how to show that power series is analytic inside the radius of convergence? 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 $R$? 
 A: 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
A: I think all you need is
Theorem: A power series converges uniformly inside its convergence interval
Proof: Follows at once from Weierstrass M-test: let $\,r>0\,$ be the convergence radius of the power series 
$$\sum_{n=1}^\infty a_n(z-a)^n$$ 
and let $\,0<\rho<r\,$, then:
$$|z-a|<\rho\Longrightarrow |a_n(z-a)^n|\leq |a_n|\rho^n=:M_n$$
and since the series $\,\displaystyle{\sum_{m=1}^\infty M_n}\,$ converges (absolutely, of course) , we're done.
A: Lemma
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 the radius of convergence of $g(z)$ is $R$.
Proof:
Let $R'$ be radius of convergence of $g(z)$.
Since $|a_nz^n| \le |na_nz^n|$, $R' \le R$.
So it suffices to prove $R \le R'$.
Let $z$ be such that $|z| < R$.
Choose $r$ such that $|z| < r < R$.
Let $\rho = \frac{|z|}{r}$.
Since $\sum_{n=1}^{\infty} a_nr^{n-1}$ converges, there exists $M > 0$ such that 
$|a_nr^{n-1}| \le M$ for all $n \ge 1$.
Then
$|na_nz^{n-1}| = n|a_n|r^{n-1}\rho^{n-1} \le nM\rho^{n-1}$
Since $0 \le \rho < 1$, $\sum_{n=1}^{\infty} nM \rho^{n-1}$ converges.
Hence $\sum_{n=1}^{\infty} na_nz^{n-1}$ converges.
Hence $R \le R'$.
QED
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:
Let $z$ be such that $|z| < R$.
Choose $r$ such that $|z| < r < R$.
Let $h$ be such that $0 < |h| \le r - |z|$
Consider $\frac{f(z+h) - f(z)}{h} = \sum_{n=0}^{\infty} a_n\frac{(z + h)^n - z^n}{h}$
By the formula $x^n - y^n = (x - y)(x^{n-1} + yx^{n-2} +\cdots+ y^{n-2}x + y^{n-1})$,
$|a_n\frac{(z + h)^n - z^n}{h}| = |a_n||(z + h)^{n-1} + z(z+h)^{n-2}+\cdots+ z^{n-2}(z+h) + z^{n-1})
\le n|a_n|r^{n-1}$
Define $\psi_n(h) = a_n\frac{(z + h)^n - z^n}{h}$ for $0 < |h| \le r - |z|$
Define $\psi_n(0) = n a_n z^{n-1}$.
Then $\psi_n(h)$ is continuous in $|h| \le r - |z|$.
Since $\sum_{n=1}^{\infty} n|a_n|r^{n-1}$ converges by the lemma,
$\Psi(h) = \sum_{n=1}^{\infty} \psi_n(h)$ converges uniformly in $|h| \le r - |z|$.
Hence $\Psi(h)$ is continuous in $|h| \le r - |z|$.
In particular $\lim_{h \to 0} \Psi(h) = \Psi(0)$.
This implies $f'(z) = g(z)$.
QED
