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Show that $sum_n=0^infty frac3^2n+1(3n)!n!(2n+1)!4^3n+1=1$
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I'm trying to show that $$sum_n=0^infty frac3^2n+1(3n)!n!(2n+1)!4^3n+1=1$$
I've tried binomially expanding different expressions to obtain this but I can't seem to find anything, though I don't really have much practice at this sort of thing.
In particular, I am aware of the identities like
$$sum_n=0^infty frac2^n+1(2n)!n!(n+1)!3^2n=1,$$
but was unable to find anything suitable.
sequences-and-series
edited Jul 23 at 4:52
John Ma
37.5k93669
add a comment |Â
up vote
2
down vote
favorite
I'm trying to show that $$sum_n=0^infty frac3^2n+1(3n)!n!(2n+1)!4^3n+1=1$$
I've tried binomially expanding different expressions to obtain this but I can't seem to find anything, though I don't really have much practice at this sort of thing.
In particular, I am aware of the identities like
$$sum_n=0^infty frac2^n+1(2n)!n!(n+1)!3^2n=1,$$
but was unable to find anything suitable.
sequences-and-series
edited Jul 23 at 4:52
John Ma
37.5k93669
Where did you find this? I "proved" it with Wolfy's help, but two of the steps are magic. See my answer.
– marty cohen
Jul 17 at 2:19
simplifize $displaystyle sum _3nC_n*frac3^2n+1(2n+1)*4^3n+1$
– Takahiro Waki
Jul 17 at 2:58
1
This sum arises combinatorially from the generating function for the Fuss-Catalan Numbers. The desired value can be obtatined by calculating $frac34 C^(3)(frac964)$, using the associated cubic equation provided in the linked answer.
– Paul LeVan
Jul 17 at 3:01
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm trying to show that $$sum_n=0^infty frac3^2n+1(3n)!n!(2n+1)!4^3n+1=1$$
I've tried binomially expanding different expressions to obtain this but I can't seem to find anything, though I don't really have much practice at this sort of thing.
In particular, I am aware of the identities like
$$sum_n=0^infty frac2^n+1(2n)!n!(n+1)!3^2n=1,$$
but was unable to find anything suitable.
sequences-and-series
edited Jul 23 at 4:52
John Ma
37.5k93669
I'm trying to show that $$sum_n=0^infty frac3^2n+1(3n)!n!(2n+1)!4^3n+1=1$$
I've tried binomially expanding different expressions to obtain this but I can't seem to find anything, though I don't really have much practice at this sort of thing.
In particular, I am aware of the identities like
$$sum_n=0^infty frac2^n+1(2n)!n!(n+1)!3^2n=1,$$
but was unable to find anything suitable.
sequences-and-series
edited Jul 23 at 4:52
John Ma
37.5k93669
edited Jul 23 at 4:52
John Ma
37.5k93669
edited Jul 23 at 4:52
John Ma
37.5k93669
edited Jul 23 at 4:52
John Ma
37.5k93669
37.5k93669
asked Jul 17 at 1:38
user544680
Where did you find this? I "proved" it with Wolfy's help, but two of the steps are magic. See my answer.
– marty cohen
Jul 17 at 2:19
simplifize $displaystyle sum _3nC_n*frac3^2n+1(2n+1)*4^3n+1$
– Takahiro Waki
Jul 17 at 2:58
1
This sum arises combinatorially from the generating function for the Fuss-Catalan Numbers. The desired value can be obtatined by calculating $frac34 C^(3)(frac964)$, using the associated cubic equation provided in the linked answer.
– Paul LeVan
Jul 17 at 3:01
add a comment |Â
Where did you find this? I "proved" it with Wolfy's help, but two of the steps are magic. See my answer.
– marty cohen
Jul 17 at 2:19
simplifize $displaystyle sum _3nC_n*frac3^2n+1(2n+1)*4^3n+1$
– Takahiro Waki
Jul 17 at 2:58
1
This sum arises combinatorially from the generating function for the Fuss-Catalan Numbers. The desired value can be obtatined by calculating $frac34 C^(3)(frac964)$, using the associated cubic equation provided in the linked answer.
