Abel Summation Formula

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Out of curiosity I wanted to calculate
$$
sum_n=1^infty fracleft(-1right)^n-1n = ln 2 approx 0.6931471806
$$
using Abel Summation formula
$$
lim_xrightarrowinfty left fracsum_nleq x left(-1right)^n-1x + int_1^x rm du , frac1-left(-1right)^lfloor u rfloor2u^2 right = int_1^infty rm du , frac1-left(-1right)^lfloor u rfloor2u^2 approx 0.6687714032 , .
$$
So any insights in why the result is different?




calculus sequences-and-series




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  • What should it yield? I used $$sum_n=1^lfloor x rfloor u^n-1 = frac1-u^lfloor x rfloor1-u , .$$ en.wikipedia.org/wiki/Abel%27s_summation_formula
    – Diger
    Jul 21 at 1:59















up vote
0
down vote

favorite












Out of curiosity I wanted to calculate
$$
sum_n=1^infty fracleft(-1right)^n-1n = ln 2 approx 0.6931471806
$$
using Abel Summation formula
$$
lim_xrightarrowinfty left fracsum_nleq x left(-1right)^n-1x + int_1^x rm du , frac1-left(-1right)^lfloor u rfloor2u^2 right = int_1^infty rm du , frac1-left(-1right)^lfloor u rfloor2u^2 approx 0.6687714032 , .
$$
So any insights in why the result is different?




calculus sequences-and-series




share|cite|improve this question



















  • What should it yield? I used $$sum_n=1^lfloor x rfloor u^n-1 = frac1-u^lfloor x rfloor1-u , .$$ en.wikipedia.org/wiki/Abel%27s_summation_formula
    – Diger
    Jul 21 at 1:59













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Out of curiosity I wanted to calculate
$$
sum_n=1^infty fracleft(-1right)^n-1n = ln 2 approx 0.6931471806
$$
using Abel Summation formula
$$
lim_xrightarrowinfty left fracsum_nleq x left(-1right)^n-1x + int_1^x rm du , frac1-left(-1right)^lfloor u rfloor2u^2 right = int_1^infty rm du , frac1-left(-1right)^lfloor u rfloor2u^2 approx 0.6687714032 , .
$$
So any insights in why the result is different?




calculus sequences-and-series




share|cite|improve this question











Out of curiosity I wanted to calculate
$$
sum_n=1^infty fracleft(-1right)^n-1n = ln 2 approx 0.6931471806
$$
using Abel Summation formula
$$
lim_xrightarrowinfty left fracsum_nleq x left(-1right)^n-1x + int_1^x rm du , frac1-left(-1right)^lfloor u rfloor2u^2 right = int_1^infty rm du , frac1-left(-1right)^lfloor u rfloor2u^2 approx 0.6687714032 , .
$$
So any insights in why the result is different?





calculus sequences-and-series





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asked Jul 21 at 1:45









Diger

56329




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  • What should it yield? I used $$sum_n=1^lfloor x rfloor u^n-1 = frac1-u^lfloor x rfloor1-u , .$$ en.wikipedia.org/wiki/Abel%27s_summation_formula
    – Diger
    Jul 21 at 1:59

















  • What should it yield? I used $$sum_n=1^lfloor x rfloor u^n-1 = frac1-u^lfloor x rfloor1-u , .$$ en.wikipedia.org/wiki/Abel%27s_summation_formula
    – Diger
    Jul 21 at 1:59
















What should it yield? I used $$sum_n=1^lfloor x rfloor u^n-1 = frac1-u^lfloor x rfloor1-u , .$$ en.wikipedia.org/wiki/Abel%27s_summation_formula
– Diger
Jul 21 at 1:59





What should it yield? I used $$sum_n=1^lfloor x rfloor u^n-1 = frac1-u^lfloor x rfloor1-u , .$$ en.wikipedia.org/wiki/Abel%27s_summation_formula
– Diger
Jul 21 at 1:59











1 Answer
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This last integral is equal to your series, but there's not much of a better way to compute it:



$$int_1^infty frac1 - (-1)^lfloor u rfloor2u^2 ,du = int_ u geq 1 : lfloor u rfloor mathrmoddfrac1u^2,du =sum_k = 0^infty int_2k+1^2k+2 frac1u^2,du = sum_k = 0^infty left(frac12k+1 - frac12k+2 right).$$



How did you calculate your approximation to this last integral?






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  • numerically in maple int((1-x^floor(u))/((1-x)*u^2), u = 1 .. infinity, numeric) where $x=-1$. You think it's a rounding issue?
    – Diger
    Jul 21 at 2:29











  • Yeah, there's something weird happening with Maple, and I'm not sure what it is. The integrand is clearly non-negative, but if you replace "infinity" with, say, 10000, then Maple gives 0.6930971831, which is closer to the correct answer. So it's doing something weird; maybe it thinks it's converging faster than it actually is?
    – Marcus M
    Jul 21 at 2:37










  • I don't really know either. Maybe I ask that question @mapleprimes.
    – Diger
    Jul 21 at 2:39










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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote













This last integral is equal to your series, but there's not much of a better way to compute it:



$$int_1^infty frac1 - (-1)^lfloor u rfloor2u^2 ,du = int_ u geq 1 : lfloor u rfloor mathrmoddfrac1u^2,du =sum_k = 0^infty int_2k+1^2k+2 frac1u^2,du = sum_k = 0^infty left(frac12k+1 - frac12k+2 right).$$



How did you calculate your approximation to this last integral?






