Maybe a trivial question, but how could I reconcile the following equation:
$$\displaystyle \prod_{n=2}^\infty \left(\frac{1}{1-\frac{1}{n^2}}\right)^{(-1)^n}=\sum_{n=1}^\infty \left(\frac{1}{(2\,n -1)^2}\right)$$
Both sides are $\frac{\pi^2}{8}$ and I can derive why the infinite sum of the odd numbers squared should be equal to that, however don't understand why it equates the infinite (alternating) product.
ADDITIONAL COMMENT
Still curious if and 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:
$\tiny{ \begin{align*} K(s) &= \frac{1}{1^2}+ \frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\frac{1}{9^2}+\dots \\ \left(\frac{1}{2^2}\right)K(s) &= \frac{1}{2^2}+ \frac{1}{6^2}+\frac{1}{10^2}+\frac{1}{14^2}+\frac{1}{18^2}+\dots \\ &\text{____________________________________________ (-)} \\ \left(1-\frac{1}{2^2}\right)K(s) &= \frac{1}{1^2}-\frac{1}{2^2}+ \frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{9^2}- \frac{1}{10^2}-\frac{1}{14^2}-\frac{1}{18^2}-\dots \\ \left(-\frac18\right)\left(1-\frac{1}{2^2}\right)K(s) &= -\frac{1}{8^2}+\frac{1}{16^2}- \frac{1}{24^2}-\frac{1}{40^2}+\frac{1}{48^2}-\frac{1}{56^2}-\frac{1}{72^2}+ \frac{1}{80^2}+\frac{1}{112^2}+\frac{1}{136^2}+\dots \\ &\text{____________________________________________________________________________________ (-)} \\ \left(1+\frac18\right) \left(1-\frac{1}{2^2}\right)K(s) &=\frac{1}{1^2}-\frac{1}{2^2}+ \frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}+\frac{1}{9^2}- \frac{1}{10^2}-\frac{1}{14^2}-\frac{1}{16^2}-\frac{1}{18^2} + \frac{1}{24^2}+\frac{1}{40^2}+\dots \\ \left(\frac{1}{4^2}\right)\left(1+\frac18\right)\left(1-\frac{1}{2^2}\right)K(s) &=\frac{1}{4^2}-\frac{1}{8^2}+ \frac{1}{12^2}+\frac{1}{20^2}-\frac{1}{24^2}+\frac{1}{28^2}+\frac{1}{32^2}+\frac{1}{36^2}- \frac{1}{40^2}-\dots \end{align*}}$
This continues as:
$\tiny{ \begin{align*} \\ \left(1-\frac{1}{4^2}\right)\left(1+\frac18\right)\left(1-\frac{1}{2^2}\right)K(s) &= \dots \\ \left(-\frac{1}{24}\right)\left(1-\frac{1}{4^2}\right)\left(1+\frac18\right)\left(1-\frac{1}{2^2}\right)K(s) &= \dots \\ \left(1+\frac{1}{24}\right)\left(1-\frac{1}{4^2}\right)\left(1+\frac18\right)\left(1-\frac{1}{2^2}\right)K(s) &= \dots \\ etc. \\ \prod_{n=2}^\infty \left(\frac{1}{1-\frac{1}{n^2}}\right)^{(-1)^{n+1}} \, K(s) &= 1 \end{align*}}$
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. Could it be that each initial term $\frac{1}{(2n-1)^2}$ is always equal to a (unique) infinite sum of the terms that are produced by this operation, i.e. that there is a function $\displaystyle \frac{1}{(2\,n-1)^2} = \sum_{k=1}^{\infty} f(k,n)$ ?
EDIT AND AFTER THOUGHT:
Another, similar relation is:
$$\displaystyle \prod_{n=2}^\infty \left(\frac{1}{1-\frac{1}{n}}\right)^{(-1)^n}=\sum_{n=1}^\infty \left(\frac{(-1)^{n-1}}{(n -\frac12)}\right)=\sum_{n=1}^\infty \left(\frac{(n-1)!}{(2\,n -1)!!}\right)$$
The value now is $\frac{\pi}{2}$ and also equal to the Wallis product. Still don't see how the alternating infinite product leads to any of the sums. The infinite products look very similar to the Euler product but 'alternate', so wonder if these products contain the unique 'factors' of the elements of their associated infinite series.
Thanks!