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Show that $$\sum_{n=1}^{\infty }\left ( \sum_{j=1}^{\infty }\frac{x^{(n-j)^2}-x^{(n+j-1)^2}}{(2n-1)(2j-1)} \right) = \frac{\pi^2}{8}$$

I liked this problem because the result is a final answer, and not a function or an answer with a variable! It's rare to find such beautiful problems

Would like to share my solution if it is not prohibited and look at other solutions. I'm more interested in knowing if there are exotic ways to solve this problem? For example, through special functions or some approximations and so on

Solution:

\begin{align*} S=&\sum_{n=1}^{\infty }\left ( \sum_{j=1}^{\infty }\frac{x^{(n-j)^2}-x^{(n+j-1)^2}}{(2n-1)(2j-1)} \right )=\sum_{n=1}^{\infty }\left [ \sum_{j=1}^{\infty }\frac{x^{(n-j)^2}}{(2n-1)(2j-1)}-\sum_{j=1}^{\infty } \frac{x^{(n+j-1)^2}}{(2n-1)(2j-1)}\right ]=\\=&\sum_{n=1}^{\infty }\left [ \frac{1}{(2n-1)^2}+\sum_{j=1}^{\infty }\left [ \frac{1}{2n-1}-\frac{1}{2j-1} \right ]\frac{x^{(n-j)^2}}{2(j-n)}-\sum_{j=1}^{\infty }\left [ \frac{1}{2n-1}+\frac{1}{2j-1} \right ]\frac{x^{(n+j-1)^2}}{2(n+j-1)} \right ]\\=&\sum_{n=1}^{\infty }\frac{1}{(2n-1)^2}-\\-&\frac{1}{2}\sum_{n=1}^{\infty }\left [ \sum_{j=1}^{\infty }\left [ \frac{1}{2n-1}-\frac{1}{2j-1} \right ]\frac{x^{(n-j)^2}}{n-j}+\sum_{j=1}^{\infty }\left [ \frac{1}{2n-1}+\frac{1}{2j-1} \right ]\frac{x^{(n+j-1)^2}}{n+j-1} \right ]=\\=&\sum_{n=1}^{\infty }\frac{1}{(2n-1)^2}-S_1 \end{align*}

\begin{align} S_1=&\sum_{n=1}^{\infty }\frac{1}{2n-1}\left [ \sum_{j=1}^{\infty }\frac{x^{(n-j)^2}}{n-j}+\sum_{j=1}^{\infty }\frac{x^{(n+j-1)^2}}{n+j-1} \right ]=\sum_{n=1}^{\infty }\frac{1}{2n-1}\left [ \sum_{k=n-1}^{\infty }\frac{x^{k^2}}{k}+\sum_{k=n}^{\infty }\frac{x^{k^2}}{k}\right ]=\\=&\sum_{n=1}^{\infty }\frac{1}{2n-1}\sum_{k=-\infty }^{\infty }\frac{x^{k^2}}{k}=0\Rightarrow S=\sum_{n=1}^{\infty }\frac{1}{(2n-1)^2}=\frac{3}{4}\zeta (2)=\frac{\pi^2}{8} \end{align}

Blue
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1 Answers1

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$$\sum_{j=1}^\infty\frac{x^{(j-n)^2}-x^{(j-1+n)^2}}{2j-1}$$ $$=\sum_{k=1-n}^\infty\frac{x^{k^2}}{2k+2n-1}-\sum_{k=n}^\infty\frac{x^{k^2}}{2k-2n+1}$$ $$=\sum_{k=1-n}^{n-1}\frac{x^{k^2}}{2k+2n-1}+\sum_{k=n}^\infty x^{k^2}\left(\frac1{2k+2n-1}-\frac1{2k-2n+1}\right)$$ $$=\frac1{2n-1}+\sum_{k=1}^\infty x^{k^2}\left(\frac1{2k+2n-1}-\frac1{2k-2n+1}\right)$$ $$=\frac1{2n-1}+2(2n-1)\sum_{k=1}^\infty\frac{x^{k^2}}{(2n-1)^2-4k^2}.$$ Therefore, $$\sum_{n=1}^\infty\frac1{2n-1}\sum_{j=1}^\infty\frac{x^{(j-n)^2}-x^{(j-1+n)^2}}{2j-1}$$ $$=\sum_{n=1}^\infty\frac1{(2n-1)^2}+2\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{x^{k^2}}{(2n-1)^2-4k^2}$$ $$=\frac{\pi^2}8+2\sum_{k=1}^\infty x^{k^2}\sum_{n=1}^\infty\frac1{(2n-1)^2-4k^2}$$ $$=\frac{\pi^2}8,$$ since $$\sum_{n=1}^\infty\frac1{(2n-1)^2-4k^2}=\frac\pi{8k}\tan(k\pi)=0.$$

Anne Bauval
  • 49,005
  • This is not what I wanted. Thanks for the solution, but it seems you are doing almost the same thing as me. I wanted to see if there was an exotic solution? Wrote in my question –  Apr 14 '24 at 18:41