$\sum_{n=1}^{\infty} \frac{n^2}{ n!}$ equals

Question

$ \sum_{n=1}^{\infty} \frac{n^2}{ n!} $ equals

I'm not able to convert in any standard series? Any hints?

Answer 1

You may write, for $n =2,3,4,...$: $$ \frac{n^2}{ n!}=\frac{n^2-n+n}{ n!}=\frac{n(n-1)}{ n!}+\frac{n}{ n!}=\frac{1}{ (n-2)!}+\frac{1}{ (n-1)!} $$ and use a change of indices in the new infinite sums.

Answer 2

HINT: First do the obvious cancellation:

$$\sum_{n\ge 1}\frac{n^2}{n!}=\sum_{n\ge 1}\frac{n}{(n-1)!}\;.$$

Now consider performing some familiar operations on the well-known power series expansion of $e^x$.

Added: Multiply

$$e^x=\sum_{n\ge 0}\frac{x^n}{n!}$$

by $x$ and differentiate:

$$\frac{d}{dx}\left(xe^x\right)=\frac{d}{dx}\left(\sum_{n\ge 0}\frac{x^{n+1}}{n!}\right)=\sum_{n\ge 0}\frac{(n+1)x^n}{n!}=\sum_{n\ge 1}\frac{nx^{n-1}}{(n-1)!}\;.$$

Answer 3

You can also use the fact that $n^2=n(n-1)+n$ and consider $$S=\sum_{n=1}^{\infty} \frac{n^2}{ n!}x^n=\sum_{n=0}^{\infty} \frac{n^2}{ n!}x^n=\sum_{n=0}^{\infty} \frac{n(n-1)}{ n!}x^n+\sum_{n=0}^{\infty} \frac{n}{ n!}x^n$$ $$S=x^2\sum_{n=0}^{\infty} \frac{n(n-1)}{ n!}x^{n-2}+x\sum_{n=0}^{\infty} \frac{n}{ n!}x^{n-1}$$ and recognize that the first summation is the second derivative of $e^x$ and that the second summation is the first derivative of $e^x$ and you know them. Then, at the end, replace $x$ by $1$.

Answer 4

In general, $~\displaystyle\sum_{n=1}^\infty\frac{n^k}{n!}~=~B_k\cdot e,~$ where $B_k$ is the $k^{th}$ Bell number, and the identity in question

is called Dobinski's formula. They are all integers for integer values of k, and the sequence they

form can be found on OEIS.