share|cite|improve this answer





















  • numerically in maple int((1-x^floor(u))/((1-x)*u^2), u = 1 .. infinity, numeric) where $x=-1$. You think it's a rounding issue?
    – Diger
    Jul 21 at 2:29











  • Yeah, there's something weird happening with Maple, and I'm not sure what it is. The integrand is clearly non-negative, but if you replace "infinity" with, say, 10000, then Maple gives 0.6930971831, which is closer to the correct answer. So it's doing something weird; maybe it thinks it's converging faster than it actually is?
    – Marcus M
    Jul 21 at 2:37










  • I don't really know either. Maybe I ask that question @mapleprimes.
    – Diger
    Jul 21 at 2:39














up vote
0
down vote













This last integral is equal to your series, but there's not much of a better way to compute it:



$$int_1^infty frac1 - (-1)^lfloor u rfloor2u^2 ,du = int_ u geq 1 : lfloor u rfloor mathrmoddfrac1u^2,du =sum_k = 0^infty int_2k+1^2k+2 frac1u^2,du = sum_k = 0^infty left(frac12k+1 - frac12k+2 right).$$



How did you calculate your approximation to this last integral?






share|cite|improve this answer





















  • numerically in maple int((1-x^floor(u))/((1-x)*u^2), u = 1 .. infinity, numeric) where $x=-1$. You think it's a rounding issue?
    – Diger
    Jul 21 at 2:29











  • Yeah, there's something weird happening with Maple, and I'm not sure what it is. The integrand is clearly non-negative, but if you replace "infinity" with, say, 10000, then Maple gives 0.6930971831, which is closer to the correct answer. So it's doing something weird; maybe it thinks it's converging faster than it actually is?
    – Marcus M
    Jul 21 at 2:37










  • I don't really know either. Maybe I ask that question @mapleprimes.
    – Diger
    Jul 21 at 2:39












up vote
0
down vote










up vote
0
down vote









This last integral is equal to your series, but there's not much of a better way to compute it:



$$int_1^infty frac1 - (-1)^lfloor u rfloor2u^2 ,du = int_ u geq 1 : lfloor u rfloor mathrmoddfrac1u^2,du =sum_k = 0^infty int_2k+1^2k+2 frac1u^2,du = sum_k = 0^infty left(frac12k+1 - frac12k+2 right).$$



How did you calculate your approximation to this last integral?






share|cite|improve this answer













This last integral is equal to your series, but there's not much of a better way to compute it:



$$int_1^infty frac1 - (-1)^lfloor u rfloor2u^2 ,du = int_ u geq 1 : lfloor u rfloor mathrmoddfrac1u^2,du =sum_k = 0^infty int_2k+1^2k+2 frac1u^2,du = sum_k = 0^infty left(frac12k+1 - frac12k+2 right).$$



How did you calculate your approximation to this last integral?







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 21 at 2:23









Marcus M

8,1731847




8,1731847











  • numerically in maple int((1-x^floor(u))/((1-x)*u^2), u = 1 .. infinity, numeric) where $x=-1$. You think it's a rounding issue?
    – Diger
    Jul 21 at 2:29











  • Yeah, there's something weird happening with Maple, and I'm not sure what it is. The integrand is clearly non-negative, but if you replace "infinity" with, say, 10000, then Maple gives 0.6930971831, which is closer to the correct answer. So it's doing something weird; maybe it thinks it's converging faster than it actually is?
    – Marcus M
    Jul 21 at 2:37










  • I don't really know either. Maybe I ask that question @mapleprimes.
    – Diger
    Jul 21 at 2:39
















  • numerically in maple int((1-x^floor(u))/((1-x)*u^2), u = 1 .. infinity, numeric) where $x=-1$. You think it's a rounding issue?
    – Diger
    Jul 21 at 2:29











  • Yeah, there's something weird happening with Maple, and I'm not sure what it is. The integrand is clearly non-negative, but if you replace "infinity" with, say, 10000, then Maple gives 0.6930971831, which is closer to the correct answer. So it's doing something weird; maybe it thinks it's converging faster than it actually is?
    – Marcus M
    Jul 21 at 2:37










  • I don't really know either. Maybe I ask that question @mapleprimes.
    – Diger
    Jul 21 at 2:39















numerically in maple int((1-x^floor(u))/((1-x)*u^2), u = 1 .. infinity, numeric) where $x=-1$. You think it's a rounding issue?
– Diger
Jul 21 at 2:29





numerically in maple int((1-x^floor(u))/((1-x)*u^2), u = 1 .. infinity, numeric) where $x=-1$. You think it's a rounding issue?
– Diger
Jul 21 at 2:29













Yeah, there's something weird happening with Maple, and I'm not sure what it is. The integrand is clearly non-negative, but if you replace "infinity" with, say, 10000, then Maple gives 0.6930971831, which is closer to the correct answer. So it's doing something weird; maybe it thinks it's converging faster than it actually is?
– Marcus M
Jul 21 at 2:37




Yeah, there's something weird happening with Maple, and I'm not sure what it is. The integrand is clearly non-negative, but if you replace "infinity" with, say, 10000, then Maple gives 0.6930971831, which is closer to the correct answer. So it's doing something weird; maybe it thinks it's converging faster than it actually is?
– Marcus M
Jul 21 at 2:37












I don't really know either. Maybe I ask that question @mapleprimes.
– Diger
Jul 21 at 2:39




I don't really know either. Maybe I ask that question @mapleprimes.
– Diger
Jul 21 at 2:39












 

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