Finding $\zeta(4)$ by Taylor series Is it possible to solve Zeta(4) function using something similar to the solution for zeta(2) as seen in this video? https://www.youtube.com/watch?v=mTPKyC3Udns
 A: Yes. This is outside of the realm of things I currently think about, but I will do my best to give a reasonable explanation.
We had that $$\frac{\sin{x}}{x} = (1 - \frac{x}{\pi})(1 + \frac{x}{\pi})\cdots ,$$ and for any integer, $n$, we also had
$$(1 - \frac{x}{\pi n})(1 + \frac{x}{\pi n}) = (1 - \frac{x^2}{\pi^2n^2}).$$ 
The question is, how to mimic this. We originally used that $(1-a)(1+a) = (1-a^2)$.  To extend this for $\zeta(4)$, consider $(1-a)(1+a)(1-ia)(1+ia) = 1 - a^4$ (we eventually want $a = x/\pi n$, as before). If you imagine the zeros of $\sin{x}$ in the complex plane, they all lie on the real line and are symmetric about the origin (a bunch of pairs of points). Now we would like the zeros to be on both the real and imaginary line (a bunch of sets of four points, forming a diamond with center at the origin). 
So we already know the  $\frac{\sin{x}}{x}$ trick. To get the imaginary zeros, consider $\frac{\sin{ix}}{ix} = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \ldots$. Then $$\frac{\sin{x} \sin{i x}}{i x^2} = (1 - \frac{x^4}{\pi^4})(1 - \frac{x^4}{2^4 \pi^4}) \ldots.$$ This time, we need to calculate the $x^4$ hand term on the left hand side, which may be done by easily by multiplying 3 terms from the power series of $\frac{sin{x}}{x}$ with 3 terms of the powers series $\frac{\sin{ix}}{(ix)} $. Then we get that $2 \frac{1}{5!} - \frac{1}{(3!)^2} = - \zeta(4)/\pi^4$.
