A sum of series problem: $\frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \cdots + \frac{2008}{2006!+2007!+2008!}$ I have a question regarding the sum of this series:
$$\frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \cdots + \frac{2008}{2006!+2007!+2008!}$$
My approach:
I found that this sum is equal to:
$$\sum_{n=3}^{2008}\frac{n}{(n-2)!+(n-1)!+(n)!}$$
I reduced it to :
$$\sum_{n=3}^{2008}\frac{1}{n(n-2)!}$$
 Please suggest how to proceed further.
 A: Hint: Apply telescoping series to
$$
\begin{align}
\sum_{k=1}^{2006}\frac{k+2}{k!+(k+1)!+(k+2)!}
&=\sum_{k=1}^{2006}\frac{k+2}{k!\,(k+2)^2}\\
&=\sum_{k=1}^{2006}\frac{k+1}{(k+2)!}\\
&=\sum_{k=1}^{2006}\left(\frac1{(k+1)!}-\frac1{(k+2)!}\right)
\end{align}
$$
A: 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}
Can you finish it off from here?
Move your mouse over the gray area below for the complete answer.

 \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.

A: $$\dfrac1{n(n-2)!}=\dfrac{n-1}{n!}$$
which is Telescoping
A: $\sum_{n=3}^{2008}\frac{1}{n(n-2)!}=\sum_{n=3}^{2008}\frac{1}{n(n-2)!}.\frac{(n-1)}{(n-1)}=\sum_{n=3}^{2008}\frac{n-1}{n!}=\sum_{n=3}^{2008}(\frac{n}{n!}-\frac{1}{n!})=\sum_{n=3}^{2008}(\frac{1}{(n-1)!}-\frac{1}{n!})=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...-\frac{1}{2007!}+\frac{1}{2007!}-\frac{1}{2008!}=\frac{1}{2!}-\frac{1}{2008!}$
