Sum of $\sum\limits_{k \geq 1} \frac{k^2}{k!}$

Question

I know the answer to the series is $2e$, but I am not sure why. I know that $\sum\limits_{k \geq 1} \frac{1}{k!} = e$, but can't figure out how to expand on that, especially given that this is a mGRE question and I have two minutes.

Answer 1

$$\sum_{k=1} \frac{k^2}{k!}=\sum_{k=1} \frac{k}{(k-1)!}=\sum_{k=0}\frac{k+1}{k!}$$ $$=\sum_{k=0} \frac{k}{k!}+\sum_{k=0}\frac{1}{k!}$$ $$=(0+1+1+\frac{1}{2!}+\frac{1}{3!}+...)+e$$ $$=2e$$

I think you mean that the answer is $2e$

Answer 2

You have $$\sum_{k\geq1}\frac{ k^2}{k!}=\sum_{k\geq1}\frac{ k}{(k-1 )!}= \sum_{k\geq0}\frac{ k+1}{k!}=\sum_{k\geq1}\frac{ 1}{(k-1)!}+\sum_{k\geq0}\frac1{k!}=e+e=2e.$$

Answer 3

Recall that $$e^{x}=\sum_{k=0}^{+\infty}\frac{x^k}{k!}$$ Differentiating once, (we can interchange summation and differentiation because of uniform convergence), $$e^{x}=\sum_{k=0}^{+\infty}\frac{kx^{k-1}}{k!}$$ $$xe^{x}=\sum_{k=0}^{+\infty}\frac{kx^{k}}{k!}$$ DIfferentiating with respect to $x$ once again, $$e^{x}(x+1)=\sum_{k=0}^{+\infty}\frac{k^2 x^{k-1}}{k!}$$ $$e^{x}(x+1)=\sum_{k=1}^{+\infty}\frac{k^2 x^{k-1}}{k!}$$ Letting $x\to 1$, $$\sum_{k=1}^{+\infty}\frac{k^2}{k!}=2e$$