– Paul LeVan
Jul 17 at 3:01
Where did you find this? I "proved" it with Wolfy's help, but two of the steps are magic. See my answer.
– marty cohen
Jul 17 at 2:19
Where did you find this? I "proved" it with Wolfy's help, but two of the steps are magic. See my answer.
– marty cohen
Jul 17 at 2:19
simplifize $displaystyle sum _3nC_n*frac3^2n+1(2n+1)*4^3n+1$
– Takahiro Waki
Jul 17 at 2:58
simplifize $displaystyle sum _3nC_n*frac3^2n+1(2n+1)*4^3n+1$
– Takahiro Waki
Jul 17 at 2:58
1
1
This sum arises combinatorially from the generating function for the Fuss-Catalan Numbers. The desired value can be obtatined by calculating $frac34 C^(3)(frac964)$, using the associated cubic equation provided in the linked answer.
– Paul LeVan
Jul 17 at 3:01
This sum arises combinatorially from the generating function for the Fuss-Catalan Numbers. The desired value can be obtatined by calculating $frac34 C^(3)(frac964)$, using the associated cubic equation provided in the linked answer.
– Paul LeVan
Jul 17 at 3:01
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
Define $, a_n := (3n)!/(n!(2n+1)!),$ which is OEIS sequence A001764 and $, b_n := a_n 3^2n+1/4^3n+1. ,$
The generating function
$$
f(x) := sum_n=0^infty a_n, x^n = frac2sqrt3x
sinBig(frac13, sin^-1Big(sqrt27x/4Big)Big) $$
and thus $$ sum_n=0^infty b_n, x^n = frac34 fBig(frac964xBig) =
frac4sqrt3x sinBig(frac13, sin^-1
Big(frac916sqrt3x Big)Big). $$
Set $, x = 1 ,$ and to prove
$$ sum_n=0^infty b_n , =,
frac4sqrt3 sinBig(frac13 sin^-1Big(frac9sqrt316Big)Big) = 1 $$
use the identity
$, sin(3theta) = 3 sin(theta) - 4sin(theta)^3 ,$
where $, sin(theta) = sqrt3/4 ,$ to find
$, sin(3theta) = 9sqrt3/16. $
In general, for $, n>1 ,$ we get that $, f(n^2/(n+1)^3) = (n+1)/n. ,$
Our example is $, f(9/64) = 4/3.,$
answered Jul 17 at 2:24
Somos
11.6k1933
How do I see that $f$ is given by this function?
– user544680
Jul 17 at 21:46
1
One is to look in OEIS which has the g.f. given. Another is to use a CAS such as Maple or Mathematica. There are other ways but they are harder and require much more knowledge about enumberative combinatorics, for example. See the almost 100 references and links in the OEIS entry
– Somos
Jul 17 at 22:38
add a comment |Â
up vote
0
down vote
Let's group things a little.
$sum_n=0^infty frac3^2n+1(3n)!n!(2n+1)!4^3n+1
=dfrac34sum_n=0^infty frac3^2n(3n)!n!(2n+1)!4^3n
=dfrac34sum_n=0^infty (3^2/4^3)^nfrac(3n)!n!(2n+1)!
$
so let
$f(x)
=sum_n=0^infty x^nfrac(3n)!n!(2n+1)!
$.
This looks sort of
like a trisection of series,
but,
being lazy,
I threw it at Wolfy and,
to my great surprise,
got
$f(x)
=dfrac2 sin(frac13 sin^-1((3 sqrt3x/2)))sqrt3x
$
which converges when
$|x| < 4/27$.
Putting $x=9/64$,
$beginarray\
f(9/64)
&=dfrac2 sin(frac13 sin^-1((3 sqrt3(9/64)/2)))sqrt3(9/64)\
&=dfrac2 sin(frac13 sin^-1((3 cdot 3sqrt3/8/2)))3sqrt3/8\
&=dfrac16 sin(frac13 sin^-1(9sqrt3/16))3sqrt3\
&=dfrac16 (sqrt3/4)3sqrt3
qquadtextagain, according to Wolfy\
&=dfrac43\
endarray
$
Muptiplying by $dfrac34$,
the result is $1$.
