We recall that $\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n+1)^{2k+1}} = \frac{\pi^{2k+1}}{2^{2k+2}(2k)!} E_{2k}$ where $E_{2k} = v^{(2k)}(0)$ and $v(x) = \frac{1}{\cos(x)}$. We also recall that $E(g(X)) = \sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n+1)^s}$ for a random variable $X$ following the zeta distribution with parameter $s > 1$, and $g(n) = r_1(n) - r_3(n)$.
Calculate $E(g(X))$ when $X$ is a random variable following the zeta distribution with parameter 3 then with parameter 5.