Find $\sum_{k=1}^\infty\frac{1} {k(k+1)(k+2)(k+3)}$

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 $\frac{1}{3\cdot 3!}=\frac{1}{18}$.

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.

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,

1. for $x=0$: $6A=1$
2. for $x=-1$: $-2B=1$
3. for $x=-2$: $2C=1$
4. 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 $$ \frac{1}{6}\cdot\frac{1}{3}=\frac{1}{18} $$