be honest: I don't really know! I guess this is quite delicate as for when you would replace $(-1)^j-1$ by $1^j-1=1$ then the entire result diverges. Also the last sum with alternating $1$s and $-1$s is not really defined, but only in the sense of analytic continuation. Therefore I guess your sum above would have to be replaced by some function $$ f(x)=sum _n=1^infty left( sum _j=2^infty frac x^j-1nj^ 2 left( 1-j^-1 right) ^n-1 right) $$ with $|x|<1$, because then the series is absolutely and uniformly convergent and can you interchange
– Diger
Jul 16 at 1:02
1
1
summation order. Then $f(x)$ is evaluated at $-1$ which amounts for, if your sum is the result of meromorphic function represented by this series, then the value of this function can be attached to it.
– Diger
Jul 16 at 1:06
summation order. Then $f(x)$ is evaluated at $-1$ which amounts for, if your sum is the result of meromorphic function represented by this series, then the value of this function can be attached to it.
– Diger
Jul 16 at 1:06
1
1
Mathematica code:
Sum[Sum[((-1)^(j - 1)*n)/j^2*(1 - j^-1)^(n - 1), n, 1, Infinity, Assumptions -> j > 1] // Simplify, j, 2, Infinity, Regularization -> "Dirichlet"] gives: $-1/2$– Mariusz Iwaniuk
Jul 18 at 14:36
Mathematica code:
Sum[Sum[((-1)^(j - 1)*n)/j^2*(1 - j^-1)^(n - 1), n, 1, Infinity, Assumptions -> j > 1] // Simplify, j, 2, Infinity, Regularization -> "Dirichlet"] gives: $-1/2$– Mariusz Iwaniuk
Jul 18 at 14:36
1
1
@crow Do you want a proof of the single identity in the title or of the double idendity below the title? Plus, what have you tried for proving them?
– Alex Francisco
Jul 18 at 15:58
@crow Do you want a proof of the single identity in the title or of the double idendity below the title? Plus, what have you tried for proving them?
– Alex Francisco
Jul 18 at 15:58
 |Â
show 8 more comments
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
In the following assume $|z|<1$ to begin with and I'll let $z to 1$ to get the OP's left hand side formula:
$$ S(z):=sum_n=1^infty z^n,n sum_j=2^infty, frac(-1)^j-1j^2big(1-frac1jbig)^n-1 .$$
$S(z)$ is solvable in closed-form in the following manner: Interchange summations and to eliminate the $n$ series use the derivative of the geometric series to obtain
$$S(z)= zsum_j=2^infty, frac(-1)^j-1j^2(1-z(1-1/j))^2=fracz(1-z)^2sum_j=2^infty, frac(-1)^j-1 (j+z/(1-z))^2 .$$
This sum is solvable in the terms of the trigamma function, once a $j=1$ term is added and subtracted. Thus
$$S(z)= fracz/4(1-z)^2Big( psi,'big(tfrac32+tfracz2(1-z)big) - psi,' big( 1+tfracz2(1-z) big) Big) .$$
To find the limit $z to 1,$ it is seen that it is equivalent to
$$S(1) = lim_y to 0 frac14y^2 Big( psi,'big(tfrac32+tfrac12y) - psi,' big(1+tfrac12y big) Big) =
lim_x to infty fracx^24 Big( psi,'big(tfrac32+tfracx2big) - psi,' big( 1+tfracx2 big) Big)$$
$$= lim_x to infty fracx^24 Big(frac2x-frac4x^2+mathitO(x^-3) - (frac2x-frac2x^2+mathitO(x^-3)) Big)= -frac12$$
where the asymptotic formula for the trigamma function has been used.
answered Jul 18 at 19:13
skbmoore
1,12026
2
Why is the interchange of a limit and a summation, licit?
– Did
Jul 18 at 19:21
@skbmoore that's a pretty cool formula. I see that the series expansion of S has the n-th term in the expansion being $chi(n)$
– crow
Jul 19 at 0:16
@Did. Assume z is 'small enough.' Do the sum and the final result tells us that there are no singularities until $|z|=1.$ Abel's theorem, I believe, lets us take the limit.
