# Find the sum of the following series: $\sum_{n=1}^{\infty}\frac{n \cdot 2^n}{(n+2)!}=?$

How can I solve this exercise:

Find the sum of : $$\sum_{n=1}^{\infty}\frac{n \cdot 2^n}{(n+2)!}=?$$

I think I should somehow bring the expression to the form of a telescopic one, and make the simplifications, but do not know what to do with that $2^n$. Can you please give me a hint. Thanks.

$$S=\frac{n2^n}{(n+2)!}=\frac{(n+2-2)2^n}{(n+2)!}=\frac{2^n}{(n+1)!}-\frac{2^{n+1}}{(n+2)!}=T(n)-T(n+1)$$

where $T(r)=\dfrac{2^r}{(r+1)!}$

Can you identify the Telescoping Series?

• Yes, thank you very much. – Ivan Gandacov Jan 19 '15 at 9:02