Assume $g(x) = \sum_{n=0}^{\infty} a_nx^n$ converges on $(-M,M)$ Assume $g(x) = \sum_{n=0}^{\infty} a_nx^n$ is convergent on $(-M,M)$
(1)Show $G(x)= \sum_{n=0}^{\infty} \frac{1}{n+1}a_nx^{n+1}$ is defined on $(-M,M)$ and satisfies $G'(x)=g(x)$
(2)If $h$ is an arbitrary function satisfying $h'(x)=g(x)$ on $(-M,M)$, find a power series representation for $h$
What I think is that given the fact that $g(x) = \sum_{n=0}^{\infty} a_nx^n$ is convergent on $(-M,M)\subseteq \mathbb{R}$, $g$ is continuous and differentiable on any $x\in (-M,M)$ and I need to do something with this. But I'm not sure whether it's the right way to approach or how I should push any further even if correct.
I guess the question is just the same as Assume that $f(x) = \sum_{n=0}^{\infty} a_n x^n $ converges on $(-R,R)$., but the problem is that I don't understand that explanation.
Can someone please interpret that answer or explain in rather a detailed, long enough proof so that I could feel kindness of him/her with definitions/theorems/proofs/properties that are used?
Thanks!
 A: (1). (Cauchy-Hadamard Radius Formula). For real or complex $a_n$ let $R=\frac {1}{\lim \sup_{n\to \infty}|a_n|^{1/n}}.$
...(i). If $D\subset \{z\in \Bbb C: |z|<R\}$ and $D$ is closed in $\Bbb C$ then $g_m(z)=\sum_{n=0}^m a_nz^n$ converges $uniformly$ on $D$ to $g(z)=\sum_{n=0}^{\infty}a_nz^n$ as $m\to \infty.$ 
...(ii). $g_m(z)$ does not converge at all if $|z|>R.$
In particular $g_m(x)\to g(x)$ uniformly on $[-A,A]$ for any $A\in (0,M)$ because by (1)(ii)  we must have $R\geq M.$
(2). For any $A\in (0,M):$
...(i). Each $g_m$ is continuous on $[-A,A]$ so the uniform convergence, on $[-A,A],$ of $g_m$ to $g$ implies  that $g$ is continuous on $[-A,A]$ and also implies that $G_m(x)=\int_0^xg_m(y)dy$ converges uniformly on $[-A,A]$ to $\int_0^xg(y)dy.$  So for any $|x|< A$ we have $G(x)=\int_0^x g(y)dy.$ 
...(ii). Since $g$  is continuous on $[-A,A],$ by the Fundmental Theorem of Calculus we have $G'(x)=\frac {d}{dx}\int_0^xg(y)dy=g(x)$ for any $|x|<A.$
...(iii). Since $A$ can be any member of $(0,M),$ therefore  $G'(x)=g(x)$ for all $x\in (-M,M).$
(3). If $h'(x)=g(x)$ for all $|x|<M$ then $h'(x)-G'(x)=0$ for all $|x|<M$ so $h(x)-G(x)$ is a constant $K$ on $(-M,M).$ So $h(x)=K+G(x)=\sum_{n=0}^{\infty}h_nx^n$ where $h_0=K$ and $h_n=a_{n-1}x^n/n$ for $n>0.$
Remark. The proof of (1) is not long nor difficult. There is a nice presentation in the book Complex Analysis by Ahlfors.