– skbmoore
Jul 19 at 17:03
2
?? "You think"? The reason that make you sure the step is legit should be written down explicitely in your answer.
– Did
Jul 19 at 17:29
1
So, each inner series converges, right (and by much simpler arguments). But this is not the problem at all...
– Did
Jul 22 at 20:38
 |Â
show 15 more comments
up vote
4
down vote
Write $chi(n) = n sum _j=2^infty frac(-1)^j-1j^2 left( 1-frac1j right)^n-1$ using OP's notation. Its partial sum up to the $(N+1)$-th term can be simplified by interchanging the order of summation:
beginalign*
sum_n=1^N+1 chi(n)
&= sum _j=2^infty frac(-1)^j-1j^2 sum_n=1^N+1 n left( 1-frac1j right)^n-1 \
&= sum_j=2^infty (-1)^j underbrace left[ left(1 + fracN+1jright)left(1 - frac1jright)^N+1 - 1 right] _=: f_N(j),
tag1
endalign*
Since $f_N(j) = mathcalO(j^-2)$ as $jtoinfty$ for each fixed $N$, $text(1)$ is indeed a convergent series. Now grouping successive terms, $text(1)$ simplifies to
beginalign*
sum_n=1^N+1 chi(n)
&= - sum_k=1^infty (f_N(2k+1) - f_N(2k)) \
&= - sum_k=1^infty int_2k^2k+1 f_N'(x) , dx \
&= - sum_k=1^infty (N+1)(N+2) int_2k^2k+1 fracleft(1 - frac1xright)^Nx^3 , dx \
&= - sum_k=1^infty frac(N+1)(N+2)N^2 int_frac2kN^frac2k+1N fracleft(1 - frac1Nuright)^Nu^3 , du tag$x=Nu$
endalign*
Then Fubini-Tonelli's theorem gives
$$
sum_n=1^N+1 chi(n)
= - frac(N+1)(N+2)N^2 int_0^infty fracleft(1 - frac1Nuright)^Nu^3 left( sum_k=1^infty mathbf1_left[frac2kN, frac2k+1Nright](u) right) , du.
$$
Since the integrand is bounded by the integrable dominating function $e^-1/u/u^3$, as $Ntoinfty$ we have
$$
lim_Ntoinfty sum_n=1^N+1 chi(n)
= - int_0^infty frace^-1/uu^3 cdot frac12 , du
= - frac12.
$$
answered Jul 20 at 8:56
Sangchul Lee
85.6k12155253
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
In the following assume $|z|<1$ to begin with and I'll let $z to 1$ to get the OP's left hand side formula:
$$ S(z):=sum_n=1^infty z^n,n sum_j=2^infty, frac(-1)^j-1j^2big(1-frac1jbig)^n-1 .$$
$S(z)$ is solvable in closed-form in the following manner: Interchange summations and to eliminate the $n$ series use the derivative of the geometric series to obtain
$$S(z)= zsum_j=2^infty, frac(-1)^j-1j^2(1-z(1-1/j))^2=fracz(1-z)^2sum_j=2^infty, frac(-1)^j-1 (j+z/(1-z))^2 .$$
This sum is solvable in the terms of the trigamma function, once a $j=1$ term is added and subtracted. Thus
$$S(z)= fracz/4(1-z)^2Big( psi,'big(tfrac32+tfracz2(1-z)big) - psi,' big( 1+tfracz2(1-z) big) Big) .$$
To find the limit $z to 1,$ it is seen that it is equivalent to
$$S(1) = lim_y to 0 frac14y^2 Big( psi,'big(tfrac32+tfrac12y) - psi,' big(1+tfrac12y big) Big) =
lim_x to infty fracx^24 Big( psi,'big(tfrac32+tfracx2big) - psi,' big( 1+tfracx2 big) Big)$$
$$= lim_x to infty fracx^24 Big(frac2x-frac4x^2+mathitO(x^-3) - (frac2x-frac2x^2+mathitO(x^-3)) Big)= -frac12$$
where the asymptotic formula for the trigamma function has been used.
answered Jul 18 at 19:13
skbmoore
1,12026
2
Why is the interchange of a limit and a summation, licit?
