Term-by-term antidifferentiation and power series representation of antiderivative I have solved the following exercise and I would like to know If I have made any mistakes (NOTE: I can't use integrals, since they are defined in a later chapter):
Assume $f(x)=\sum_{n=0}^{\infty} a_n x^n$ converges on $(-R,R)$.
(a) $F(x)=\sum_{n=0}^{\infty}\frac{a_n}{n+1}x^{n+1}$ is defined on $(-R,R)$ and satisfies $F'(x)=f(x)$.
(b) Antiderivatives are not unique. If $g$ is an arbitrary function satisfying $g'(x)=f(x)$ on $(-R,R)$ find a power series representation for $g$.
My solution:
(a) Let $x\in (-R,R)$: then the series  $\sum_{n=0}^{\infty} a_n x^{n+1}$ is convergent in $(-R,R)$ since $\sum_{n=0}^{\infty} a_n x^n$ is convergent there by hypothesis and since $(\frac{1}{n+1})_{n=0}^{\infty}$ is a non-negative and decreasing sequence by Abel's Test we have that $\sum_{n=0}^{\infty}\frac{a_n}{n+1}x^{n+1}$ is convergent in $(-R,R)$ too so $F(x)$ is defined on $(-R,R)$. Since by Theorem 6.5.2 * $\sum_{n=0}^{\infty} a_n x^n$ converges uniformly on every interval $[-c,c]\subseteq (-R,R)$, $0\in [-c,c]$ and $F(0)=\sum_{n=0}^{\infty}\frac{a_n}{n+1}0^{n+1}=0$ is convergent, by Term-by-Term Differentiability Theorem we have $F'(x)=\sum_{n=0}^{\infty} F'_n(x)=\sum_{n=0}^{\infty} a_n x^n=f(x)$ on each $[-c,c]\subseteq (-R,R)$.
(b) $g(x)=C +\sum_{n=0}^{\infty}\frac{a_n}{n+1}x^{n+1}$ ($C$ costant) since $g'(x)=0+(\sum_{n=0}^{\infty}\frac{a_n}{n+1}x^{n+1})'\stackrel{\text{(a)}}{=}\sum_{n=0}^{\infty}a_n x^n=f(x)$, as desired.

*Theorem 6.5.2 If a power series converges absolutely at $x_0$ then it converges uniformly on the interval $[-|x_0|,|x_0|]$
 A: In order to apply the term-by-term differentiation theorem, you need uniform convergence for $\sum_{n=1}^{\infty}n a_{n} x^{n-1}$ as well. But you only show that it converges on $(-R,R)$.
I would do (a) as follows.
Let $x\in(-R,R)$ be arbitrary and pick $t$ to satisfy $|x|<t<R$. Observe that
$$
\sum_{n=1}^{N}\left|n a_{n} x^{n-1}\right|=\sum_{n=1}^{N} \frac{1}{t}\left(n\left|\frac{x}{t}\right|^{n-1}\right)\left|a_{n} t^{n}\right|
$$
Because $|x/t|<1$, we can pick$\dagger$ $L$ such that for all $n$,
$$
n|\frac{x}{t}|^{n-1}\le L
$$
Now you have
$$
\sum_{n=1}^{N}\left|n a_{n} x^{n-1}\right|=\sum_{n=1}^{N} \frac{1}{t}\left(n\left|\frac{x}{t}\right|^{n-1}\right)\left|a_{n} t^{n}\right| \leq \frac{L}{t} \sum_{n=1}^{N}\left|a_{n} t^{n}\right|
$$
where the last sum converges because $t\in(-R,R)$.
For (b), use the fact that if $g'=f'$ on $(-R,R)$, you must have $g=f+C$ for some constant $C$.

$\dagger$ Notes.
If $0<s<1$, you can show that $ns^{n-1}$ is a bounded sequence. This can be seen by applying the ratio test to the sequence $a_n=ns^{n-1}$.