As to how I could
prove it by my self,
I don't know.
answered Jul 17 at 2:17
marty cohen
69.3k446122
2
Amazing ! I directly gave a CAS $$f(x)=sum_n=0^infty frac3^2n+1(3n)!n!(2n+1)!4^3n+1x^n$$ and received your answer.
– Claude Leibovici
Jul 17 at 2:50
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Define $, a_n := (3n)!/(n!(2n+1)!),$ which is OEIS sequence A001764 and $, b_n := a_n 3^2n+1/4^3n+1. ,$
The generating function
$$
f(x) := sum_n=0^infty a_n, x^n = frac2sqrt3x
sinBig(frac13, sin^-1Big(sqrt27x/4Big)Big) $$
and thus $$ sum_n=0^infty b_n, x^n = frac34 fBig(frac964xBig) =
frac4sqrt3x sinBig(frac13, sin^-1
Big(frac916sqrt3x Big)Big). $$
Set $, x = 1 ,$ and to prove
$$ sum_n=0^infty b_n , =,
frac4sqrt3 sinBig(frac13 sin^-1Big(frac9sqrt316Big)Big) = 1 $$
use the identity
$, sin(3theta) = 3 sin(theta) - 4sin(theta)^3 ,$
where $, sin(theta) = sqrt3/4 ,$ to find
$, sin(3theta) = 9sqrt3/16. $
In general, for $, n>1 ,$ we get that $, f(n^2/(n+1)^3) = (n+1)/n. ,$
Our example is $, f(9/64) = 4/3.,$
answered Jul 17 at 2:24
Somos
11.6k1933
How do I see that $f$ is given by this function?
– user544680
Jul 17 at 21:46
1
One is to look in OEIS which has the g.f. given. Another is to use a CAS such as Maple or Mathematica. There are other ways but they are harder and require much more knowledge about enumberative combinatorics, for example. See the almost 100 references and links in the OEIS entry
– Somos
Jul 17 at 22:38
add a comment |Â
up vote
3
down vote
Define $, a_n := (3n)!/(n!(2n+1)!),$ which is OEIS sequence A001764 and $, b_n := a_n 3^2n+1/4^3n+1. ,$
The generating function
$$
f(x) := sum_n=0^infty a_n, x^n = frac2sqrt3x
sinBig(frac13, sin^-1Big(sqrt27x/4Big)Big) $$
and thus $$ sum_n=0^infty b_n, x^n = frac34 fBig(frac964xBig) =
frac4sqrt3x sinBig(frac13, sin^-1
Big(frac916sqrt3x Big)Big). $$
Set $, x = 1 ,$ and to prove
$$ sum_n=0^infty b_n , =,
frac4sqrt3 sinBig(frac13 sin^-1Big(frac9sqrt316Big)Big) = 1 $$
use the identity
$, sin(3theta) = 3 sin(theta) - 4sin(theta)^3 ,$
where $, sin(theta) = sqrt3/4 ,$ to find
$, sin(3theta) = 9sqrt3/16. $
In general, for $, n>1 ,$ we get that $, f(n^2/(n+1)^3) = (n+1)/n. ,$
Our example is $, f(9/64) = 4/3.,$
answered Jul 17 at 2:24
Somos
11.6k1933
How do I see that $f$ is given by this function?
