For every integer $k \geqslant 2$, we set $\zeta(k) = \sum_{n=1}^{+\infty} n^{-k}$. Let $h$ be the function from $\mathbb{R}$ to $\mathbb{R}$ defined by $$\forall x \in \mathbb{R}, \quad h(x) = \begin{cases} \frac{x}{e^x - 1} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases}$$ We define a sequence of real numbers $(b_n)_{n \in \mathbb{N}}$ by setting $b_0 = 1$, $b_1 = -\frac{1}{2}$, then $$\forall n \in \mathbb{N}^*, \quad b_{2n+1} = 0 \quad \text{and} \quad b_{2n} = \frac{(-1)^{n-1}(2n)! \zeta(2n)}{2^{2n-1}\pi^{2n}}.$$ Show that for all $n \in \mathbb{N}$: $$\sum_{k=0}^{n} \frac{b_k}{k!(n+1-k)!} = \begin{cases} 1 & \text{if } n = 0 \\ 0 & \text{if } n \geqslant 1 \end{cases}.$$
For every integer $k \geqslant 2$, we set $\zeta(k) = \sum_{n=1}^{+\infty} n^{-k}$.
Let $h$ be the function from $\mathbb{R}$ to $\mathbb{R}$ defined by
$$\forall x \in \mathbb{R}, \quad h(x) = \begin{cases} \frac{x}{e^x - 1} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases}$$
We define a sequence of real numbers $(b_n)_{n \in \mathbb{N}}$ by setting $b_0 = 1$, $b_1 = -\frac{1}{2}$, then
$$\forall n \in \mathbb{N}^*, \quad b_{2n+1} = 0 \quad \text{and} \quad b_{2n} = \frac{(-1)^{n-1}(2n)! \zeta(2n)}{2^{2n-1}\pi^{2n}}.$$
Show that for all $n \in \mathbb{N}$:
$$\sum_{k=0}^{n} \frac{b_k}{k!(n+1-k)!} = \begin{cases} 1 & \text{if } n = 0 \\ 0 & \text{if } n \geqslant 1 \end{cases}.$$