For every integer $k \geqslant 2$, we set $\zeta(k) = \sum_{n=1}^{+\infty} n^{-k}$. We define a sequence of real numbers $(b_n)_{n \in \mathbb{N}}$ by setting $b_0 = 1$, $b_1 = -\frac{1}{2}$, then
$$\forall n \in \mathbb{N}^*, \quad b_{2n+1} = 0 \quad \text{and} \quad b_{2n} = \frac{(-1)^{n-1}(2n)!\zeta(2n)}{2^{2n-1}\pi^{2n}}$$
Using the relation $\sum_{k=0}^{n} \frac{b_k}{k!(n+1-k)!} = 0$ for $n \geqslant 1$, calculate $b_2$, $b_4$ and $b_6$, then $\zeta(2)$, $\zeta(4)$ and $\zeta(6)$.