<s>[INST] Infinite Series $\sum\limits_{n=1}^{\infty}\frac{1}{\prod\limits_{k=1}^{m}(n+k)}$ How to prove the following equality?
$$\sum_{n=1}^{\infty}\frac{1}{\prod\limits_{k=1}^{m}(n+k)}=\frac{1}{(m-1)m!}.$$
 [/INST] Addendum. The partial fraction identity at the top can be shown by induction on $m$. The case $m=1$ is trivial. The induction step gives
$$\frac{1}{(m-1)!} \sum_{k=0}^{m-1} (-1)^k \binom{m-1}{k} \frac{1}{n+k+1} \frac{1}{n+m+1}
\\ = \frac{1}{(m-1)!} \sum_{k=0}^{m-1} (-1)^k \binom{m-1}{k} 
\frac{1}{m-k} \left( \frac{1}{n+k+1} -  \frac{1}{n+m+1} \right).$$
The first part of the sum is
$$\frac{1}{(m-1)!} \sum_{k=0}^{m-1} (-1)^k \binom{m-1}{k}
 \frac{1}{m-k} \frac{1}{n+k+1} = 
\frac{1}{m!} \sum_{k=0}^{m-1} (-1)^k \binom{m}{k} \frac{1}{n+k+1}.$$
The second part is
$$-  \frac{1}{n+m+1} \frac{1}{(m-1)!} \sum_{k=0}^{m-1} (-1)^k \binom{m-1}{k} \frac{1}{m-k}
 = -  \frac{1}{n+m+1} \frac{1}{m!} \sum_{k=0}^{m-1} (-1)^k \binom{m}{k} \\
= -  \frac{1}{n+m+1} \frac{1}{m!} (0 - (-1)^m)
= \frac{1}{n+m+1} (-1)^m \frac{1}{m!}.$$
Putting these two together completes the induction.
This partial fraction identity was also used here at MSE.
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