Find the sum to n terms of the series $\frac{1} {1\cdot2\cdot3\cdot4} + \frac{1} {2\cdot3\cdot4\cdot5} + \frac{1} {3\cdot4\cdot5\cdot6}\ldots $ Find the  sum to n terms of the series $\frac{1} {1\cdot2\cdot3\cdot4} + \frac{1} {2\cdot3\cdot4\cdot5} + \frac{1} {3\cdot4\cdot5\cdot6}\ldots $
Please suggest an approach for this task.
 A: This is similar to what has been said by Branimir, but shows how we can extend the result to
$$\sum_{k=1}^n {1 \over k(k+1) \cdots (k+m)}, \qquad m \in \mathbb{N}.$$ 
We can build up the result from the identities
$${1 \over k(k+1)} = {1 \over k} - { 1 \over k+1}, \qquad (1)$$
$${1 \over k(k+1)(k+2)} = {1 \over 2} \left( {1 \over k(k+1)} - { 1 \over (k+1)(k+2)} \right),$$
$${1 \over k(k+1)(k+2)(k+3)} = 
{1 \over 3} \left( {1 \over k(k+1)(k+2)} - { 1 \over (k+1)(k+2)(k+3)} \right), \quad \textrm{ etc...}$$
Write $S_1 = \sum_{k=1}^n {1 \over k(k+1)},$
$S_2 = \sum_{k=1}^n {1 \over k(k+1)(k+2)},$ etc
Summing for $S_1$ using (1) all terms on RHS cancel to get the classic
$$S_1 = \sum_{k=1}^n {1 \over k(k+1)} = 1 – {1 \over n+1} = {n \over n+1}.$$
We then sum the series for $S_2$ using this result obtained for $S_1,$ and so on.
A: HINT $\rm\displaystyle\ \frac{1}{(k+1)(k+2)(k+3)(k+4)} = \frac{1}{6(k+1)} - \frac{1}{2(k+2)}+\frac{1}{2(k+3)}-\frac{1}{6(k+4)}$
$\rm\ f(k+1)-f(k)\: = $ above $\rm\displaystyle\  \Rightarrow\  f(k) \:=\: c_0 + \frac{c_1}{k+1}\ \:+\:\ \frac{c_2}{k+2}\ \:+\:\ \frac{c_3}{k+3}$
Calculating yields $\rm\ c_0,c_1,c_2,c_3 \ =\ 1/18,\ -1/6,\ 1/3,\ -1/6$.
For remarks on the group theory behind rational indefinite summation see my post here
A: $\dfrac{1}{k(k+1)(k+2)(k+3)} = \dfrac{1}{3} (\dfrac{k+3}{k(k+1)(k+2)(k+3)} - \dfrac{k}{k(k+1)(k+2)(k+3)})$
$ = \dfrac{1}{3}(\dfrac{1}{k(k+1)(k+2)} - \dfrac{1}{(k+1)(k+2)(k+3)})$
$\sum_{k=1}^{\infty}\dfrac{1}{k(k+1)(k+2)(k+3)} = \dfrac{1}{3} \dfrac{1}{2*3} = \dfrac{1}{18}$
@moron Yes, you are right. For the first n terms:
$\sum_{k=1}^{n}\dfrac{1}{k(k+1)(k+2)(k+3)} = \dfrac{1}{3} (\dfrac{1}{1*2*3} - \dfrac{1}{(n+1)(n+2)(n+3)})$
[edit] more details
$\sum_{k=1}^{n}\dfrac{1}{k(k+1)(k+2)(k+3)} = \sum_{k=1}^{n} \dfrac{1}{3} (\dfrac{1}{k(k+1)(k+2)} - \dfrac{1}{(k+1)(k+2)(k+3)})$
$= \dfrac{1}{3} [\sum_{k=1}^{n} \dfrac{1}{k(k+1)(k+2)} - \sum_{k=1}^{n} \dfrac{1}{(k+1)(k+2)(k+3)}]$
$= \dfrac{1}{3} [\sum_{k=0}^{n-1} \dfrac{1}{(k+1)(k+2)(k+3)} - \sum_{k=1}^{n} \dfrac{1}{(k+1)(k+2)(k+3)}]$
$= \dfrac{1}{3} (\dfrac{1}{1*2*3} - \dfrac{1}{(n+1)(n+2)(n+3)})$

