# How do I prove that: $\sum_{i=0}^{\infty} \frac{i^2}{i!}=2e$ [duplicate]

I've seen in Wolfram Alpha that $$\sum_{i=0}^{\infty} \frac{i^2}{i!}=2e$$ but I have no idea how to prove that.

Can anyone help me?

Thanks.

## marked as duplicate by Workaholic, Community♦Nov 30 '16 at 9:31

Consider $$S=\sum_{i=0}^{\infty} \frac{i^2}{i!}x^i=\sum_{i=0}^{\infty} \frac{i(i-1)+i}{i!}x^i=\sum_{i=0}^{\infty} \frac{i(i-1)}{i!}x^i+\sum_{i=0}^{\infty} \frac{i}{i!}x^i$$ $$S=x^2\sum_{i=0}^{\infty} \frac{i(i-1)}{i!}x^{i-2}+x\sum_{i=0}^{\infty} \frac{i}{i!}x^{i-1}$$ $$S=x^2\left(\sum_{i=0}^{\infty} \frac{x^i}{i!}\right)''+x\left(\sum_{i=0}^{\infty} \frac{x^i}{i!}\right)'=x^2e^x+xe^x$$ Now, use $x=1$.

Edit

Suppose that, instead of $i^2$, you faced $i^3$. It would be the same approach using $$i^3=i(i-1)(i-2)+3i(i-1)+i$$ Similarly $$i^4=i(i-1)(i-2)(i-3)+6i(i-1)(i-2)+7i(i-1)+i$$ Sooner or later, you will lear that $$S_k=\sum_{i=0}^{\infty} \frac{i^k}{i!}=B_k\, e$$ where appear Bell numbers. These grow extremely fast $(B_{20}=51724158235372)$.

\eqalign{\sum_{i=0}^\infty \dfrac{i^2 z^i}{i!} &= \sum_{i=0}^\infty z \dfrac{d}{dz} z \dfrac{d}{dz} \dfrac{z^i}{i!}\cr &= z \dfrac{d}{dz} z \dfrac{d}{dz} e^z\cr &= z (z+1) e^z}

Now substitute $z=1$.

The short way:

$$\sum_{i=0}^{\infty} \frac{i^2}{i!}=\sum_{i=1}^{\infty} \frac{i}{(i-1)!}=\sum_{i=0}^{\infty} \frac{i+1}{i!}=\sum_{i=1}^{\infty} \frac1{(i-1)!}+\sum_{i=0}^{\infty} \frac1{i!}=2\sum_{i=0}^{\infty} \frac1{i!}.$$