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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.

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

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

Can you identify the Telescoping Series?

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

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