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I suspect the following identity is valid, but I can not prove it. I just calculate it numerically.

$\sum_{m=0}^\infty\left[\sum_{n=0}^\infty\frac{(-1)^{n+m}}{(n+1)(n+m+1)}\right]=\sum_{m=0}^\infty\left[\sum_{n=0}^\infty\frac{(-1)^{m}}{(n+1)(n+m+2)}\right]$

I would appreciate any idea on how to prove it. Thanks.

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    $\begingroup$ In what order are you taking the summation? Neither of the sum is absolutely convergent, so it can equal anything by the Riemann rearrangement theorem. $\endgroup$
    – Fan Zheng
    Commented Mar 8, 2015 at 3:49
  • $\begingroup$ By following the usual order (First n, then m). $\endgroup$
    – cd14
    Commented Mar 8, 2015 at 4:02
  • $\begingroup$ Usual typographically, or with respect to the operations? It'd be much better to write the formula so that there is just one interpretation. By the way, where does this problem come from? $\endgroup$
    – Włodzimierz Holsztyński
    Commented Mar 8, 2015 at 4:06
  • $\begingroup$ The same: could you provide any background information on that identity. I hope it's not a homework problem you're asked to solve. $\endgroup$
    – Fan Zheng
    Commented Mar 8, 2015 at 4:09
  • $\begingroup$ This came from two different integral, which may have connections. This is the simplist case I suspect. $\endgroup$
    – cd14
    Commented Mar 8, 2015 at 4:14

1 Answer 1

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Both are equal to $(\pi^2-6\log^22)/12$. The inner sum on the right $$\sum_{n=0}^\infty\frac{(-1)^m}{(n+1)(n+m+2)}=\frac{(-1)^m}{m+1}\sum_{n=0}^\infty \Bigl(\frac{1}{n+1}-\frac{1}{n+m+2}\Bigr)= \frac{(-1)^m}{m+1}\Bigl(1+\frac12+\frac13+\cdots+\frac{1}{m+1}\Bigr)$$ Therefore the sum on the right is equal to $$R:=\sum_{m=0}^\infty \int_0^1(-1)^m x^m\Bigl(1+\frac12+\frac13+\cdots+\frac{1}{m+1}\Bigr)\,dx=$$ For $0<x<1$ we have $$\sum_{m=0}^{2M-1} (-1)^m x^m\Bigl(1+\frac12+\frac13+\cdots+\frac{1}{m+1}\Bigr)= \frac{1-x^{2M}}{1+x}-\frac{1}{2}\frac{x-x^{2M}}{1+x}+\cdots - \frac{1}{2M}\frac{x^{2M-1}-x^{2M}}{1+x}$$ So $R$ is the limit for $M\to\infty$ of $$R_M=\int_0^1\frac{1}{1+x}\Bigl(1-\frac{x}{2}+\frac{x^2}{3}-\cdots-\frac{x^{2M-1}}{2M} \Bigr)\,dx-\int_0^1\frac{x^{2M}}{1+x}\Bigl(1+\frac12+\frac13+\cdots +\frac{1}{2M}\Bigr)\,dx$$ The limit is easily seen to be $$R=\int_0^1\frac{\log(1+x)}{x(1+x)}\,dx=\frac{\pi^2-6\log^22}{12}.$$

The inner sum in the left hand side is for $m>0$ $$I=\sum_{n=0}^\infty \frac{(-1)^{n+m}}{(n+1)(n+m+1)}=\frac{(-1)^m}{m}\sum_{n=0}^\infty(-1)^n\Bigl(\frac{1}{n+1}-\frac{1}{n+m+1}\Bigr)$$ When $m=2k$ this is equal to $$I=\frac{1}{2k}\Bigl(1-\frac12+\frac13-\cdots-\frac{1}{2k}\Bigr).$$ When $m=2k-1$ $$ I=-\frac{1}{2k-1}\Bigl\{-\Bigl(1-\frac12+\frac13-\cdots-\frac{1}{2k-1}\Bigr)+2\sum_{j=0}^\infty \frac{(-1)^{j}}{j+1}\Bigr\}= -\frac{1}{2k-1}\Bigl\{2\log2-\Bigl(1-\frac12+\frac13-\cdots-\frac{1}{2k-1}\Bigr)\Bigr\}.$$ The terms with $m=0$ add to $$\sum_{n=0}^\infty\frac{(-1)^n}{(n+1)^2}=\frac{\pi^2}{12}.$$ Therefore the left hand side is $$L=\frac{\pi^2}{12}+\sum_{k=1}^\infty \Bigl(\frac{1}{2k}\Bigl(1-\frac12+\frac13-\cdots-\frac{1}{2k}\Bigr)+\frac{1}{2k-1}\Bigl\{-2\log2+\Bigl(1-\frac12+\frac13-\cdots-\frac{1}{2k-1}\Bigr)\Bigr)=$$ $$=\frac{\pi^2}{12}+\sum_{k=1}^\infty \Bigl(\frac{1}{2k}\Bigl\{-\log 2+\Bigl(1-\frac12+\frac13-\cdots-\frac{1}{2k}\Bigr)\Bigr\}+$$ $$+\frac{1}{2k-1}\Bigl\{-\log2+\Bigl(1-\frac12+\frac13-\cdots-\frac{1}{2k-1}\Bigr)\Bigr\}+\Bigl(\frac{1}{2k}-\frac{1}{2k-1}\Bigr)\log 2\Bigr)$$

$$L=\frac{\pi^2}{12}-\log^22+\sum_{k=1}^\infty\Bigl(\frac{1}{2k}\sum_{\ell={2k+1}}^\infty \frac{(-1)^\ell}{\ell}+\frac{1}{2k-1}\sum_{\ell=2k}^\infty\frac{(-1)^\ell}{\ell}\Bigr)=$$ $$= \frac{\pi^2}{12}-\log^22+\sum_{m=1}^\infty \frac{1}{m}\sum_{\ell=m+1}^\infty\frac{(-1)^\ell}{\ell}.$$ This is $$L=\frac{\pi^2}{12}-\log^22+\sum_{m=1}^\infty\frac{1}{m}\sum_{\ell=m+1}^\infty(-1)^\ell\int_0^1x^{\ell-1}\,dx= \frac{\pi^2}{12}-\log^22+\sum_{m=1}^\infty\frac{1}{m}\int_0^1\frac{(-1)^{m+1}x^m}{1+x}\,dx$$ $$= \frac{\pi^2}{12}-\log^22+\int_0^1\frac{\log(1+x)}{1+x}\,dx=\frac{\pi^2}{12}-\log^22+\frac12\log^22= \frac{\pi^2-6\log^22}{12}.$$

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  • $\begingroup$ 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.$\sum_{n=0}^\infty\frac{(-1)^n}{(n+1)^2}$. $\endgroup$
    – cd14
    Commented Mar 8, 2015 at 23:18
  • $\begingroup$ @cd14 With respect your first point I checked my formula,I think it is correct. And some numerical checks with Mathematica also agree. $\endgroup$
    – juan
    Commented Mar 9, 2015 at 9:11
  • $\begingroup$ @cd14 Your second observation is true. I have to think about it. Numerically you appear to be true and my value correct. $\endgroup$
    – juan
    Commented Mar 9, 2015 at 11:37
  • $\begingroup$ @cd14 I have corrected the second part of my answer. The missing terms was not the only error in my previous version. $\endgroup$
    – juan
    Commented Mar 9, 2015 at 16:32
  • $\begingroup$ 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. $\endgroup$
    – cd14
    Commented Mar 10, 2015 at 21:06

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