Power series representation of the sum of $\sum_{k=0}^\infty(k^3+1) x^k/k!$ So, recently I was searching for, what seemed like a fairly simple one, what function does this sum represent:
$\displaystyle f(x)=\sum_{k=0}^\infty\left(k^3+1\right)\frac{x^k}{k!}$
So the natural thing would be to separate both equations:
$\displaystyle f(x)=\sum_{k=0}^\infty k^3\frac{x^k}{k!}+e^x$
So, with further identification $f(x)$ seemed like a simple answer:
$f(x)=T_3(x)e^x+e^x$
where $T_n$ is the $\,n^{\mathrm{th}}$ Touchard polynomial.
$f(x)=\left(x+3x^2+x^3+1\right)e^x$
What bugged me is the fact that neither I do I understand Touchard Polynomials nor does the question fit the level of knowing them, this is intended as a Second Semester undergrad Calculus II question, so my question is simple, is there a method of finding the sum without having to deal with the Touchard polynomials, is there a way of arriving to the same (or different, maybe I’m wrong) answer without dealing with them?
 A: You start with the series for $e^x$, then take derivative and then multiply by $x$, and so on as follows:
\begin{align*}
e^x&=\sum_{k=0}^{\infty}\frac{x^k}{k!}\\
\color{red}{\frac{d}{dx}}e^x&=\sum_{k=0}^{\infty}\color{red}{k}\frac{x^{k-1}}{k!}\\
\color{red}{x}e^x&=\sum_{k=0}^{\infty}k\frac{x^{\color{red}{k}}}{k!}\\
\color{red}{\frac{d}{dx}}(xe^x)&=\sum_{k=0}^{\infty}\color{red}{k^2}\frac{x^{k-1}}{k!}\\
\color{red}{x}(xe^x+e^x)&=\sum_{k=0}^{\infty}k^2\frac{x^{\color{red}{k}}}{k!}
\end{align*}
Can you take it from here?
A: $$
\begin{align}
\sum_{k=0}^\infty k^3 \frac{x^k}{k!} &= x\sum_{k=1}^\infty k^3 \frac{x^{k-1}}{k!} \\
&=x \sum_{k=0}^\infty (k+1)^2 \frac{x^k}{k!} \\
&=x\left(\sum_{k=0}^\infty (k+1)\frac{x^{k+1}}{k!}  \right)^\prime \\
&=x\left(\sum_{k=1}^\infty k \frac{x^{k+1}}{k!} +xe^x\right)^\prime \\
&=x\left(x^2e^x + xe^x\right)^\prime \\
&=x(x^2 + 3x +1)e^x
\end{align}
$$
A: Note that
$$\begin{aligned}\frac{k^3+1}{k!}&=
\frac{k^2}{(k-1)!}+\frac1{k!}\\&
=\frac{(k+1)(k-1)+1}{(k-1)!}+\frac1{k!}\\&
=\frac{k+1}{(k-2)!}+\frac1{(k-1)!}+\frac1{k!}\\&
=\frac1{(k-3)!}+\frac3{(k-2)!}+\frac1{(k-1)!}+\frac1{k!}\end{aligned}$$
and that $\sum \frac{x^k}{(k-r)!}=\sum\frac{x^{k+r}}{k!}=x^re^x$. (Well, one has to be careful with the first summands of the series, but everything works out nicely)
A: Hint
Use
$$k^3=k(k-1)(k-2)+3k(k-1)+k$$ and you will face simple expansions if you properly change the powers of $x$ as required to face $e^x$.
