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Proving $$1+\frac{4}{2^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{4}{6^2}+\frac{1}{7^2}+\frac{1}{9^2}+\frac{4}{10^2}+\frac{1}{11^2}+\cdots=\frac{\pi ^2}{4}$$ Firstly, I thought to prove it by comparison the terms with the terms of $1/n^2$ , but the problem with the missing terms, so I couldn't reach to the proving . Can anybody help? Best regards.

  • What are your attempts? – Jack D'Aurizio Feb 07 '15 at 15:45
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    Hint: $\sum_{i=1}^\infty 1/i^2 = \pi^2/6$ – Simon S Feb 07 '15 at 15:45
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    What's the pattern? Would you care to show more terms so that it is exhibited explicitly? – Vim Feb 07 '15 at 15:45
  • I Checked it numerically only –  Feb 07 '15 at 15:49
  • The pattern appears to be this: Start with $\sum_{n=1}^\infty 1/n^2$, then delete all the terms in which $n$ is a multiple of $4$, then multiply by $4$ all the terms in which $n$ is congruent to $2$ mod $4$. To do it from scratch would be a substantial problem, but if you know the result that Simon S posted in his comment, that should reduce the whole thing to simple algebra. ${}\qquad{}$ – Michael Hardy Feb 07 '15 at 15:56
  • Why this question is closed. I tried to prove it but I couldn't do. –  Feb 07 '15 at 16:07
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    @Easy - the question was closed because you didn't make any efforts you made clear in the question. If you edit the question and add a few words, people may be inclined to reopen it. – Nathaniel Bubis Feb 07 '15 at 16:25
  • @nbubisf thanks for your advising –  Feb 07 '15 at 16:37
  • @Easy: there is only one vote left for reopen, so if you haven't voted yet you can reopen it yourselves by doing so. – Lehs Feb 07 '15 at 21:58

1 Answers1

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This sum is explicitly: $$\sum_{n=1}^\infty\frac{1}{n^2}+3\frac{1}{(2n)^2}-4\frac{1}{(4n)^2}$$ You can check for yourself that this eliminates $n$'s which are a multiple of four, while multiplying by four terms with even $n$ but which are not a multiple of four. In other terms: $$\sum_{n=1}^\infty\frac{1}{n^2}\left(1+\frac{3}{4} - \frac{1}{4}\right)=\frac{\pi^2}{6}\cdot\frac{3}{2}=\frac{\pi^2}{4}$$

Nathaniel Bubis
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  • Brilliant! I am just stuck here about how to derive it from the famous $$\sum_{n=1}^\infty\frac{1}{n^2}$$ and you went ahead of me before I even got a thread – Vim Feb 07 '15 at 16:01
  • Hey! Your post just made me want to proceed further: Is it possible to derive simply from $\sum_{i=1}{\infty}\frac{1}{i^2}$ the explicit value of such general cases as $\sum_{i=1}{\infty}\frac{1}{(ki+b)^2}$ where $(ki+b)$ is a positive integer arithmetic sequence? – Vim Feb 07 '15 at 16:10
  • @Vim - the general case involves the Polygamma function, so I don't think so. – Nathaniel Bubis Feb 07 '15 at 16:17
  • Oh that's something I have never dealt with. but thanks anyway. – Vim Feb 07 '15 at 16:22