TITLE: Calculate the sum of three series which may be telescoping
QUESTION [2 upvotes]: 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.

REPLY [1 votes]: 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:
$$\sum_{n=1}^\infty \frac{n-2}{n!} = -e + 2$$
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