My Math Forum Relation between an infinite product and an infinite sum.

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 March 8th, 2014, 05:25 AM #1 Senior Member   Joined: Jan 2011 Posts: 120 Thanks: 2 Relation between an infinite product and an infinite sum. In this question http://math.stackexchange.com/questi...an-i-reconcile, I observed that: $\displaystyle \prod_{n=2}^\infty \left(\frac{1}{1-\frac{1}{n^2}}\right)^{(-1)^n}=\sum_{n=1}^\infty \left(\frac{1}{(2\,n -1)^2}\right)=\dfrac{\pi^2}{8}$ Got curious how the product and the sum relate to each other and whether some form of factorization similar to the Euler product of primes exists. So I tried: \begin{align*} K(s) &= \frac{1}{1^2}+ \frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\frac{1} {9^2}+\dots \\ (\frac{1}{2^2})K(s) &= \frac{1}{2^2}+ \frac{1}{6^2}+\frac{1}{10^2}+\frac{1}{14^2}+\frac{ 1}{18^2}+\dots \\ (1-\frac{1}{2^2})K(s) &= \frac{1}{1^2}-\frac{1}{2^2}+ \frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{9^2}- \frac{1}{10^2}-\frac{1}{14^2}-\frac{1}{18^2}-\dots \\ (-\frac1(1-\frac{1}{2^2})K(s) &= -\frac{1}{8^2}+\frac{1}{16^2}- \frac{1}{24^2}-\frac{1}{40^2}+\frac{1}{48^2}-\frac{1}{56^2}-\frac{1}{72^2}+ \frac{1}{80^2}+\frac{1}{112^2}+\frac{1}{136^2}+\do ts \\ (1+\frac1(1-\frac{1}{2^2})K(s) &=\frac{1}{1^2}-\frac{1}{2^2}+ \frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}+\frac{1} {9^2}- \frac{1}{10^2}-\frac{1}{14^2}-\frac{1}{16^2}-\frac{1}{18^2} + \frac{1}{24^2}+\frac{1}{40^2}+\dots \\ (\frac{1}{4^2})(1+\frac1(1-\frac{1}{2^2})K(s) &=\frac{1}{4^2}-\frac{1}{8^2}+ \frac{1}{12^2}+\frac{1}{20^2}-\frac{1}{24^2}+\frac{1}{28^2}+\frac{1}{32^2}+\frac {1}{36^2}- \frac{1}{40^2}-\dots \end{align*}" /> This continues as: \begin{align*} \\ (1-\frac{1}{4^2})(1+\frac1(1-\frac{1}{2^2})K(s) &= \dots \\ (-\frac{1}{24})(1-\frac{1}{4^2})(1+\frac1(1-\frac{1}{2^2})K(s) &= \dots \\ (1+\frac{1}{24})(1-\frac{1}{4^2})(1+\frac1(1-\frac{1}{2^2})K(s) &= \dots \\ etc. \end{align*}" /> Note that there doesn't seem to be a single term that gets annihilated by this operation, however the end result of continued multiplying and subtracting must be 1. I therefore suspect that each term $\left(\frac{1}{(2k-1)^2}\right)$ must be equal to a (unique) infinite sum of the terms that are produced by this operation, i.e. : $\left(\frac{1}{(2k-1)^2}\right)= \sum_{n=1}^{\infty} f(n,k)$. Is my reasoning correct and if so, what is this $f(n,k)$ ?

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