Show $\sum_{k=1}^\infty \frac{k^2}{k!} = 2\mathrm{e}$ [duplicate]

I stumpled upon the equation

$$\sum_{k=1}^\infty \frac{k^2}{k!} = 2\mathrm{e}$$

and was just curious how to deduce the right hand side of the eqution - which identities could be of use here? Trying to simplify the partial sums to deduce the value of the series itself didn't help too much thus far.

Edit:

The only obvious transformation is $$\sum_{k=1}^\infty \frac{k^2}{k!} = \sum_{k=0}^\infty \frac{k+1}{k!}$$ but there was nothing more I came up with.

Solution 1:

The most obvious way is to look at a function that gets that value at some $x$.

Start with $e^x=\sum_{k=0}^\infty \frac{x^k}{k!}$. Then try differentiating over $x$ to get those $k$ at the right places.

In particular... differentiating once gives you $$e^x=\sum_{k=1}^\infty k\frac{x^{k-1}}{k!}$$ To get another $k$, multiply by $x$... $$x e^x=\sum_{k=1}^\infty k\frac{x^k}{k!}$$ and differentiate again $$(x e^x)'=(x+1)e^x=\sum_{k=1}^\infty k^2\frac{x^{k-1}}{k!}$$ Now plug in $1$ and you're done.

Solution 2:

As $k^2=k(k-1)+k$

$$\dfrac{k^2}{k!}=\dfrac1{(k-2)!}+\dfrac1{(k-1)!}$$

$$\sum_{k=1}^\infty\dfrac{k^2}{k!}=\sum_{k=1}^\infty\dfrac1{(k-2)!}+\sum_{k=1}^\infty\dfrac1{(k-1)!} =2\sum_{r=0}^\infty\dfrac1{r!}$$

as $\dfrac1{r!}=0$ for integer $r<0$

Now $e^x=\sum_{r=0}^\infty\dfrac{x^r}{r!}$

Solution 3:

Recall that $$\sum_{k=1}^\infty \frac{k^2}{k!} = \sum_{k=1}^\infty \frac{k}{(k-1)!} = \sum_{k=0}^\infty \frac{k+1}{k!} $$ Now write $$ \sum_{k=0}^\infty \frac{k+1}{k!} = \sum_{k=0}^\infty \frac{k}{k!} + \sum_{k=0}^\infty \frac{1}{k!} = \sum_{k=1}^\infty \frac{1}{(k-1)!} +e = 2e $$