Zeta

The sum is well known and equals (pi^2)/6. Ed Gray

First of all, write the Taylor Expansion for $\sin(s)$

$\sin(s) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1} s^{2k+1}}{(2k+1)!}$

Then we can verify that

$\frac{\sin(\pi s)}{\pi s} = \sum_{k=1}^{\infty} \frac{(-1)^{k+1} (\pi s)^{2k}}{(2k+1)!} = \prod_{k=1}^{\infty} \left ( 1 - \frac{s^2}{k^2} \right )$

And expanding the product...

$\prod_{k=1}^{\infty} \left ( 1 - \frac{s^2}{k^2} \right ) = 1 - \zeta(2) \; s^2 + \left ( \frac{1}{1^2 \times 2^2} + \frac{1}{1^2 \times 3^2} + \cdots \right )s^4 + \cdots$

But the Taylor Expansion is unique, which means that all the coefficients are the same (in both expansions). Therefore

$\zeta(2) = \frac{\pi^{2}}{3!} = \frac{\pi^{2}}{6} \approx 1.644$

Where the left-hand side came from the infinite product and the right-hand side came from the infinite sum.