1
$\begingroup$

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.

  • 1
    I assume you mean sum, not some – 2017-01-24

6 Answers 6

3

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.

1

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

0

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

  • 0
    Did you mean $xe^x$ or $e^x/x$ instead of $e^x$? – 2017-01-24
  • 0
    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 ... – 2017-01-24
  • 0
    Oh! I didn't consider splitting the series after the second derivative, neat! My apologizes! – 2017-01-24
  • 0
    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) – 2017-01-24
0

With convergence use ration criterion $$\lim_\infty\frac{a_{n+1}}{a_{n}}$$

  • 0
    That proves convergence, but doesn't help with finding the sum. – 2017-01-24
  • 0
    Of course.you said that before me. – 2017-01-24
0

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 $x=1$.

0

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 $$e=\sum_{m=0}^{+\infty}\frac{1}{m!}.$$