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.

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.
