TITLE: What's the value of $\sum\limits_{k=1}^{\infty}\frac{k^2}{k!}$?
QUESTION [40 upvotes]: For some series, it is easy to say whether it is convergent or not by the "convergence test", e.g., ratio test. However, it is nontrivial to calculate the value of the sum when the series converges. The question is motivated from the simple exercise to determining whether the series $\sum\limits_{k=1}^{\infty}\frac{k^2}{k!}$ is convergent. One may immediately get that it is convergent by the ratio test. So here is my question:

What's the value of $$\sum_{k=1}^{\infty}\frac{k^2}{k!}?$$

REPLY [3 votes]: Just to give a slightly different approach,
$$\sum_{n=1}^\infty{n^2\over n!}=\sum_{n=1}^\infty{n\over(n-1)!}=\sum_{m=0}^\infty{m+1\over m!}=\sum_{m=0}^\infty{m\over m!}+e=\sum_{m=1}^\infty{m\over m!}+e=\sum_{m=1}^\infty{1\over(m-1)!}+e=\sum_{k=0}^\infty{1\over k!}+e=e+e$$
The trick in reading this is to note what changes across each equal sign as you proceed from left to right and understand what justifies the equality for each change.