I have two infinite sums that forms an equality:
$$\sum_{n=1}^{\infty} \left(\zeta(2n)\frac{x^{2n}}{\pi^{2n}}\right) = \sum_{n=1}^\infty \left(\frac{B_{2n}}{(2n)!}\left(-\frac{1}{2}\right)(2ix)^{2n}\right)$$
I know that the entire sum of the left side is equals to the entire sum of the right side. But how do I know that if I expand the left side, the coefficient of index 1 is gonna be equals to the coefficient of index 1 in the right side? In other words, proof that:
$$\zeta(2n)\frac{x^{2n}}{\pi^{2n}} = \frac{B_{2n}}{(2n)!}\left(-\frac{1}{2}\right)(2ix)^{2n}$$
If a smooth function $f$ can be expressed as a power series around $0$: $$ f(x) = \sum_{n=0}^\infty a_n x^n, $$ then $a_n = \frac{1}{n!}f^{(n)}(0)$, where $f^{(n)}$ is the $n$-th derivative of $f$.