Finding $f^{(n)}(0)$ of $1+\sum\limits_{k=1}^{\infty}\frac{x^{2k}}{k^2}$.

Question

Maclaurin series for function $f$ is $1+x^2+\frac{x^4}{2^2}+\frac{x^6}{3^2}+...=1+\sum\limits_{k=1}^{\infty}\frac{x^{2k}}{k^2}$.

How to find $f^{(n)}(0)$ for all $n\in\mathbb{N}$?

Answer 1

Hint:

$$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n$$

Particularly, all the odd coefficients are $0$ and all the even coefficients are $1/k^2$ (except for $f(0)$), hence,

$$f^{(2n)}(0)=\frac{(2n)!}{n^2}$$