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. 
 A: 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 $x$
$$\frac{d}{dx}[xe^x] = \sum_{n=0}^{\infty}{\frac{n+1}{n!}{x^{n}}}  $$
$$e^x(1+x) = \sum_{n=0}^{\infty}{\frac{n+1}{n!}{x^{n}}} $$
$$2e = \sum_{n=0}^{\infty}{\frac{n+1}{n!}{1^{n}}} $$
$$2e-1 = \sum_{n=1}^{\infty}{\frac{n+1}{n!} }$$
A: $$\frac{\sqrt{n+1}-\sqrt n}{\sqrt{n+n^2}}=\frac1{\sqrt n}-\frac1{\sqrt{n+1}}$$
A: There's actually a general approach for formulas of the form:
$$\sum_{n=0}^\infty \frac{f(n)}{n!}$$ where $f$ is a polynomial.It is determined by writing $f(x)$ in terms of the basis of $p_n(x)=x(x-1)\cdots(x-(n-1))$ (the so-called falling-factorial functions.)
$$f(x)=\sum_{k=0}^m a_k p_k(x)$$
Then:
$$\sum_{n=0}^\infty \frac{f(n)}{n!} = \left(\sum_{i=0}^m a_i\right)e$$
So $n-2= p_1(n)-2p_0(n)$ so:
$$\sum_{n=0}^\infty \frac{n-2}{n!} = (1-2)e$$
and thus:
$$\sum_{n=1}^\infty \frac{n-2}{n!} = -e + 2$$
A: The second one seems to be easy if we can use $$\sum_{n=0}^{\infty }\frac{1}{%
n!}=e$$ as well known 
\begin{eqnarray*}
\sum_{n=1}^{\infty }\frac{n+1}{n!} &=&\sum_{n=1}^{\infty }(\frac{n}{n!}+%
\frac{1}{n!}) \\
&=&\sum_{n=1}^{\infty }(\frac{1}{(n-1)!}+\frac{1}{n!}) \\
&=&\left( \sum_{m=0}^{\infty }\frac{1}{m!}\right) +\left( \sum_{n=0}^{\infty
}\frac{1}{n!}\right) -1 \\
&=&e+e-1 \\
&=&2e-1.
\end{eqnarray*}
A: The first one can also be computed as the second one so easly
\begin{eqnarray*}
\sum_{n=1}^{\infty }\frac{n-2}{n!} &=&\sum_{n=1}^{\infty }(\frac{n}{n!}+%
\frac{-2}{n!}) \\
&=&\sum_{n=1}^{\infty }(\frac{1}{(n-1)!}-2\frac{1}{n!}) \\
&=&\left( \sum_{m=0}^{\infty }\frac{1}{m!}\right) +(-2)\left(
\sum_{n=0}^{\infty }\frac{1}{n!}-1\right)  \\
&=&e+(-2)(e-1) \\
&=&2-e.
\end{eqnarray*}
A: We can compute the first series by splitting it into two parts.  The first part can be rearranged into a telescoping series while the second one is easily recognized.  To that end we write
$$\sum_1^{\infty} \frac{n-2}{n!}=\sum_1^{\infty} \frac{n-1}{n!}-\sum_1^{\infty} \frac{1}{n!}$$
The first series on the right can be written as the telescoping series 
$$\sum_1^{\infty} \frac{n-1}{n!}=\sum_1^{\infty} \left(\frac{1}{(n-1)!}-\frac{1}{n!}\right)=1$$
The second series is simply the Taylor Series for $e^1-1=e-1$.
Putting it all together we have 
$$\sum_1^{\infty} \frac{n-2}{n!}=2-e$$
