Is the following series $\sum_{n=0}^\infty \frac{n^3}{n!}$ convergent and if so what is the sum? I started by converting the following $\sum_{n=0}^\infty \frac{n^3}{n!}$ into 
$$\sum_{n=1}^\infty \frac{n(n-1)(n-2)}{n!} + \frac{3n(n-1)}{n!} + \frac{n}{n!}.$$ Did i get it right and where should I go from here ?
 A: Note that by limit comparison test with $\sum \frac 1{n^2}$
$$\frac{\frac{n^3}{n!}}{\frac 1{n^2}}=\frac{n^5}{n!}\to 0$$
therefore the series converges.
For the sum we can use that
$$\sum_{n=0}^\infty \frac{n^3}{n!}=\sum_{n=1}^\infty \frac{n^2}{(n-1)!}=\sum_{n=0}^\infty \frac{n^2-1+1}{(n-1)!}=\sum_{n=2}^\infty \frac{n+1}{(n-2)!}+\sum_{n=1}^\infty \frac{1}{(n-1)!}=$$
$$= \sum_{n=2}^\infty \frac{n-2+3}{(n-2)!}+\sum_{n=1}^\infty \frac{1}{(n-1)!}=\sum_{n=3}^\infty \frac{1}{(n-3)!}+3\sum_{n=2}^\infty \frac{1}{(n-2)!}+\sum_{n=1}^\infty \frac{1}{(n-1)!}=5e$$
A: Just look at each individual term:
$$
\sum_{n=1}^{\infty} \frac{n(n-1)(n-2)}{n!}=\sum_{n=3}^{\infty}\frac{n(n-1)(n-2)}{n!} = \sum_{n=3}^{\infty}\frac{1}{(n-3)!} = \sum_{n=0}^{\infty}\frac{1}{n!} = e
$$
$$
\sum_{n=1}^{\infty} \frac{3n(n-1)}{n!}= 3 \sum_{n=2}^{\infty}\frac{1}{(n-2)!} = 3e
$$
$$
\sum_{n=1}^{\infty}\frac{n}{n!} = \sum_{n=0}^{\infty} \frac{1}{n!} = e
$$
Therefore, the series is convergent and the sum is $5e$.
A: Let $a_{n}=\frac{n^{3}}{n!}$. Consider the power series 
$$
\sum_{n=1}^{\infty}a_{n}x^{n-1}.
$$
Firstly, we show that the power series converges for $x=2$. By ratio
test, 
\begin{eqnarray*}
\lim_{n}\frac{a_{n+1}x^{n}}{a_{n}x^{n-1}} & = & \lim_{n}2\cdot\left(\frac{n+1}{n}\right)^{3}\cdot\frac{1}{n+1}\\
 & = & 0.
\end{eqnarray*}
Therefore, the series converges for $x=2$ and hence the radius of
convergence $R$ is at least $2$.
Define $f:(-2,2)\rightarrow\mathbb{R}$ by $f(x)=\sum_{n=1}^{\infty}a_{n}x^{n-1}$.
Then, $f$ is analytic. Define $f_{1}:(-2,2)\rightarrow\mathbb{R}$
by $f_{1}(x)=\int_{0}^{x}f(t)dt$. Since the series converges uniformly
over any compact subset of $(-2,2)$, termwise integration is valid.
We have that
\begin{eqnarray*}
f_{1}(x) & = & \sum_{n=1}^{\infty}\int_{0}^{x}a_{n}t^{n-1}dt\\
 & = & \sum_{n=1}^{\infty}\frac{a_{n}}{n}x^{n}.\\
 & = & \sum_{n=1}^{\infty}\frac{n^{2}}{n!}x^{n}.
\end{eqnarray*}
Define $f_{2}:(-2,2)\rightarrow\mathbb{R}$ by 
$$
f_{2}(x)=\sum_{n=1}^{\infty}\frac{n^{2}}{n!}x^{n-1}.
$$
Note that $f_{2}$ is well-defined and $f_{2}(x)=\begin{cases}
f_{1}(x)/x, & \mbox{if }x\neq0\\
0, & \mbox{if }x=0
\end{cases}.$
Define $f_{3}:(-2,2)\rightarrow\mathbb{R}$ by $f_{3}(x)=\int_{0}^{x}f_{2}(t)dt$,
we have $f_{3}(x)=\sum_{n=1}^{\infty}\frac{n}{n!}x^{n}=x\sum_{n=1}^{\infty}\frac{x^{n-1}}{(n-1)!}=xe^{x}.$
On the other hand, $f_{2}(x)=f_{3}'(x)=e^{x}(1+x)$. $f_{1}(x)=xf_{2}(x)=e^{x}(x+x^{2})$.
Finally, $f(x)=f_{1}'(x)=e^{x}(1+3x+x^{2})$. Put $x=1$, then we obtain
$f(1)=5e$. Hence 
$$
\sum_{n=1}^{\infty}a_{n}=5e.
$$