– Did
Jul 18 at 19:21
@skbmoore that's a pretty cool formula. I see that the series expansion of S has the n-th term in the expansion being $chi(n)$
– crow
Jul 19 at 0:16
@Did. Assume z is 'small enough.' Do the sum and the final result tells us that there are no singularities until $|z|=1.$ Abel's theorem, I believe, lets us take the limit.
– skbmoore
Jul 19 at 17:03
2
?? "You think"? The reason that make you sure the step is legit should be written down explicitely in your answer.
– Did
Jul 19 at 17:29
1
So, each inner series converges, right (and by much simpler arguments). But this is not the problem at all...
– Did
Jul 22 at 20:38
 |Â
show 15 more comments
up vote
2
down vote
accepted
In the following assume $|z|<1$ to begin with and I'll let $z to 1$ to get the OP's left hand side formula:
$$ S(z):=sum_n=1^infty z^n,n sum_j=2^infty, frac(-1)^j-1j^2big(1-frac1jbig)^n-1 .$$
$S(z)$ is solvable in closed-form in the following manner: Interchange summations and to eliminate the $n$ series use the derivative of the geometric series to obtain
$$S(z)= zsum_j=2^infty, frac(-1)^j-1j^2(1-z(1-1/j))^2=fracz(1-z)^2sum_j=2^infty, frac(-1)^j-1 (j+z/(1-z))^2 .$$
This sum is solvable in the terms of the trigamma function, once a $j=1$ term is added and subtracted. Thus
$$S(z)= fracz/4(1-z)^2Big( psi,'big(tfrac32+tfracz2(1-z)big) - psi,' big( 1+tfracz2(1-z) big) Big) .$$
To find the limit $z to 1,$ it is seen that it is equivalent to
$$S(1) = lim_y to 0 frac14y^2 Big( psi,'big(tfrac32+tfrac12y) - psi,' big(1+tfrac12y big) Big) =
lim_x to infty fracx^24 Big( psi,'big(tfrac32+tfracx2big) - psi,' big( 1+tfracx2 big) Big)$$
$$= lim_x to infty fracx^24 Big(frac2x-frac4x^2+mathitO(x^-3) - (frac2x-frac2x^2+mathitO(x^-3)) Big)= -frac12$$
where the asymptotic formula for the trigamma function has been used.
answered Jul 18 at 19:13
skbmoore
1,12026
2
Why is the interchange of a limit and a summation, licit?
– Did
Jul 18 at 19:21
@skbmoore that's a pretty cool formula. I see that the series expansion of S has the n-th term in the expansion being $chi(n)$
– crow
Jul 19 at 0:16
@Did. Assume z is 'small enough.' Do the sum and the final result tells us that there are no singularities until $|z|=1.$ Abel's theorem, I believe, lets us take the limit.
– skbmoore
Jul 19 at 17:03
2
?? "You think"? The reason that make you sure the step is legit should be written down explicitely in your answer.
– Did
Jul 19 at 17:29
1
So, each inner series converges, right (and by much simpler arguments). But this is not the problem at all...