– user544680
Jul 17 at 21:46
1
One is to look in OEIS which has the g.f. given. Another is to use a CAS such as Maple or Mathematica. There are other ways but they are harder and require much more knowledge about enumberative combinatorics, for example. See the almost 100 references and links in the OEIS entry
– Somos
Jul 17 at 22:38
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Define $, a_n := (3n)!/(n!(2n+1)!),$ which is OEIS sequence A001764 and $, b_n := a_n 3^2n+1/4^3n+1. ,$
The generating function
$$
f(x) := sum_n=0^infty a_n, x^n = frac2sqrt3x
sinBig(frac13, sin^-1Big(sqrt27x/4Big)Big) $$
and thus $$ sum_n=0^infty b_n, x^n = frac34 fBig(frac964xBig) =
frac4sqrt3x sinBig(frac13, sin^-1
Big(frac916sqrt3x Big)Big). $$
Set $, x = 1 ,$ and to prove
$$ sum_n=0^infty b_n , =,
frac4sqrt3 sinBig(frac13 sin^-1Big(frac9sqrt316Big)Big) = 1 $$
use the identity
$, sin(3theta) = 3 sin(theta) - 4sin(theta)^3 ,$
where $, sin(theta) = sqrt3/4 ,$ to find
$, sin(3theta) = 9sqrt3/16. $
In general, for $, n>1 ,$ we get that $, f(n^2/(n+1)^3) = (n+1)/n. ,$
Our example is $, f(9/64) = 4/3.,$
answered Jul 17 at 2:24
Somos
11.6k1933
Define $, a_n := (3n)!/(n!(2n+1)!),$ which is OEIS sequence A001764 and $, b_n := a_n 3^2n+1/4^3n+1. ,$
The generating function
$$
f(x) := sum_n=0^infty a_n, x^n = frac2sqrt3x
sinBig(frac13, sin^-1Big(sqrt27x/4Big)Big) $$
and thus $$ sum_n=0^infty b_n, x^n = frac34 fBig(frac964xBig) =
frac4sqrt3x sinBig(frac13, sin^-1
Big(frac916sqrt3x Big)Big). $$
Set $, x = 1 ,$ and to prove
$$ sum_n=0^infty b_n , =,
frac4sqrt3 sinBig(frac13 sin^-1Big(frac9sqrt316Big)Big) = 1 $$
use the identity
$, sin(3theta) = 3 sin(theta) - 4sin(theta)^3 ,$
where $, sin(theta) = sqrt3/4 ,$ to find
$, sin(3theta) = 9sqrt3/16. $
In general, for $, n>1 ,$ we get that $, f(n^2/(n+1)^3) = (n+1)/n. ,$
Our example is $, f(9/64) = 4/3.,$
answered Jul 17 at 2:24
Somos
11.6k1933
edited Jul 17 at 11:49
answered Jul 17 at 2:24
Somos
11.6k1933
answered Jul 17 at 2:24
Somos
11.6k1933
answered Jul 17 at 2:24
Somos
11.6k1933
11.6k1933
How do I see that $f$ is given by this function?
– user544680
Jul 17 at 21:46
1
One is to look in OEIS which has the g.f. given. Another is to use a CAS such as Maple or Mathematica. There are other ways but they are harder and require much more knowledge about enumberative combinatorics, for example. See the almost 100 references and links in the OEIS entry
– Somos
Jul 17 at 22:38
add a comment |Â
How do I see that $f$ is given by this function?
– user544680
Jul 17 at 21:46
1
One is to look in OEIS which has the g.f. given. Another is to use a CAS such as Maple or Mathematica. There are other ways but they are harder and require much more knowledge about enumberative combinatorics, for example. See the almost 100 references and links in the OEIS entry
– Somos
Jul 17 at 22:38
How do I see that $f$ is given by this function?
– user544680
Jul 17 at 21:46
How do I see that $f$ is given by this function?
– user544680
Jul 17 at 21:46
1
1
One is to look in OEIS which has the g.f. given. Another is to use a CAS such as Maple or Mathematica. There are other ways but they are harder and require much more knowledge about enumberative combinatorics, for example. See the almost 100 references and links in the OEIS entry
– Somos
Jul 17 at 22:38
One is to look in OEIS which has the g.f. given. Another is to use a CAS such as Maple or Mathematica. There are other ways but they are harder and require much more knowledge about enumberative combinatorics, for example. See the almost 100 references and links in the OEIS entry
– Somos
Jul 17 at 22:38
add a comment |Â
up vote
0
down vote
Let's group things a little.
