Find a function $f(x)$ sucht that $\sum_{n=0}^{\infty} \frac{n+1}{n!}x^n$ is the Taylor series of $f(x)$.

Question

Find a function $f(x)$ sucht that $\sum_{n=0}^{\infty} \frac{n+1}{n!}x^n$ is the Taylor series of $f(x)$.

I wanted to split the sum in two parts, so you get $\sum _{n=0}^{\infty} x^n -\sum_{n=0}^{\infty}....$. Because $x^n$ is the geometric series. But I'm having problems with finding something to put on the dots. I don't really know how to approach such a problem.

Answer 1

Hint: $$\sum_{n=0}^{\infty} \frac{n+1}{n!}x^n=\sum_{n=0}^{\infty} \frac{n}{n!}x^n+\sum_{n=0}^{\infty} \frac{1}{n!}x^n=x\sum_{n=1}^{\infty} \frac{1}{(n-1)!}x^{n-1}+\sum_{n=0}^{\infty} \frac{1}{n!}x^n$$

Answer 2

According to hint of @Arthur, let $F(x)=xe^x=x\sum_{n=0}^\infty \frac{x^n}{n!}=\sum_{n=0}^\infty \frac{x^{n+1}}{n!}.$ Taking the derivative both sides $$(xe^x)'=\sum_{n=0}^\infty \frac{n+1}{n!}x^n,$$ and hence the desired function $f(x)=(xe^x)' = e^x +xe^x$.