Calculating the sum of $\sum\frac{n^2-2}{n!}$ Calculating the sum of $\sum\frac{n^2-2}{n!}$
I want to calculate the sum of $\sum_{n=0}^{+\infty}\frac{n^2-2}{n!}$. 
This is what I have done so far:
$$ \sum_{n=0}^{+\infty}\frac{n^2-2}{n!}=\sum_{n=0}^{+\infty}\frac{n^2}{n!}-2\sum_{n=0}^{+\infty}\frac{1}{n!}=\sum_{n=0}^{+\infty}\frac{n}{(n-1)!}-2e$$
And here I don't know how to deal with the $\frac{n}{(n-1)!} $. Any tips?
EDIT:
One of the answers recommends to write down the sum as follows:
$$\sum_{n=0}^{+\infty}\frac{n^2-2}{n!}=\sum_{n=0}^{+\infty}\frac{n(n-1)}{n!} + \sum_{n=0}^{+\infty}\frac{n}{n!}-2\sum_{n=0}^{+\infty}\frac{1}{n!}$$
Which equals to:
$$\sum_{n=0}^{+\infty}\frac{n(n-1)}{n!} + \sum_{n=0}^{+\infty}\frac{n}{n!}-2\sum_{n=0}^{+\infty}\frac{1}{n!}=\sum_{n=0}^{+\infty}\frac{(n-1)}{(n-1)!}+\sum_{n=0}^{+\infty}\frac{1}{(n-1)!} -2e$$
But here I have negative factorials. What should I do next? Or can I just say that $\sum_{n=0}^{+\infty}\frac{1}{(n-1)!}=e$?
 A: Hint: Write
$$
\frac{n^2-2}{n!}=\frac{n(n-1)}{n!}+\frac{n}{n!}-\frac2{n!}
$$

Note that by throwing out terms which are zero,
$$
\sum_{n=0}^\infty\frac{n(n-1)}{n!}=\sum_{n=2}^\infty\frac{n(n-1)}{n!}
$$
and
$$
\sum_{n=0}^\infty\frac{n}{n!}=\sum_{n=1}^\infty\frac{n}{n!}
$$
A: 
I thought it might be instructive to present a way forward that can be applied to a wide class of problems.


The Taylor series representation of he exponential function is given by

$$\bbox[5px,border:2px solid #C0A000]{e^x=\sum_{n=0}^\infty \frac{x^n}{n!}} \tag 1$$  

Differentiating $(1)$ term-by-term, we see that 
$$\frac{d\,e^x}{dx}=e^x=\sum_{n=0}^\infty \frac{n x^{n-1}}{n!} \tag2$$
whereby multiplying $(2)$ by $x$ reveals
$$xe^x =\sum_{n=0}^\infty \frac{nx^n}{n!} \tag 3$$

Differentiating $(3)$, multiplying by $x$, and subtracting $2e^x$, we obtain
$$(x^2+x-2)e^x=\sum_{n=0}^\infty \frac{(n^2-2)\,x^{n}}{n!} \tag 4$$
Finally, setting $x=1$ in $(4)$ yields

$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=0}\frac{n^2-2}{n!}=0}$$

and we are done!
A: As per this post, we have
$$\sum_{n=0}^\infty\frac{n^p}{n!}=B_pe$$
where $B_n$ is the $n$th Bell number.
A: Adjust indices so that you have $\frac {n+1}{n!}$ and separate that out into $\frac n{n!}+\frac1{n!}$.