– Did
Jul 22 at 20:38
 |Â
show 15 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
In the following assume $|z|<1$ to begin with and I'll let $z to 1$ to get the OP's left hand side formula:
$$ S(z):=sum_n=1^infty z^n,n sum_j=2^infty, frac(-1)^j-1j^2big(1-frac1jbig)^n-1 .$$
$S(z)$ is solvable in closed-form in the following manner: Interchange summations and to eliminate the $n$ series use the derivative of the geometric series to obtain
$$S(z)= zsum_j=2^infty, frac(-1)^j-1j^2(1-z(1-1/j))^2=fracz(1-z)^2sum_j=2^infty, frac(-1)^j-1 (j+z/(1-z))^2 .$$
This sum is solvable in the terms of the trigamma function, once a $j=1$ term is added and subtracted. Thus
$$S(z)= fracz/4(1-z)^2Big( psi,'big(tfrac32+tfracz2(1-z)big) - psi,' big( 1+tfracz2(1-z) big) Big) .$$
To find the limit $z to 1,$ it is seen that it is equivalent to
$$S(1) = lim_y to 0 frac14y^2 Big( psi,'big(tfrac32+tfrac12y) - psi,' big(1+tfrac12y big) Big) =
lim_x to infty fracx^24 Big( psi,'big(tfrac32+tfracx2big) - psi,' big( 1+tfracx2 big) Big)$$
$$= lim_x to infty fracx^24 Big(frac2x-frac4x^2+mathitO(x^-3) - (frac2x-frac2x^2+mathitO(x^-3)) Big)= -frac12$$
where the asymptotic formula for the trigamma function has been used.
answered Jul 18 at 19:13
skbmoore
1,12026
In the following assume $|z|<1$ to begin with and I'll let $z to 1$ to get the OP's left hand side formula:
$$ S(z):=sum_n=1^infty z^n,n sum_j=2^infty, frac(-1)^j-1j^2big(1-frac1jbig)^n-1 .$$
$S(z)$ is solvable in closed-form in the following manner: Interchange summations and to eliminate the $n$ series use the derivative of the geometric series to obtain
$$S(z)= zsum_j=2^infty, frac(-1)^j-1j^2(1-z(1-1/j))^2=fracz(1-z)^2sum_j=2^infty, frac(-1)^j-1 (j+z/(1-z))^2 .$$
This sum is solvable in the terms of the trigamma function, once a $j=1$ term is added and subtracted. Thus
$$S(z)= fracz/4(1-z)^2Big( psi,'big(tfrac32+tfracz2(1-z)big) - psi,' big( 1+tfracz2(1-z) big) Big) .$$
To find the limit $z to 1,$ it is seen that it is equivalent to
$$S(1) = lim_y to 0 frac14y^2 Big( psi,'big(tfrac32+tfrac12y) - psi,' big(1+tfrac12y big) Big) =
lim_x to infty fracx^24 Big( psi,'big(tfrac32+tfracx2big) - psi,' big( 1+tfracx2 big) Big)$$
$$= lim_x to infty fracx^24 Big(frac2x-frac4x^2+mathitO(x^-3) - (frac2x-frac2x^2+mathitO(x^-3)) Big)= -frac12$$
where the asymptotic formula for the trigamma function has been used.
answered Jul 18 at 19:13
skbmoore
1,12026
answered Jul 18 at 19:13
skbmoore
1,12026
answered Jul 18 at 19:13
skbmoore
1,12026
answered Jul 18 at 19:13
skbmoore
1,12026
1,12026
2
Why is the interchange of a limit and a summation, licit?
– Did
Jul 18 at 19:21
@skbmoore that's a pretty cool formula. I see that the series expansion of S has the n-th term in the expansion being $chi(n)$
– crow
Jul 19 at 0:16
@Did. Assume z is 'small enough.' Do the sum and the final result tells us that there are no singularities until $|z|=1.$ Abel's theorem, I believe, lets us take the limit.
– skbmoore
Jul 19 at 17:03
2
?? "You think"? The reason that make you sure the step is legit should be written down explicitely in your answer.
– Did
Jul 19 at 17:29
1
So, each inner series converges, right (and by much simpler arguments). But this is not the problem at all...
– Did
Jul 22 at 20:38
 |Â
show 15 more comments
2
Why is the interchange of a limit and a summation, licit?
– Did
Jul 18 at 19:21
@skbmoore that's a pretty cool formula. I see that the series expansion of S has the n-th term in the expansion being $chi(n)$
– crow
Jul 19 at 0:16
@Did. Assume z is 'small enough.' Do the sum and the final result tells us that there are no singularities until $|z|=1.$ Abel's theorem, I believe, lets us take the limit.
