TITLE: How to prove that $\sum\limits_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)...(n+k)} = \frac{1}{kk!}$ for every $k\geqslant1$
QUESTION [6 upvotes]: Does anyone have any idea how to prove that
$$\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)...(n+k)} = \frac{1}{kk!}$$

REPLY [6 votes]: Here is a way to solve this without telescoping.  Define $a_n=\displaystyle\frac{k!}{n(n+1)(n+2)\cdots(n+k)}$.  The partial fraction expansion of $a_n$ is  $\displaystyle \sum_{m=0}^k\frac{(-1)^m{k\choose m}}{n+m}.$  Let $\displaystyle f(x)=\sum_{m=0}^k\frac{(-1)^m{k\choose m}}{n+m}x^{n+m},$ so that $f(0)=0$, $f(1) = a_n$, and $f'(x)=x^{n-1}(1-x)^k$, by the binomial formula.  Thus $a_n=f(1)=\int_0^1f'(x)\,dx=\int_0^1 x^{n-1}(1-x)^k\, dx$.  Then 
$$
\begin{align*}
\sum_{n=1}^\infty \frac{k!}{n(n+1)(n+2)\cdots(n+k)} &= \sum_{n=1}^\infty \int_0^1 x^{n-1}(1-x)^k\,dx.\\
&= \int_0^1 \left(\sum_{n=1}^\infty x^{n-1}(1-x)^{k}\right)\,dx\\
&=\int_0^1(1-x)^{k-1}\,dx\\
&=\frac{1}{k}.
\end{align*} 
$$
Interchanging the summation and integration is valid because of monotone convergence.