Using the result that for $x \in ]-\frac{1}{2}, \frac{1}{2}[$: $$\frac{\pi}{\cos(\pi x)} = \sum_{k=0}^{+\infty} \left(\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n+1)^{2k+1}}\right) 2^{2k+2} x^{2k},$$ deduce that the function $$v : \begin{array}{ccc} ]-\frac{\pi}{2}, \frac{\pi}{2}[ & \longrightarrow & \mathbb{R} \\ x & \longmapsto & \frac{1}{\cos(x)} \end{array}$$ is expandable as a power series and that, for all $k \in \mathbb{N}$, $$\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n+1)^{2k+1}} = \frac{\pi^{2k+1}}{2^{2k+2}(2k)!} E_{2k}$$ where, for all $k \in \mathbb{N}$, $E_{2k} = v^{(2k)}(0)$.
Using the result that for $x \in ]-\frac{1}{2}, \frac{1}{2}[$:
$$\frac{\pi}{\cos(\pi x)} = \sum_{k=0}^{+\infty} \left(\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n+1)^{2k+1}}\right) 2^{2k+2} x^{2k},$$
deduce that the function
$$v : \begin{array}{ccc} ]-\frac{\pi}{2}, \frac{\pi}{2}[ & \longrightarrow & \mathbb{R} \\ x & \longmapsto & \frac{1}{\cos(x)} \end{array}$$
is expandable as a power series and that, for all $k \in \mathbb{N}$,
$$\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n+1)^{2k+1}} = \frac{\pi^{2k+1}}{2^{2k+2}(2k)!} E_{2k}$$
where, for all $k \in \mathbb{N}$, $E_{2k} = v^{(2k)}(0)$.