– skbmoore
Jul 19 at 17:03
2
?? "You think"? The reason that make you sure the step is legit should be written down explicitely in your answer.
– Did
Jul 19 at 17:29
1
So, each inner series converges, right (and by much simpler arguments). But this is not the problem at all...
– Did
Jul 22 at 20:38
2
2
Why is the interchange of a limit and a summation, licit?
– Did
Jul 18 at 19:21
Why is the interchange of a limit and a summation, licit?
– Did
Jul 18 at 19:21
@skbmoore that's a pretty cool formula. I see that the series expansion of S has the n-th term in the expansion being $chi(n)$
– crow
Jul 19 at 0:16
@skbmoore that's a pretty cool formula. I see that the series expansion of S has the n-th term in the expansion being $chi(n)$
– crow
Jul 19 at 0:16
@Did. Assume z is 'small enough.' Do the sum and the final result tells us that there are no singularities until $|z|=1.$ Abel's theorem, I believe, lets us take the limit.
– skbmoore
Jul 19 at 17:03
@Did. Assume z is 'small enough.' Do the sum and the final result tells us that there are no singularities until $|z|=1.$ Abel's theorem, I believe, lets us take the limit.
– skbmoore
Jul 19 at 17:03
2
2
?? "You think"? The reason that make you sure the step is legit should be written down explicitely in your answer.
– Did
Jul 19 at 17:29
?? "You think"? The reason that make you sure the step is legit should be written down explicitely in your answer.
– Did
Jul 19 at 17:29
1
1
So, each inner series converges, right (and by much simpler arguments). But this is not the problem at all...
– Did
Jul 22 at 20:38
So, each inner series converges, right (and by much simpler arguments). But this is not the problem at all...
– Did
Jul 22 at 20:38
 |Â
show 15 more comments
up vote
4
down vote
Write $chi(n) = n sum _j=2^infty frac(-1)^j-1j^2 left( 1-frac1j right)^n-1$ using OP's notation. Its partial sum up to the $(N+1)$-th term can be simplified by interchanging the order of summation:
beginalign*
sum_n=1^N+1 chi(n)
&= sum _j=2^infty frac(-1)^j-1j^2 sum_n=1^N+1 n left( 1-frac1j right)^n-1 \
&= sum_j=2^infty (-1)^j underbrace left[ left(1 + fracN+1jright)left(1 - frac1jright)^N+1 - 1 right] _=: f_N(j),
tag1
endalign*
Since $f_N(j) = mathcalO(j^-2)$ as $jtoinfty$ for each fixed $N$, $text(1)$ is indeed a convergent series. Now grouping successive terms, $text(1)$ simplifies to
beginalign*
sum_n=1^N+1 chi(n)
&= - sum_k=1^infty (f_N(2k+1) - f_N(2k)) \
&= - sum_k=1^infty int_2k^2k+1 f_N'(x) , dx \
&= - sum_k=1^infty (N+1)(N+2) int_2k^2k+1 fracleft(1 - frac1xright)^Nx^3 , dx \
&= - sum_k=1^infty frac(N+1)(N+2)N^2 int_frac2kN^frac2k+1N fracleft(1 - frac1Nuright)^Nu^3 , du tag$x=Nu$
endalign*
Then Fubini-Tonelli's theorem gives
$$
sum_n=1^N+1 chi(n)
= - frac(N+1)(N+2)N^2 int_0^infty fracleft(1 - frac1Nuright)^Nu^3 left( sum_k=1^infty mathbf1_left[frac2kN, frac2k+1Nright](u) right) , du.
$$
Since the integrand is bounded by the integrable dominating function $e^-1/u/u^3$, as $Ntoinfty$ we have
$$
lim_Ntoinfty sum_n=1^N+1 chi(n)
= - int_0^infty frace^-1/uu^3 cdot frac12 , du
= - frac12.
