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What would be the sum of the series $\dfrac{n^2}{n!}$ ?

I don't even know where to start with. It's nothing like telescopic. I tried to compare with some known series but that doesn't seems to work.

Widawensen
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Nitish
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6 Answers6

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Note that $\frac {n^2}{n!}=\frac n{(n-1)!}=\frac 1{(n-2)!}+\frac 1{(n-1)!}$ You didn't say what the lower limit of $n$ is, so you may need some correction at the bottom end.

Ross Millikan
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$$ 2e=(xe^x)'(1)=\left(\sum_{n=0}^\infty\frac{x^{n+1}}{n!}\right)'(1)=\left(\sum_{n=0}^\infty\frac{(n+1)x^n}{n!}\right)(1)=\sum_{n=0}^\infty\frac{n+1}{n!}=\sum_{n=1}^\infty\frac{n}{(n-1)!}=\sum_{n=1}^\infty\frac{n^2}{n!} $$

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Hint: start with the Maclaurin series for $e^x$, and consider first and second derivatives with respect to $x$.

Robert Israel
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  • Did you mean $xe^x$ or $e^x/x$ instead of $e^x$? –  Jan 24 '17 at 04:33
  • No, I really mean $e^x$. And yes, the derivative is $e^x$. But it is also $\sum_{n=0}^\infty n x^{n-1}/n!$. And the second derivative is ... – Robert Israel Jan 24 '17 at 08:36
  • Oh! I didn't consider splitting the series after the second derivative, neat! My apologizes! –  Jan 24 '17 at 08:47
  • It doesn't let me turn my downvote into an upvote unless you edit your answer. Maybe you can turn that 'start' into 'Start' or something and I'll be more than happy to do so... (I will wait longer in the future before downvoting an answer, my apologies again) –  Jan 24 '17 at 09:09
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With convergence use ration criterion $$\lim_\infty\frac{a_{n+1}}{a_{n}}$$

Nosrati
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Consider $$A=\sum_{n=0}^\infty \frac {n^2}{n!}x^n=\sum_{n=0}^\infty \frac {n(n-1)+n)}{n!}x^n=\sum_{n=0}^\infty \frac {n(n-1)}{n!}x^n+\sum_{n=0}^\infty \frac {n}{n!}x^n$$ $$A=x^2\sum_{n=0}^\infty \frac {n(n-1)}{n!}x^{n-2}+x\sum_{n=0}^\infty \frac {n}{n!}x^{n-1}$$ $$A=x^2\left(\sum_{n=0}^\infty \frac {x^n}{n!} \right)''+x\left(\sum_{n=0}^\infty \frac {x^n}{n!} \right)'=(x^2+x)e^x$$ Now, make $x=1$.

Claude Leibovici
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In general: $$\sum_{n=1}^{+\infty}\frac{P(n)}{(n+a)!}\text{ with }a\in \mathbb{N} \text{ and }P\text{ polynomial of degree }k.$$ Express$$P(n)=A_k\underbrace{(n+a)(n+a-1)\ldots}_{k\text{ factors}}+A_{k-1}\underbrace{(n+a)(n+a-1)\ldots}_{k-1\text{ factors}}$$ $$+\cdots+A_2\underbrace{(n+a)(n+a-1)}_{2\text{ factors}}+A_1\underbrace{(n+a)}_{1\text{ factor}}+A_0.$$ The indetermined coefficients $A_0,A_1,\ldots,A_k$ can be found giving to $n$ the values $-a,$ $-(a-1),$ etc. Decompose $P(n)$ in the mentioned form and use $$e=\sum_{m=0}^{+\infty}\frac{1}{m!}.$$

Fernando Revilla
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