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!
 A: 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 $\frac{1}{3\cdot 3!}=\frac{1}{18}$.
A: Hint:
Look at $$\frac{1}{k(k+1)(k+2)} - \frac{1}{(k+1)(k+2)(k+3)}$$
It is not exactly the general term of your sum, but you only need to divide by a constant such that it is.
A: You can also go the hard way: write
$$
\frac{1}{x(x+1)(x+2)(x+3)}=
\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x+2}+\frac{D}{x+3}
$$
so
$$
A(x+1)(x+2)(x+3)+Bx(x+2)(x+3)+Cx(x+1)(x+3)+Dx(x+1)(x+2)=1
$$
Now,


*

*for $x=0$: $6A=1$

*for $x=-1$: $-2B=1$

*for $x=-2$: $2C=1$

*for $x=-3$: $-6D=1$


Thus
$$
\frac{1}{k(k+1)(k+2)(k+3)}=
\frac{1}{6}\left(\frac{1}{k}-\frac{3}{k+1}+\frac{3}{k+2}-\frac{1}{k+3}\right)
$$
Hence your summation is $1/6$ of
\begin{align}
\sum_{k=1}^\infty \frac{1}{k}
-3\sum_{k=1}^\infty \frac{1}{k+1}
+3\sum_{k=1}^{\infty} \frac{1}{k+2}
-\sum_{k=1}^{\infty} \frac{1}{k+3}
&
=\sum_{k=1}^\infty \frac{1}{k}
-3\sum_{k=2}^\infty \frac{1}{k}
+3\sum_{k=3}^{\infty} \frac{1}{k}
-\sum_{k=4}^{\infty} \frac{1}{k}
\\[6px]
&=
\left(1+\frac{1}{2}+\frac{1}{3}\right)-3\left(\frac{1}{2}+\frac{1}{3}\right)
+3\frac{1}{3}
\\[6px]
&=\frac{1}{3}
\end{align}
Therefore the sum of your series is
$$
\frac{1}{6}\cdot\frac{1}{3}=\frac{1}{18}
$$