$$
answered Jul 20 at 8:56
Sangchul Lee
85.6k12155253
add a comment |Â
up vote
4
down vote
Write $chi(n) = n sum _j=2^infty frac(-1)^j-1j^2 left( 1-frac1j right)^n-1$ using OP's notation. Its partial sum up to the $(N+1)$-th term can be simplified by interchanging the order of summation:
beginalign*
sum_n=1^N+1 chi(n)
&= sum _j=2^infty frac(-1)^j-1j^2 sum_n=1^N+1 n left( 1-frac1j right)^n-1 \
&= sum_j=2^infty (-1)^j underbrace left[ left(1 + fracN+1jright)left(1 - frac1jright)^N+1 - 1 right] _=: f_N(j),
tag1
endalign*
Since $f_N(j) = mathcalO(j^-2)$ as $jtoinfty$ for each fixed $N$, $text(1)$ is indeed a convergent series. Now grouping successive terms, $text(1)$ simplifies to
beginalign*
sum_n=1^N+1 chi(n)
&= - sum_k=1^infty (f_N(2k+1) - f_N(2k)) \
&= - sum_k=1^infty int_2k^2k+1 f_N'(x) , dx \
&= - sum_k=1^infty (N+1)(N+2) int_2k^2k+1 fracleft(1 - frac1xright)^Nx^3 , dx \
&= - sum_k=1^infty frac(N+1)(N+2)N^2 int_frac2kN^frac2k+1N fracleft(1 - frac1Nuright)^Nu^3 , du tag$x=Nu$
endalign*
Then Fubini-Tonelli's theorem gives
$$
sum_n=1^N+1 chi(n)
= - frac(N+1)(N+2)N^2 int_0^infty fracleft(1 - frac1Nuright)^Nu^3 left( sum_k=1^infty mathbf1_left[frac2kN, frac2k+1Nright](u) right) , du.
$$
Since the integrand is bounded by the integrable dominating function $e^-1/u/u^3$, as $Ntoinfty$ we have
$$
lim_Ntoinfty sum_n=1^N+1 chi(n)
= - int_0^infty frace^-1/uu^3 cdot frac12 , du
= - frac12.
$$
answered Jul 20 at 8:56
Sangchul Lee
85.6k12155253
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Write $chi(n) = n sum _j=2^infty frac(-1)^j-1j^2 left( 1-frac1j right)^n-1$ using OP's notation. Its partial sum up to the $(N+1)$-th term can be simplified by interchanging the order of summation:
beginalign*
sum_n=1^N+1 chi(n)
&= sum _j=2^infty frac(-1)^j-1j^2 sum_n=1^N+1 n left( 1-frac1j right)^n-1 \
&= sum_j=2^infty (-1)^j underbrace left[ left(1 + fracN+1jright)left(1 - frac1jright)^N+1 - 1 right] _=: f_N(j),
tag1
endalign*
Since $f_N(j) = mathcalO(j^-2)$ as $jtoinfty$ for each fixed $N$, $text(1)$ is indeed a convergent series. Now grouping successive terms, $text(1)$ simplifies to
beginalign*
sum_n=1^N+1 chi(n)
&= - sum_k=1^infty (f_N(2k+1) - f_N(2k)) \
&= - sum_k=1^infty int_2k^2k+1 f_N'(x) , dx \
&= - sum_k=1^infty (N+1)(N+2) int_2k^2k+1 fracleft(1 - frac1xright)^Nx^3 , dx \
&= - sum_k=1^infty frac(N+1)(N+2)N^2 int_frac2kN^frac2k+1N fracleft(1 - frac1Nuright)^Nu^3 , du tag$x=Nu$
endalign*
Then Fubini-Tonelli's theorem gives
$$
sum_n=1^N+1 chi(n)
= - frac(N+1)(N+2)N^2 int_0^infty fracleft(1 - frac1Nuright)^Nu^3 left( sum_k=1^infty mathbf1_left[frac2kN, frac2k+1Nright](u) right) , du.
$$
Since the integrand is bounded by the integrable dominating function $e^-1/u/u^3$, as $Ntoinfty$ we have
$$
lim_Ntoinfty sum_n=1^N+1 chi(n)
= - int_0^infty frace^-1/uu^3 cdot frac12 , du
= - frac12.
