Timeline for How to prove this identity on double summation series?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Mar 12, 2015 at 19:19 | vote | accept | cd14 | ||
| Mar 11, 2015 at 20:17 | comment | added | juan | @cd14 Yes I corrected the signs. Of course the limit is zero, as i said before. | |
| Mar 11, 2015 at 20:14 | history | edited | juan | CC BY-SA 3.0 |
Correct several sign.
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| Mar 10, 2015 at 21:13 | comment | added | cd14 | I would be grateful if you let me know whether you agree or not. | |
| Mar 10, 2015 at 21:06 | comment | added | cd14 | Now I feel your results are correct except the first point in my last reply. But that term doesn't contribute. Actually, $\lim_{M\rightarrow\infty}\int_0^1\frac{x^{2M}}{1+x}\Bigl(1+\frac1 2+\frac1 3+\cdots+\frac{1}{2M}\Bigr)dx$ $=\lim_{M\rightarrow\infty}\int_0^1\frac{x^{2M}}{1+x}\Bigl(ln2M+\gamma\Bigr)$ $=\lim_{M\rightarrow\infty}\Bigl(ln2M+\gamma\Bigr)\int_0^1\Bigl(x^{2M}-\frac{x^{2M+1}}{1+x}\Bigr)dx$. Then we can prove this term disappears. Thank you very much. | |
| Mar 9, 2015 at 16:32 | comment | added | juan | @cd14 I have corrected the second part of my answer. The missing terms was not the only error in my previous version. | |
| Mar 9, 2015 at 16:29 | history | edited | juan | CC BY-SA 3.0 |
I corrected some wrong indices and signs in the previous version. The idea is the same
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| Mar 9, 2015 at 11:37 | comment | added | juan | @cd14 Your second observation is true. I have to think about it. Numerically you appear to be true and my value correct. | |
| Mar 9, 2015 at 9:11 | comment | added | juan | @cd14 With respect your first point I checked my formula,I think it is correct. And some numerical checks with Mathematica also agree. | |
| Mar 8, 2015 at 23:18 | comment | added | cd14 | Thanks a lot. I try to follow your answer. But I have some doubts. As for the right hand, $\sum_{m=0}^{2M-1} (-1)^m x^m\Bigl(1+\frac12+\frac13+\cdots+\frac{1}{m+1}\Bigr)= \frac{1-x^{2M}}{1+x}-\frac{1}{2}\frac{x-x^{2M}}{1+x}+\cdots - \frac{1}{2M}\frac{x^{2M-1}-x^{2M}}{1+x}$ may be not right. It should be $\sum_{m=0}^{2M-1} (-1)^m x^m\Bigl(1+\frac12+\frac13+\cdots+\frac{1}{m+1}\Bigr)= \frac{1-x^{2M}}{1+x}-\frac{1}{2}\frac{x+x^{2M}}{1+x}+\cdots - \frac{1}{2M}\frac{x^{2M-1}+x^{2M}}{1+x}.$ As for the left hand, you didn't add the term for $m=0$, i.e.$\sum_{n=0}^\infty\frac{(-1)^n}{(n+1)^2}$. | |
| Mar 8, 2015 at 10:33 | history | edited | juan | CC BY-SA 3.0 |
added 1 character in body
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| Mar 8, 2015 at 10:28 | history | answered | juan | CC BY-SA 3.0 |