$$ \sum_{n=1}^{\infty} {n2^n\over(n+2)!} $$ The exercise mentions that this can be written as a telescopic series; I've been trying to write it in such a way but I'm stuck, can't seem to find one! Any help is much appreciated!
Solution:
So n*2^n can be written as (n+2)*2^n-2^(n+1); from there the partial sum is reduced to 1 - 2^(k+1)/(k+2)! and the sum is 1.
Note that $e^{x} = \sum_{n = 0}^{\infty}\frac{x^{n}}{n!}$. Hence $$ e^{x} - 1 - \frac{x}{1!} - \frac{x^{2}}{2!} = \sum_{n = 3}^{\infty}\frac{x^{n}}{n!} = \sum_{n = 1}^{\infty}\frac{x^{n+2}}{(n+2)!}. $$
$$ \frac{e^{x} - 1 - \frac{x}{1!} - \frac{x^{2}}{2!}}{x^{2}} = \sum_{n = 1}^{\infty}\frac{x^{n}}{(n+2)!}. $$ Now take the derivative with respect to $x$ on both sides, then multiply both sides by $x$ and plug in $2$ for $x$.