TITLE: how to show that power series is analytic inside the radius of convergence?
QUESTION [6 upvotes]: Let $f(z) = \sum a_n z^n$ be a power series with radius of convergence $R$. How do we show that $f$ is analytic in the circular region of radius $R$?

REPLY [5 votes]: I will show another proof.
Proposition
Let $R$ be the radius of convergence of $f(z) = \sum_{n=0}^{\infty} a_nz^n$.
Let $g(z) = \sum_{n=1}^{\infty} na_nz^{n-1}$.
Then $f'(z) = g(z)$ for every $|z| < R$.
Proof:
By the lemma of my previous answer, the radius of convergence of $g(z)$ is $R$.
Let $z$ be such that $|z| < R$.
Since $\sum_{n=1}^{\infty} na_nz^{n-1}$ converges uniformly in every compact subset of $\{z; |z| < R\}$,
$\int_{0}^{z} g(\zeta) d\zeta = \sum_{n=1}^{\infty} \int_{0}^{z} na_n\zeta^{n-1} d\zeta = \sum_{n=1}^{\infty} a_nz^n = f(z) - a_0$,
where the integral path is any smooth curve inside the domain $\{z; |z| < R\}$ starting from $0$ to $z$.
Hence $f'(z) = g(z)$.
QED