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Assuming we know that : $$\sum_{n=1}^{+\infty}{\frac{1}{n^2}} = \frac{\pi^2}{6}$$

How do you find the sum of a series where all terms are in this one ?

For instance, how do you prove that ?$$\sum_{n=1}^{+\infty}{\frac{1}{(2n-1)^2}} = \frac{\pi^2}{8}$$

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  • $\begingroup$ The question has no answer, except that the sums one can realize are some (but not all) numbers in $[0,\pi^2/6]$. The specific instance in the second part of the question is answered below. $\endgroup$
    – Did
    Oct 14, 2012 at 9:17
  • $\begingroup$ The first sum is absolutely convergent, so you may reorder the terms. Consider splitting into even and odd terms and the second sum follows almost instantly. $\endgroup$
    – fretty
    Oct 14, 2012 at 9:30
  • $\begingroup$ By the way, $\pi^2/8$ in the RHS of the second identity should be replaced by $(\pi^2/8)-1$. $\endgroup$
    – Did
    Oct 14, 2012 at 9:57
  • $\begingroup$ I changed the LHS. $\endgroup$
    – Cydonia7
    Oct 14, 2012 at 11:54

1 Answer 1

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Observe that:

$$\sum_{n=1}^{\infty} \frac{1}{n^2} = \sum_{n=1}^{\infty}\frac{1}{(2n)^2} + \sum_{n=1}^{\infty}\frac{1}{(2n-1)^2} = \frac{\pi^2}{6}$$

and

$$\sum_{n=1}^{\infty} \frac{1}{(2n)^2} = \frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{24} $$

therefore \begin{eqnarray*} \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} & = & \sum_{n=1}^{\infty} \frac{1}{n^2} - \sum_{n=1}^{\infty} \frac{1}{(2n)^2} \\ & = & \frac{\pi^2}{6} - \frac{\pi^2}{24} \\ & = & \frac{\pi^2}{8} \end{eqnarray*}

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