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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.


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.


marked as duplicate by Watson, Shahab, user228113, Daniel W. Farlow, gebruiker Mar 25 '16 at 13:18

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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.


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


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

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

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


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


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


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