Calculate the sum of three series which may be telescoping Let $$\sum_{n=1}^\infty \frac{n-2}{n!}$$
 $$\sum_{n=1}^\infty \frac{n+1}{n!}$$
 $$\sum_{n=1}^\infty \frac{\sqrt{n+1} -\sqrt n}{\sqrt{n+n^2}}$$
I have to calculate their sums. So I guess they are telescoping. However, I've no idea about how to make the telescopy emerge. 

Lets begin with the first two since they represent a very simple class. Consider the function
$$ e^x = 1 + x + \frac{1}{2}x^2 + \frac{1}{3!}x^3 + ...  = \sum_{n=0}^{\infty}{\frac{1}{n!}{x^n}} $$
From here it suffices to consider
$$ \frac{e^x}{x^2} = \sum_{n=0}^{\infty}{\frac{1}{n!}{x^{n-2}}} $$
And therefore
$$ \frac{d}{dx}\left[\frac{e^x}{x^2} \right] = \frac{e^x}{x^2}-2\frac{e^x}{x^3}  = \sum_{n=0}^{\infty}{\frac{n-2}{n!}{x^{n-3}}}  $$ 
Let $x=  1$
$$ \frac{e^1}{1^2}-2\frac{e^1}{1^3} = -e =  \sum_{n=0}^{\infty}{\frac{n-2}{n!}{1^{n-3}}} = \sum_{n=0}^{\infty}{\frac{n-2}{n!}}$$
But your sum starts from one so,
$$ -e - \frac{0-2}{0!} = \sum_{n=1}^{\infty}{\frac{n-2}{n!}} $$
Thus:
$$ 2 - e = \sum_{n=1}^{\infty}{\frac{n-2}{n!}}$$

For the second one we pull the same technique instead by multiplying by $x$
$$\frac{d}{dx}[xe^x] = \sum_{n=0}^{\infty}{\frac{n+1}{n!}{x^{n}}}  $$
$$e^x(1+x) = \sum_{n=0}^{\infty}{\frac{n+1}{n!}{x^{n}}} $$
$$2e = \sum_{n=0}^{\infty}{\frac{n+1}{n!}{1^{n}}} $$
$$2e-1 = \sum_{n=1}^{\infty}{\frac{n+1}{n!} }$$
