finding a suitable region for a series of functions to be analytic Let $f_n(z) = \frac{(2z-1)^n}{n} $. I want to find some suitable $A \subset \mathbb{C}$ such that $\sum_{n=1}^{\infty} f_n(z) $ is analytic on $A$. 
Well, first of all $\sum f_n (z) $ must be convergent on that region. We can apply ratio test to obtain 
$$ \left| \frac{ f_{n+1}}{f_n} \right| = \frac{n}{n+1} \cdot |2z-1| \to |2z-1|$$
Hence, $\sum f_n $ converges if $|2z - 1 | < 1 $
Is there other ways to find suitable regions ? I mean, do we have that $\sum f_n$ converges uniformly on $|2z-1|< 1 $ ?
 A: Given any power series $\sum_{k=0} a_k (z-c)^k$ there exists an $R \in [0,+\infty]$ (the radius of convergence) with the following properties:


*

*The series converges (absolutely) for all $z$ with $|z-c|<R$.

*The series diverges for all $z$ with $|z-c| > R$.

*For every $r < R$, the series converges uniformly on $|z-c| \le r$.


The proof of the above should be in every textbook on complex analysis.
Your series can be written as
$$
\sum_{k=1}^\infty \frac{2^n(z-\frac12)^n}{n}
$$
and you have correctly computed the radius of convergence as $\frac12$. By the above theorem, the series converges uniformly on $|2z-1| \le r$ for every $r < 1$. Thus the sum is holomorphic (analytic) on $|2z-1| < 1$, being the locally uniform limit of polynomials (the partial sums of the series). 
Since the series diverges for $|2z-1| > 1$, there is no larger open set on which the series defines a holomorphic function. (But, to complicate things further, it's possible to find an analytic function $g$ defined on a larger open set than $|2z-1| < 1$ that agrees with $f$ on the disc in question.)
