$\lim\limits_{n\to\infty}\sum_{k=1}^n\frac{1}{k(k+1)…(k+m+1)}=\frac{1}{(m+1)\cdot(m+1)!}$ Can I proof this identity somehow without using the Riemann integral?
I have tried Stolz–Cesàro Theorem and sums of valuations' progressions.

Note that
$$
\frac{1}{k(k+1)\cdots(k+m+1)}=\frac{1}{(m+1)}
\left(\frac{1}{k(k+1)\cdots(k+m)}-\frac{1}{(k+1)(k+2)\cdots(k+m+1)}\right).
$$
Thus
$$
\sum_{k=1}^n \frac{1}{k(k+1)\cdots(k+m+1)}=\frac{1}{(m+1)}
\left(\frac{1}{(m+1)!}-\frac{1}{(n+1)(n+2)\cdots(n+m+1)}\right),
$$
and hence
$$
\lim_{n\to\infty}\sum_{k=1}^n \frac{1}{k(k+1)\cdots(k+m+1)}=\frac{1}{(m+1)\cdot (m+1)!}.
$$
