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Evaluate the limit of:

$$\sum_{n=1}^{\infty}\frac{n^2}{n!}$$

Hints I am given:

1- The exponential series of type $\sum_{n=0}^{\infty}\frac{x^n}{n!}$ converge into $e^x$ for any $x\in R$

2- Any series of type $\sum_{n=0}^{\infty}\frac{P(n)}{n!}x^n$ , being $P(n)$ a polynomial of any level and $\forall x\in R$, it also converges.

3- $\sum_{n=1}^{\infty}a_{n-1}=\sum_{n=0}^{\infty}a_{n}$

What is the limit of this series?

According to the hints it is convergent, this is what I tried so far:

$\sum_{n=1}^{\infty}\frac{n^2}{n!}=\sum_{n=1}^{\infty}\frac{n^2}{n(n-1)!}=\sum_{n=1}^{\infty}\frac{n}{(n-1)!}$

Applying hint 3:

$\sum_{n=0}^{\infty}\frac{n+1}{n!}$

Split the series in 2:

$\sum_{n=0}^{\infty}\frac{n+1}{n!}=\sum_{n=0}^{\infty}\frac{n}{n!}+\sum_{n=0}^{\infty}\frac{1}{n!}$

Applying hint 1, with $x=1$ to the second part:

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

I dont know how to keep going, I know that the series $\sum_{n=0}^{\infty}\frac{n}{n!}$ converges into $e$ therefore the result of the whole series is 2 times $e$ but I dont understand why!

user605734 MBS
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    $$\sum_{n\geq 0}\frac{n}{n!}=\sum_{n\geq 1}\frac{n}{n!}=\sum_{n\geq 1}\frac{1}{(n-1)!}=\sum_{m\geq 0}\frac{1}{m!}=e.$$ – Jack D'Aurizio Oct 31 '18 at 21:13
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    Additionally, "the limit of $\sum_{n\geq 1}f(n)$" is redundant since $\sum_{n\geq 1}f(n)$ is already a limit, $\lim_{N\to +\infty}\sum_{n=1}^{N}f(n)$. So "the limit of $\sum_{n=1}^{N}f(n)$ as $N\to +\infty$" or just "$\sum_{n\geq 1}f(n)$". – Jack D'Aurizio Oct 31 '18 at 21:15
  • Note that $n^2=n(n-1)+n$ - that's how I would begin this. (Polynomials in $n$ can be expressed in a similar way) – Mark Bennet Oct 31 '18 at 21:23
  • https://math.stackexchange.com/questions/1711318/show-sum-k-1-infty-frack2k-2-mathrme/1711325#1711325 – lab bhattacharjee Nov 01 '18 at 00:51

6 Answers6

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$S_n = \sum_{k=1}^{n} \frac{k^2}{k!}$

Writing $k! = k(k-1)!$, we get

$S_n = \sum_{k=1}^{n} \frac{k}{(k-1)!} = \sum_{k=1}^{n} \frac{k-1}{(k-1)!} +\sum_{k=1}^{n} \frac{1}{(k-1)!}$.

Using $(k-1)! = (k-1)(k-2)!$, this can be further simplified to:

$S_n = \sum_{k=2}^{n} \frac{1}{(k-2)!} + \sum_{k=1}^{n} \frac{1}{(k-1)!}$.

Each of the above summations converges to $e$ as $n \rightarrow \infty$, so we get $\text{lim}_{n \rightarrow \infty}S_n = 2e$.

Godfather
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  • You go from $\sum_{n=1}^{\infty} \frac{n-1}{(n-1)!}$ to $\sum_{n=2}^{\infty} \frac{1}{(n-2)!}$ and change the summation's first item to 2, why can you do this? – user605734 MBS Oct 31 '18 at 21:44
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    @user605734MBS, because ${n-1\over(n-1)!}=0$ for $n=1$ and $1\over(n-2)!$ for $n\ge2$. – Barry Cipra Oct 31 '18 at 21:49
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    It's because the first term of the summation is 0 ($\frac{n-1}{(n-1)!} = 0$ for $n=1$), so changing the starting index to 2 won't make a difference. – Godfather Oct 31 '18 at 21:50
  • Okay I'll make sure to change the index first from now on, I was stuck there because I simplified first and then tried to change the inex, but I got $\sum_{n=1}^{\infty} \frac{1}{(n-2)!}$ then the first item would be $\frac{1}{(-1)!}$ Which is undefined. – user605734 MBS Oct 31 '18 at 21:54
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    $S_n$ is not a function of $n$. – marty cohen Nov 01 '18 at 01:16
  • @martycohen Edited my answer. Better? – Godfather Nov 01 '18 at 21:26
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$$ f(x) = \sum_{n=0}^\infty \frac{n^2 x^n}{n!} $$ can be expressed as $$ f(x) = x \frac{d}{dx}\left(x\frac{d}{dx}e^x\right) = x \frac{d}{dx}(xe^x) = e^x(x+x^2). $$ Therefore your sum is $f(1)=2e$.

