Closed form of a power series

Question

Find the function that represents the following sum: $\sum\limits_{k=0} ^\infty \frac{(n^2)}{n!} x^n$. Can't find the function that represents this.

Answer 1

Hint: If $f(x) = \sum \limits_{n = 0}^\infty a_n x^n$ is a convergent power series, then $x \cdot \frac{d}{dx}f(x) = \sum \limits_{n = 0}^\infty a_n n x^n$.

Answer 2

Note: I suppose you mean $n$ instead of $k$ when writing that the sum starts at $k=0$.

First note that $\frac{n^2}{n!}x^n=\frac{n}{(n-1)!}x^n=\frac{(n-1)+1}{(n-1)!}x^n=\frac{1}{(n-2)!}x^n+\frac{1}{(n-1)!}x^n$ So you can split your problem into two seperate (and in fact quite similar) problems (Divide and rule!).

You have to find $$\sum_{n=1}^{\infty} \frac{x^n}{(n-1)!}=x \sum_{n=0}^{\infty} \frac{x^n}{n!}$$ and $$\sum_{n=2}^{\infty} \frac{x^n}{(n-2)!}=x^2 \sum_{n=0}^{\infty} \frac{x^n}{n!}$$ I.e. the total sum amounts to $$(x^2+x)\sum_{n=0}^{\infty} \frac{x^n}{n!}$$ And if you are at least somewhat familiar with power series, the latter one should really be approachable for you...