Show $\sum_{k=1}^\infty \frac{k^2}{k!} = 2\mathrm{e}$ 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.
 A: 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.
A: 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 $e^x=\sum_{r=0}^\infty\dfrac{x^r}{r!}$
A: 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 $$
 \sum_{k=0}^\infty \frac{k+1}{k!} = \sum_{k=0}^\infty \frac{k}{k!} + \sum_{k=0}^\infty \frac{1}{k!} = \sum_{k=1}^\infty \frac{1}{(k-1)!} +e = 2e $$
A: First of all,
$$
e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}.
$$
Now, multiply by $x$ and differentiate with respect to $x$:
$$
\frac{d}{dx}(x e^x) = \frac{d}{dx} \sum_{k=0}^{\infty} \frac{x^{k+1}}{k!}
$$
yielding
$$
(1+x)e^x = \sum_{k=0}^{\infty} \frac{(k+1)x^k}{k!}
= \sum_{k=1}^{\infty} \frac{kx^{k-1}}{(k-1)!}
= \sum_{k=1}^{\infty} \frac{k^2x^{k-1}}{k!}.
$$
Now, evaluate at $x=1$.  (All of these series converge everywhere.)
$$
\sum_{k=1}^{\infty} \frac{k^2}{k!} = (1+1)e^1 = 2e.
$$