Federico
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4

$$\begin{matrix} \\&\dfrac1{0!}&+&\dfrac1{1!}&+&\dfrac1{2!}&+&\dfrac1{3!}&+&\dfrac1{4!}&+&\cdots \\+&&&\dfrac1{0!}&+&\dfrac1{1!}&+&\dfrac1{2!}&+&\dfrac1{3!}&+&\cdots \\\hline =&\dfrac1{0!}&+&\dfrac{1+1}{1!}&+&\dfrac{1+2}{2!}&+&\dfrac{1+3}{3!}&+&\dfrac{1+4}{4!}&+&\cdots \\=&\dfrac1{0!}&+&\dfrac{2}{1!}&+&\dfrac{3}{2!}&+&\dfrac{4}{3!}&+&\dfrac{5}{4!}&+&\cdots \\=&\dfrac{1^2}{1!}&+&\dfrac{2^2}{2!}&+&\dfrac{3^2}{3!}&+&\dfrac{4^2}{4!}&+&\dfrac{5^2}{5!}&+&\cdots \end{matrix}$$

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As an alternative

$$\sum_{n=1}^{N}\frac{n^2}{n!}=\sum_{n=1}^{N}\frac{n}{(n-1)!}=\sum_{n=0}^{N-1}\frac{(n+1)}{n!}=\sum_{n=0}^{N-1}\frac{n}{n!}+\sum_{n=0}^{N-1}\frac{1}{n!}=\sum_{n=1}^{N-1}\frac{1}{(n-1)!}+\sum_{n=0}^{N-1}\frac{1}{n!}=$$$$=\sum_{n=0}^{N-2}\frac{1}{n!}+\sum_{n=0}^{N-1}\frac{1}{n!} \to e+e=2e$$

user
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$$e^{x}=\sum_{n=0}^{\infty }\frac{x^n}{n!}$$ differentiate it $$e^{x}=\sum_{n=1}^{\infty }\frac{nx^{n-1}}{n!}$$ multiply by $x$ $$xe^{x}=\sum_{n=1}^{\infty }\frac{nx^{n}}{n!}$$ differentiate it again $$xe^{x}+e^x=\sum_{n=1}^{\infty }\frac{n^2x^{n-1}}{n!}$$ let $x=1$ $$1e^{1}+e^1=\sum_{n=1}^{\infty }\frac{n^2}{n!}$$ so $$\sum_{n=1}^{\infty }\frac{n^2}{n!}=2e$$

E.H.E
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1

If $S_k =\sum_{n=1}^{\infty} \dfrac{n^k}{n!} $, then $S_0 =e-1$ and, for $k \ge 0$,

$\begin{array}\\ S_{k+1} &=\sum_{n=1}^{\infty} \dfrac{n^{k}}{(n-1)!}\\ &=\sum_{n=0}^{\infty} \dfrac{(n+1)^{k}}{n!}\\ &=\sum_{n=0}^{\infty} \dfrac1{n!}(n+1)^{k}\\ &=\sum_{n=0}^{\infty} \dfrac1{n!}\sum_{j=0}^k \binom{k}{j}n^j\\ &=\sum_{j=0}^k \binom{k}{j}\sum_{n=0}^{\infty} \dfrac{n^j}{n!}\\ &=\sum_{n=0}^{\infty} \dfrac{1}{n!}+\sum_{j=1}^k \binom{k}{j}\sum_{n=0}^{\infty} \dfrac{n^j}{n!}\\ &=e+\sum_{j=1}^k \binom{k}{j}\sum_{n=1}^{\infty} \dfrac{n^j}{n!}\\ &=e+\sum_{j=1}^k \binom{k}{j}S_j\\ \end{array} $

Setting $k=0$, $S_1 = e$.

Setting $k=1$, $S_2 = e+S_1 =2e$.

Setting $k=2$, $S_3 = e+2S_1+S_2 =5e$.

If $S_k =et_k$ then $t_1 = 1$ and $t_k =1+\sum_{j=1}^k \binom{k}{j}t_j $.

The first few $t_k$ are $1, 2, 5$ and these turn out to be the well-studied Bell numbers - see https://en.wikipedia.org/wiki/Bell_number for a reasonable discussion.

marty cohen
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