Sum of $\sum_{n=1}^{\infty} \frac {x^n}{(n-1)!} $

Question

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.

Answer 1

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*}

Answer 2

By reindexing, we see that $$\sum_{n=1}^{\infty} \frac {{x}^{n-1}}{(n-1)!}=\sum_{n=0}^{\infty} \frac {{x}^{n}}{(n)!}$$