Power series and computation

Question

Let $F\colon \mathbb{C}\to\mathbb{C}$ and let $f\colon \mathbb{R}\to\mathbb{R}$ be the functions defined by:

$$ F(z) = \sum_{n=0}^{\infty} \frac{z^n}{(n+1)!}$$

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

Show that $$ F(z) = \begin{cases} \frac{e^z-1}{z} & \text{for $z\in\mathbb{C}\setminus \{0\}$} \\ 1 & \text{if $z=0$} \end{cases}$$

and show that $xf'(x) + f(x) = e^x$

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It is trivial that for $F(z) = 0\ \text{if $z=0$}$ However, I am not sure how to show that for all $z\in\mathbb{C}$ $F(z) = \frac{e^z-1}{z}$.

In order to show that $xf'(x) + f(x) = e^x$ I know that $e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$ and $xf'(x) = \sum_{n=0}^{\infty}n\cdot \frac{x^{n-1}\cdot x}{(n+1)!}=\sum_{n=0}^{\infty}n\cdot \frac{x^{n}}{(n+1)!}$

$x'f(x) + f(x) = \sum_{n=0}^{\infty}n\cdot \frac{x^{n}}{(n+1)!}+\sum_{n=0}^{\infty} \frac{x^{n}}{(n+1)!} = \sum_{n=0}^{\infty}n\cdot \frac{x^{n}}{(n+1)!}$

Where do I go from here to get $e^x$?

Answer 1

$$F(z) = \sum_{n=0}^{\infty} \frac{z^n}{(n+1)!}=\frac{1}{z}\sum_{n=0}^{\infty} \frac{z^{n+1}}{(n+1)!}=\frac{1}{z}\sum_{\color{blue}{n=1}}^{\infty} \frac{z^{n}}{n!}=\frac{1}{z}(\color{blue}{e^z-1})$$ edit: For the second part,

$$xf'(x) + f(x)=x\sum_{n=0}^{\infty} \frac{nx^{n-1}}{(n+1)!}+\sum_{n=0}^{\infty} \frac{x^n}{(n+1)!}=\sum_{n=0}^{\infty} \frac{(n+1)x^{n}}{(n+1)!}=e^x$$ by cancelling $(n+1)$ from the fraction.

Answer 2

Observe that $$ zF(z)=\sum_{n=0}^\infty \frac{z^{n+1}}{(n+1)!}=\mathrm{e}^z-1. $$ Then $$ \mathrm{e}^z=(\mathrm{e}^z-1)'=\big(zF(z)\big)'=F(z)+zF'(z) $$