$sum_n=0^infty frac3^2n+1(3n)!n!(2n+1)!4^3n+1
=dfrac34sum_n=0^infty frac3^2n(3n)!n!(2n+1)!4^3n
=dfrac34sum_n=0^infty (3^2/4^3)^nfrac(3n)!n!(2n+1)!
$
so let
$f(x)
=sum_n=0^infty x^nfrac(3n)!n!(2n+1)!
$.
This looks sort of
like a trisection of series,
but,
being lazy,
I threw it at Wolfy and,
to my great surprise,
got
$f(x)
=dfrac2 sin(frac13 sin^-1((3 sqrt3x/2)))sqrt3x
$
which converges when
$|x| < 4/27$.
Putting $x=9/64$,
$beginarray\
f(9/64)
&=dfrac2 sin(frac13 sin^-1((3 sqrt3(9/64)/2)))sqrt3(9/64)\
&=dfrac2 sin(frac13 sin^-1((3 cdot 3sqrt3/8/2)))3sqrt3/8\
&=dfrac16 sin(frac13 sin^-1(9sqrt3/16))3sqrt3\
&=dfrac16 (sqrt3/4)3sqrt3
qquadtextagain, according to Wolfy\
&=dfrac43\
endarray
$
Muptiplying by $dfrac34$,
the result is $1$.
As to how I could
prove it by my self,
I don't know.
answered Jul 17 at 2:17
marty cohen
69.3k446122
2
Amazing ! I directly gave a CAS $$f(x)=sum_n=0^infty frac3^2n+1(3n)!n!(2n+1)!4^3n+1x^n$$ and received your answer.
– Claude Leibovici
Jul 17 at 2:50
add a comment |Â
up vote
0
down vote
Let's group things a little.
$sum_n=0^infty frac3^2n+1(3n)!n!(2n+1)!4^3n+1
=dfrac34sum_n=0^infty frac3^2n(3n)!n!(2n+1)!4^3n
=dfrac34sum_n=0^infty (3^2/4^3)^nfrac(3n)!n!(2n+1)!
$
so let
$f(x)
=sum_n=0^infty x^nfrac(3n)!n!(2n+1)!
$.
This looks sort of
like a trisection of series,
but,
being lazy,
I threw it at Wolfy and,
to my great surprise,
got
$f(x)
=dfrac2 sin(frac13 sin^-1((3 sqrt3x/2)))sqrt3x
$
which converges when
$|x| < 4/27$.
Putting $x=9/64$,
$beginarray\
f(9/64)
&=dfrac2 sin(frac13 sin^-1((3 sqrt3(9/64)/2)))sqrt3(9/64)\
&=dfrac2 sin(frac13 sin^-1((3 cdot 3sqrt3/8/2)))3sqrt3/8\
&=dfrac16 sin(frac13 sin^-1(9sqrt3/16))3sqrt3\
&=dfrac16 (sqrt3/4)3sqrt3
qquadtextagain, according to Wolfy\
&=dfrac43\
endarray
$
Muptiplying by $dfrac34$,
the result is $1$.
As to how I could
prove it by my self,
I don't know.
answered Jul 17 at 2:17
marty cohen
69.3k446122
2
Amazing ! I directly gave a CAS $$f(x)=sum_n=0^infty frac3^2n+1(3n)!n!(2n+1)!4^3n+1x^n$$ and received your answer.
– Claude Leibovici
Jul 17 at 2:50
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let's group things a little.
$sum_n=0^infty frac3^2n+1(3n)!n!(2n+1)!4^3n+1
=dfrac34sum_n=0^infty frac3^2n(3n)!n!(2n+1)!4^3n
=dfrac34sum_n=0^infty (3^2/4^3)^nfrac(3n)!n!(2n+1)!
$
so let
$f(x)
=sum_n=0^infty x^nfrac(3n)!n!(2n+1)!
$.
This looks sort of
like a trisection of series,
but,
being lazy,
I threw it at Wolfy and,
to my great surprise,
got
$f(x)
=dfrac2 sin(frac13 sin^-1((3 sqrt3x/2)))sqrt3x
$
which converges when
$|x| < 4/27$.
