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March 8th, 2014, 04:25 AM   #1
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Relation between an infinite product and an infinite sum.

In this question, I observed that:

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:

(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:
(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 must be equal to a (unique) infinite sum of the terms that are produced by this operation, i.e. : .

Is my reasoning correct and if so, what is this ?
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