finding $\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!} +\frac{5}{3!+4!+5!}+\cdots+\frac{2008}{2006!+2007!+2008!}$

How to find :

$$\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!} +\frac{5}{3!+4!+5!}+\cdots+\frac{2008}{2006!+2007!+2008!}$$

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What we want is $$\sum_{n=1}^{N} \dfrac{n+2}{n! + (n+1)! +(n+2)!}$$ \begin{align} \dfrac{n+2}{n! + (n+1)! +(n+2)!} & = \dfrac{n+2}{n! \left( 1 + (n+1) + (n+1)(n+2) \right)}\\ & = \dfrac{n+2}{n! \left( n^2 + 4n + 4 \right)}\\ & = \dfrac1{n! \left( n+2 \right)}\\ & = \dfrac{n+1}{(n+2)!}\\ & = \dfrac{n+2}{(n+2)!} - \dfrac1{(n+2)!}\\ & = \dfrac1{(n+1)!} - \dfrac1{(n+2)!} \end{align}
\begin{align}\sum_{n=1}^{N} \dfrac{n+2}{n! + (n+1)! +(n+2)!} & = \sum_{n=1}^{N} \left( \dfrac1{(n+1)!} - \dfrac1{(n+2)!}\right)\\ & = \left( \dfrac1{2!} - \dfrac1{3!} + \dfrac1{3!} - \dfrac1{4!} + \dfrac1{4!} - \dfrac1{5!} + \cdots + \dfrac1{(N+1)!} - \dfrac1{(N+2)!}\right)\\ & = \dfrac1{2!} - \dfrac1{(N+2)!}\end{align} Set $N=2006$ to get the answer to your question.
What should $n$ replaced by ? – spa Sep 26 '12 at 17:16
@spa $N = 2006$ – user17762 Sep 26 '12 at 17:16