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March 8th, 2014, 04:25 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...anireconcile, 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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infinite, product, relation, sum 
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