TITLE: A sum of series problem: $\frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \cdots + \frac{2008}{2006!+2007!+2008!}$
QUESTION [5 upvotes]: 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.

REPLY [4 votes]: Hint: Apply telescoping series to
$$
\begin{align}
\sum_{k=1}^{2006}\frac{k+2}{k!+(k+1)!+(k+2)!}
&=\sum_{k=1}^{2006}\frac{k+2}{k!\,(k+2)^2}\\
&=\sum_{k=1}^{2006}\frac{k+1}{(k+2)!}\\
&=\sum_{k=1}^{2006}\left(\frac1{(k+1)!}-\frac1{(k+2)!}\right)
\end{align}
$$