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.

$\sum_{n=3}^{2008}\frac{1}{n(n-2)!}=\sum_{n=3}^{2008}\frac{1}{n(n-2)!}.\frac{(n-1)}{(n-1)}=\sum_{n=3}^{2008}\frac{n-1}{n!}=\sum_{n=3}^{2008}(\frac{n}{n!}-\frac{1}{n!})=\sum_{n=3}^{2008}(\frac{1}{(n-1)!}-\frac{1}{n!})=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...-\frac{1}{2007!}+\frac{1}{2007!}-\frac{1}{2008!}=\frac{1}{2!}-\frac{1}{2008!}$
