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This question already has an answer here:

How to compute the value of $\sum\limits_{n= 0}^\infty \frac{n^2}{n!}$ ?

I started with the ratio test which told me that it converges but I don't know to what value it converges. I realized I only know how to calculate the limit of a power/geometric series.

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marked as duplicate by Najib Idrissi, Daniel Fischer Sep 30 '15 at 10:02

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.

  • $\begingroup$ Try to find a closed expression for the power series $f(x) = \sum_n (n^2/n!) x^n$, then evaluate $f(1)$. $\endgroup$ – Ulrik Feb 17 '14 at 18:11
  • $\begingroup$ These sums are known as Bell numbers. $\endgroup$ – Lucian Feb 17 '14 at 23:11
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\begin{eqnarray} \sum_{n=0}^\infty\frac{n^2}{n!}&=&\sum_{n=1}^\infty\frac{n^2}{n!}=\sum_{n=1}^\infty\frac{n}{(n-1)!}=\sum_{n=0}^\infty\frac{n+1}{n!} =\sum_{n=0}^\infty\frac{n}{n!}+\sum_{n=0}^\infty\frac{1}{n!}=\sum_{n=1}^\infty\frac{n}{n!}+e\\ &=&e+\sum_{n=1}^\infty\frac{1}{(n-1)!}=e+\sum_{n=0}^\infty\frac{1}{n!}=e+e=2e \end{eqnarray}

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  • $\begingroup$ $\sum_{n=0}^\infty \frac{n^2}{n!} = \sum_{n=0}^\infty \frac{n+1}{n!}$ seems a little bit weird to me. Are you sure that this is correct? $\endgroup$ – Maik Klein Feb 17 '14 at 18:31
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    $\begingroup$ What happened there is that the first term in the series on the LHS is actually 0. So we can ignore it. Notice the change of index after the first equality. $\endgroup$ – Joel Feb 17 '14 at 19:08
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    $\begingroup$ Equivalently, $\sum_{n=0}^\infty \frac{n^2}{n!} = \sum_{n=0}^\infty \frac{n(n-1)}{n!} + \sum_{n=0}^\infty \frac{n}{n!}$. And for any $k\ge1$,$$\sum_{n=0}^\infty \frac{n(n-1)\dots(n-k+1)}{n!} = \sum_{n=k}^\infty \frac{n(n-1)\dots(n-k+1)}{n!} = \sum_{n=k}^\infty \frac1{(n-k)!} = \sum_{m=0}^\infty \frac1{m!} = e.$$ $\endgroup$ – Greg Martin Feb 17 '14 at 19:13
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Note that, for every $n\geqslant2$, $$\frac{n^2}{n!}=\frac{n(n-1)+n}{n!}=\frac1{(n-2)!}+\frac1{(n-1)!},$$ ...and watch out for the small $n$ cases.

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Hint:

  • The exponential generating function for $a_n = 1$ is $\sum_n a_n\frac{z^n}{n!} = e^z$.
  • Differentiating both sides by $z$ and then multiplying by $z$ gets you $\sum_n n \frac{z^{n}}{n!} = ze^z$.
  • Doing it again will get you $ze^z+z^2e^z = e^z(z+z^2)$.
  • To eliminate $z$'s, substitute $z=1$.

I hope this helps $\ddot\smile$

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The idea for $$\frac{n^r+b_{r-1}n^{r-1}+\cdots+b_1\cdot n}{n!},$$ we can set this to $$\frac{n(n-1)\cdots(n-r+1)+a_{r-1}\cdot n(n-1)\cdots(n-r+2)+\cdots+a_2n(n-1)+a_1\cdot n+a_0}{n!}$$

$$\frac1{(n-r)!}+\frac{a_{r-2}}{(n-r+1)!}+\frac{a_1}{(n-2)!}+\frac{a_1}{(n-1)!}+\frac{a_0}{(n)!}$$

where the arbitrary constants $a_is,0\le i\le r-2$ can be found comapring the coefficients of the different powers of $n$

Here let $$\frac{n^2}{n!}=\frac{a_0+a_1n+a_2n(n-1)}{n!}=\frac{a_0}{n!}+\frac{a_1}{(n-1)!}+\frac{a_2}{(n-2)!}$$

$$\implies n^2=a_0+n(a_1-a_2)+a_2n^2$$

$$\implies a_2=1,a_1-a_2=0\iff a_1=a_2,a_0=0$$

So, we have $$\sum_{n=0}^{\infty}\frac{n^2}{n!}=\sum_{n=0}^{\infty}\frac1{(n-1)!}+\sum_{n=0}^{\infty}\frac1{n!}=\sum_{n=1}^{\infty}\frac1{(n-1)!}+\sum_{n=0}^{\infty}\frac1{n!}$$ as $\displaystyle\frac1{(-1)!}=0$

Now, observe that each summand is $e$(exponential 'e')

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