<s>[INST] Sum of $\sum_{n=1}^{\infty} \frac {x^n}{(n-1)!} $ the task is to find sum of $\sum_{n=1}^{\infty} \frac {x^n}{(n-1)!} $. I've tried to factor out $x$ to get $x\sum_{n=1}^{\infty} \frac {{x}^{n-1}}{(n-1)!}$  and integrate the sum, which yields $x\sum_{n=1}^{\infty} \frac {x^n}{n!} $ which I can sum as $xe^x$. Then I need to derivate it which yields $e^x+xe^x$. The correct answer is supposed to be $xe^x$. What step am I doing wrong?
Thanks for your help.
 [/INST] Note that you have to integrate the complete expression $x\sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!}$ if you want to use integration. We then obtain by using integration by parts
\begin{align*}
\int \underbrace{x}_{u}\underbrace{\sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!}}_{v^\prime}\,dx
&= x\sum_{n=1}^\infty \frac{x^{n}}{n!}-\int  \sum_{n=1}^\infty \frac{x^{n}}{n!}\,dx\\
&=x\left(e^x-1\right)-\int \left(e^x-1\right)\,dx\\
&=x\left(e^x-1\right)-e^x+x+C\\
&=xe^x-e^x+C\tag{1}
\end{align*}
and differentiation of (1) gives
\begin{align*}
\frac{d}{dx}\left(xe^x-e^x+C\right)&=\left(e^x+xe^x\right)-e^x\\
&=xe^x
\end{align*}
as wanted.

More convenient is reindexing the series. This way we obtain
  \begin{align*}
\color{blue}{\sum_{n=1}^\infty\frac{x^n}{(n-1)!}}&=\sum_{n=0}^\infty \frac{x^{n+1}}{n!}=x\sum_{n=0}^\infty\frac{x^n}{n!}
\color{blue}{=xe^x}
\end{align*}

</s>