$$
answered Jul 20 at 8:56
Sangchul Lee
85.6k12155253
Write $chi(n) = n sum _j=2^infty frac(-1)^j-1j^2 left( 1-frac1j right)^n-1$ using OP's notation. Its partial sum up to the $(N+1)$-th term can be simplified by interchanging the order of summation:
beginalign*
sum_n=1^N+1 chi(n)
&= sum _j=2^infty frac(-1)^j-1j^2 sum_n=1^N+1 n left( 1-frac1j right)^n-1 \
&= sum_j=2^infty (-1)^j underbrace left[ left(1 + fracN+1jright)left(1 - frac1jright)^N+1 - 1 right] _=: f_N(j),
tag1
endalign*
Since $f_N(j) = mathcalO(j^-2)$ as $jtoinfty$ for each fixed $N$, $text(1)$ is indeed a convergent series. Now grouping successive terms, $text(1)$ simplifies to
beginalign*
sum_n=1^N+1 chi(n)
&= - sum_k=1^infty (f_N(2k+1) - f_N(2k)) \
&= - sum_k=1^infty int_2k^2k+1 f_N'(x) , dx \
&= - sum_k=1^infty (N+1)(N+2) int_2k^2k+1 fracleft(1 - frac1xright)^Nx^3 , dx \
&= - sum_k=1^infty frac(N+1)(N+2)N^2 int_frac2kN^frac2k+1N fracleft(1 - frac1Nuright)^Nu^3 , du tag$x=Nu$
endalign*
Then Fubini-Tonelli's theorem gives
$$
sum_n=1^N+1 chi(n)
= - frac(N+1)(N+2)N^2 int_0^infty fracleft(1 - frac1Nuright)^Nu^3 left( sum_k=1^infty mathbf1_left[frac2kN, frac2k+1Nright](u) right) , du.
$$
Since the integrand is bounded by the integrable dominating function $e^-1/u/u^3$, as $Ntoinfty$ we have
$$
lim_Ntoinfty sum_n=1^N+1 chi(n)
= - int_0^infty frace^-1/uu^3 cdot frac12 , du
= - frac12.
$$
answered Jul 20 at 8:56
Sangchul Lee
85.6k12155253
edited Jul 22 at 2:58
answered Jul 20 at 8:56
Sangchul Lee
85.6k12155253
answered Jul 20 at 8:56
Sangchul Lee
85.6k12155253
answered Jul 20 at 8:56
Sangchul Lee
85.6k12155253
85.6k12155253
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2852832%2fproof-that-sum-limits-n-1-infty-n-sum-limits-j-2-infty-frac%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
A confirmation about the numeric evaluation being so slow would be nice though.
– Diger
Jul 15 at 22:03
1
To be honest: I don't really know! I guess this is quite delicate as for when you would replace $(-1)^j-1$ by $1^j-1=1$ then the entire result diverges. Also the last sum with alternating $1$s and $-1$s is not really defined, but only in the sense of analytic continuation. Therefore I guess your sum above would have to be replaced by some function $$ f(x)=sum _n=1^infty left( sum _j=2^infty frac x^j-1nj^ 2 left( 1-j^-1 right) ^n-1 right) $$ with $|x|<1$, because then the series is absolutely and uniformly convergent and can you interchange
– Diger
Jul 16 at 1:02
1
summation order. Then $f(x)$ is evaluated at $-1$ which amounts for, if your sum is the result of meromorphic function represented by this series, then the value of this function can be attached to it.
– Diger
Jul 16 at 1:06
1
Mathematica code:
Sum[Sum[((-1)^(j - 1)*n)/j^2*(1 - j^-1)^(n - 1), n, 1, Infinity, Assumptions -> j > 1] // Simplify, j, 2, Infinity, Regularization -> "Dirichlet"]gives: $-1/2$– Mariusz Iwaniuk
Jul 18 at 14:36
1
@crow Do you want a proof of the single identity in the title or of the double idendity below the title? Plus, what have you tried for proving them?
– Alex Francisco
Jul 18 at 15:58