Series: prove that odd terms are null Please find below a seies defined as:
$$U_0=1$$
$$
U_n=-\sum_{0 \le i \le n - 1} \frac{U_i}{(n+1-i)!}
$$
How to prove that, for all $n>1$, $U_{2n+1}=0$?
 A: The recursion yields that the generating function $U(s)=\sum\limits_{n\geqslant0}U_ns^n$ solves
$$
U(s)=1-\sum_{i\geqslant0}\sum_{n\geqslant i+1}U_is^i\frac{s^{n-i}}{(n+1-i)!}=1-U(s)V(s),
$$
where
$$
V(s)=\sum_{k\geqslant1}\frac{s^k}{(k+1)!}=\frac{\mathrm e^s-1-s}s,
$$
hence
$$
U(s)=\frac{s}{\mathrm e^s-1}.
$$
Now, the fact that $U_{2n+1}=0$ for every $n\geqslant1$ means that $$U(s)=1-\tfrac12s+W(s),$$ where $W$ is even, hence the result follows from the fact, easy to prove, that
$$
U(-s)=s+U(s).
$$
A: Define the generating function $u(z) = \sum_{n \ge 0} U_n z^n$, write the recurrence as:
$$
U_{n + 1} = \sum_{0 \le k \le n} \frac{1}{(n + 2 - k)!} \cdot U_k
$$
Multiply by $z^n$, sum over $n \ge 0$ and recognize some sums:
\begin{align}
\frac{u(z) - U_0}{z}
  &= - \sum_{n \ge 0} z^n \sum_{0 \le k \le n} \frac{1}{(n + 2 - k)!} \cdot U_k \\
\frac{u(z) - 1}{z}
  &= - \left( \sum_{n \ge 0} \frac{z^n}{(n + 2)!} \right)
         \cdot \left( \sum_{n \ge 0} U_n z^n \right) \\
  &= - \frac{\mathrm{e}^z - 1 - z}{z^2} \cdot u(z)
\end{align}
This gives:
$$
u(z) = \frac{z}{\mathrm{e}^z - 1}
$$
If the odd terms are zero, this function is even, i.e., $u(z) = u(-z)$. You get:
$$
u(z) - u(-z) = -z
$$
So for $k > 1$, $U_{2 k + 1} = 0$.