Putting $x=9/64$,
$beginarray\
f(9/64)
&=dfrac2 sin(frac13 sin^-1((3 sqrt3(9/64)/2)))sqrt3(9/64)\
&=dfrac2 sin(frac13 sin^-1((3 cdot 3sqrt3/8/2)))3sqrt3/8\
&=dfrac16 sin(frac13 sin^-1(9sqrt3/16))3sqrt3\
&=dfrac16 (sqrt3/4)3sqrt3
qquadtextagain, according to Wolfy\
&=dfrac43\
endarray
$
Muptiplying by $dfrac34$,
the result is $1$.
As to how I could
prove it by my self,
I don't know.
answered Jul 17 at 2:17
marty cohen
69.3k446122
Let's group things a little.
$sum_n=0^infty frac3^2n+1(3n)!n!(2n+1)!4^3n+1
=dfrac34sum_n=0^infty frac3^2n(3n)!n!(2n+1)!4^3n
=dfrac34sum_n=0^infty (3^2/4^3)^nfrac(3n)!n!(2n+1)!
$
so let
$f(x)
=sum_n=0^infty x^nfrac(3n)!n!(2n+1)!
$.
This looks sort of
like a trisection of series,
but,
being lazy,
I threw it at Wolfy and,
to my great surprise,
got
$f(x)
=dfrac2 sin(frac13 sin^-1((3 sqrt3x/2)))sqrt3x
$
which converges when
$|x| < 4/27$.
Putting $x=9/64$,
$beginarray\
f(9/64)
&=dfrac2 sin(frac13 sin^-1((3 sqrt3(9/64)/2)))sqrt3(9/64)\
&=dfrac2 sin(frac13 sin^-1((3 cdot 3sqrt3/8/2)))3sqrt3/8\
&=dfrac16 sin(frac13 sin^-1(9sqrt3/16))3sqrt3\
&=dfrac16 (sqrt3/4)3sqrt3
qquadtextagain, according to Wolfy\
&=dfrac43\
endarray
$
Muptiplying by $dfrac34$,
the result is $1$.
As to how I could
prove it by my self,
I don't know.
answered Jul 17 at 2:17
marty cohen
69.3k446122
answered Jul 17 at 2:17
marty cohen
69.3k446122
answered Jul 17 at 2:17
marty cohen
69.3k446122
answered Jul 17 at 2:17
marty cohen
69.3k446122
69.3k446122
2
Amazing ! I directly gave a CAS $$f(x)=sum_n=0^infty frac3^2n+1(3n)!n!(2n+1)!4^3n+1x^n$$ and received your answer.
– Claude Leibovici
Jul 17 at 2:50
add a comment |Â
2
Amazing ! I directly gave a CAS $$f(x)=sum_n=0^infty frac3^2n+1(3n)!n!(2n+1)!4^3n+1x^n$$ and received your answer.
– Claude Leibovici
Jul 17 at 2:50
2
2
Amazing ! I directly gave a CAS $$f(x)=sum_n=0^infty frac3^2n+1(3n)!n!(2n+1)!4^3n+1x^n$$ and received your answer.
– Claude Leibovici
Jul 17 at 2:50
Amazing ! I directly gave a CAS $$f(x)=sum_n=0^infty frac3^2n+1(3n)!n!(2n+1)!4^3n+1x^n$$ and received your answer.
– Claude Leibovici
Jul 17 at 2:50
add a comment |Â
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Where did you find this? I "proved" it with Wolfy's help, but two of the steps are magic. See my answer.
– marty cohen
Jul 17 at 2:19
simplifize $displaystyle sum _3nC_n*frac3^2n+1(2n+1)*4^3n+1$
– Takahiro Waki
Jul 17 at 2:58
1
This sum arises combinatorially from the generating function for the Fuss-Catalan Numbers. The desired value can be obtatined by calculating $frac34 C^(3)(frac964)$, using the associated cubic equation provided in the linked answer.
– Paul LeVan
Jul 17 at 3:01