Find $\sum_{k=1}^\infty\frac{1} {k(k+1)(k+2)(k+3)}$ I have to solve this series transforming it into a telescopic sequence
$$\sum_{k=1}^\infty\frac{1}
{k(k+1)(k+2)(k+3)}$$
But I'm lost in the calculation!

There is a general solution for this type of problem: $$\sum_{k=1}^{\infty} \frac{1}{k(k+1)\cdot\dotso\cdot (k+N)}.$$
Observe that
$$\sum_{k\geq 1} \frac{1}{k(k+1)\cdot\dotso\cdot (k+N)}=\sum_{k\geq 1} \frac{(k-1)!}{(k+N)!}=\frac{1}{N!}\sum_{k\geq 1}\frac{\Gamma(k)\Gamma(N+1)}{\Gamma(k+N+1)}=\frac{1}{N!}\sum_{k\geq 1} B(k,N+1).$$
Consequently,
\begin{align}
\sum_{k=1}^{\infty} \frac{1}{k(k+1)\cdot\dotso\cdot (k+N)} & = \frac{1}{N!}\sum_{k\geq 1}\int_0^1 x^{k-1}(1-x)^{N}\,dx \\
& =\frac{1}{N!}\int_0^1\left(\sum_{k\geq 1}x^{k-1}\right)(1-x)^N\, dx \\
&=\frac{1}{N!}\int_0^1 (1-x)^{N-1}\, dx \\
& =\frac{1}{N\cdot N!}.
\end{align}
Here we have $N=3$. Hence, the answer is $\frac{1}{3\cdot 3!}=\frac{1}{18}$.
