mathimatikoi.org

Prove that:\( \displaystyle \prod_{k=1}^{m}\left ( n+k \right )=\frac{\left ( n+m \right )!}{n!} \)

Use the above result to evaluate the series: $$\sum_{n=1}^{\infty}\frac{1}{ \prod \limits_{k=1}^{m}\left ( n+k \right )}$$

Ok, we have already seen a way of computing this sum because It telescopes. Let's see another one.
Note: It should be pointed out that \(m\geq 2\) otherwise the sum diverges.

Successively we have:
$$\sum_{n=1}^{\infty}\frac{1}{\prod \limits_{k=1}^{m}\left ( n+k \right )}=\sum_{n=1}^{\infty}\frac{n!}{\left ( n+m \right )!}=\frac{1}{m!}\sum_{n=1}^{\infty}\frac{1}{\binom{m+n}{n}}$$ whereas in the second step we used the sentence above which can be easily be proved by induction.

This results in: $$\begin{aligned}
\frac{1}{m!}\sum_{n=1}^{\infty}\frac{1}{\binom{m+n}{n}} &=\frac{m}{m!}\sum_{n=1}^{\infty}\int_{0}^{1}x^{m-1}\left ( 1-x \right )^n\,dx \\
&= \frac{m}{m!}\int_{0}^{1}x^{m-1}\sum_{n=1}^{\infty}\left ( 1-x \right )^n\,dx\\
&=\frac{m}{m!}\int_{0}^{1}x^{m-1}\left ( \frac{1-x}{x} \right )\,dx \\
&= \frac{m}{m!}\int_{0}^{1}\left ( x^{m-2}-x^{m-1} \right )\,dx\\
&= \frac{1}{\left ( m-1 \right )m!}
\end{aligned}$$ which completes the proof!