Evaluate the limit of $\sum_{n=1}^{\infty}\frac{n^2}{n!}$ 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! 
 A: $S_n = \sum_{k=1}^{n} \frac{k^2}{k!}$
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 $\text{lim}_{n \rightarrow \infty}S_n = 2e$.
A: $$\begin{matrix}
\\&\dfrac1{0!}&+&\dfrac1{1!}&+&\dfrac1{2!}&+&\dfrac1{3!}&+&\dfrac1{4!}&+&\cdots
\\+&&&\dfrac1{0!}&+&\dfrac1{1!}&+&\dfrac1{2!}&+&\dfrac1{3!}&+&\cdots
\\\hline
=&\dfrac1{0!}&+&\dfrac{1+1}{1!}&+&\dfrac{1+2}{2!}&+&\dfrac{1+3}{3!}&+&\dfrac{1+4}{4!}&+&\cdots
\\=&\dfrac1{0!}&+&\dfrac{2}{1!}&+&\dfrac{3}{2!}&+&\dfrac{4}{3!}&+&\dfrac{5}{4!}&+&\cdots
\\=&\dfrac{1^2}{1!}&+&\dfrac{2^2}{2!}&+&\dfrac{3^2}{3!}&+&\dfrac{4^2}{4!}&+&\dfrac{5^2}{5!}&+&\cdots
\end{matrix}$$
A: $$
f(x) = \sum_{n=0}^\infty \frac{n^2 x^n}{n!}
$$
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 $f(1)=2e$.
A: As an alternative
$$\sum_{n=1}^{N}\frac{n^2}{n!}=\sum_{n=1}^{N}\frac{n}{(n-1)!}=\sum_{n=0}^{N-1}\frac{(n+1)}{n!}=\sum_{n=0}^{N-1}\frac{n}{n!}+\sum_{n=0}^{N-1}\frac{1}{n!}=\sum_{n=1}^{N-1}\frac{1}{(n-1)!}+\sum_{n=0}^{N-1}\frac{1}{n!}=$$$$=\sum_{n=0}^{N-2}\frac{1}{n!}+\sum_{n=0}^{N-1}\frac{1}{n!} \to e+e=2e$$
A: $$e^{x}=\sum_{n=0}^{\infty }\frac{x^n}{n!}$$
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
$$\sum_{n=1}^{\infty }\frac{n^2}{n!}=2e$$
A: 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.
