What sum to $\sum_{n=0}^\infty\frac{x^n}{(n+1)!}$ in its convergence radus? My task is this:
Find the sum to $$\sum_{n=0}^\infty\frac{x^n}{(n+1)!}.$$ in its convergence radus.
My work so far:
By ration test we get $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\left|\frac{x^{n+1}}{(n+2)!}*\frac{(n+1)!}{x^n}\right|= \lim_{n\to\infty}\left|\frac{x}{n+2}\right|=0$. So the series converges $\forall x\in \mathbb{R}$. Let $$\begin{align}S(x)=\sum_{n=0}^\infty\frac{x^n}{(n+1)!}\to xS(x)=\sum_{n=0}^\infty\frac{x^{n+1}}{(n+1)!}\to&\\ \left(xS(x)\right)'=\sum_{n=0}^\infty\frac{x^n}{n!}=e^x\to xS(x)=\int e^xdx=e^x+c.\\ \end{align}$$
Now I'm thinking that $\forall n\to 0*S(0)= e^0+c=1+c\implies c = 0\implies S(x)=\frac{e^x}{x}.$ However the answer is $S(x)=\frac{e^x-1}{x}$ for $x\neq0$
Can someone explain the last thing? Thanks in advance!

From
$$
xS(x)=\sum_{n=0}^\infty\frac{x^{n+1}}{(n+1)!},
$$ just by a change of index, you get
$$
xS(x)=\sum_{p=1}^\infty\frac{x^p}{p!}=\sum_{p=0}^\infty\frac{x^p}{p!}-1=e^x-1,
$$ that is

$$
S(x)=\frac{e^x-1}x, \quad x \neq0.
$$

