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This question already has an answer here:

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.

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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.

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

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

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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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