The answer to Question 2 is no. The essential reason is the following: Du Bois - Reymond constructed a continuous function $f$ on the unit circle whose Fourier series, $\sum_n a_n e^{i n \theta}$, does not converge at $\theta=0$. If you look at the construction, you'll see that it is easy to arrange that $a_n=0$ for $n<0$. Then $\sum_{n=0}^{\infty} a_n z^n$ defines an analytic function inside the unit disc, which extends continuously to the boundary, but such that $\sum a_n$ is divergent.
Below, I give the details. I used Pinsky's book as my reference for the construction of du Bois - Reymond.
Define $S_M(z) = \sum_{r=1}^M \frac{z^r-z^{-r}}{r}$. Our function will be of the form
$$f(z) := \sum_{k=1}^{\infty} \frac{z^{N_k}}{k^2} S_{M_k}(z) \quad (1)$$
where $M_k$ and $N_k$ are sequences of positive integers chosen such that
$$0 < N_1 - M_1 < N_1 + M_1 < N_2 - M_2 < N_2 + M_2 < N_3 - M_3 < N_3+ M_3 < \cdots \quad (2)$$
and
$$\frac{\log M_k}{k^2} \to \infty \ \mbox{as} \ k \to \infty. \quad (3)$$
We will show below that the sum (1) is uniformly convergent in the closed unit disc. Hence, it defines a continuous function on the closed disc and an analytic function in the interior. Condition (2) forces that the polynomials $z^{N_k} s_{M_k}(z)$ have no overlapping terms, so the Taylor series of $f$ just looks like blocks of $-1/M_k$, $-1/(M_k-1)$, ..., $-1$, $0$, $1$, $1/2$, ..., $1/M_k$, separated by long blocks of zeroes. Define $a_n$ to be the coefficients of $f(z) = \sum_{n=0}^{\infty} a_n z^n$.
We now check that $\sum a_n$ is divergent. We have
$$\sum_{n=0}^{N_k} a_n = \sum_{j=1}^{k-1} \frac{1}{j^2} \left( \frac{-1}{M_j} + \cdots + \frac{-1}{1} + \frac{1}{1} + \cdots + \frac{1}{M_j} \right) - \frac{1}{k^2} \left( \frac{1}{M_k} + \frac{1}{M_k -1} + \cdots + \frac{1}{2} + \frac{1}{1} \right)$$
$$=0+0+\cdots + 0 - \frac{1}{k^2} \left( \frac{1}{M_k} + \frac{1}{M_k -1} + \cdots + \frac{1}{2} + \frac{1}{1} \right) \approx - \frac{\log M_k}{k^2}.$$
Using condition (3), this goes to $- \infty$. So there is a subsequence of partial sums of $\sum a_n$ which goes to $- \infty$ and $\sum a_n$ diverges.
We now must prove that (1) is uniformly convergent. We need
Lemma There is an absolute constant $C$ so that, for any real angle $\theta$, and any positive integer $M$, we have
$$\left| \sum_{r=1}^M \frac{\sin (r \theta)}{r} \right| \leq C.$$
Given the lemma, uniform convergence is easy. By the maximum modulus principle, $\left| z^{N_k} s_{M_k}(z) \right|$ is maximized on the boundary of the unit disc.
On that boundary, $|e^{i N_k \theta} s_{M_k}(e^{i \theta} )| = 2 \left| \sum_{r=1}^{M_k} \frac{\sin (r \theta)}{r} \right|$ and so, by the lemma, is bounded independently of $\theta$. The $\frac{1}{k^2}$ factor in front then forces uniform convergence.
We now prove the lemma. This is the only part I'm not adapting from Pinsky, because he treats this as obvious.
Proof of lemma: Since the sum of sines is clearly periodic modulo $2 \pi$, and is clearly odd, we may assume that $\theta \in (0,\pi)$. We break the sum at $r=K$, for a parameter $K$ to be chosen later. For the first part of the sum,
$$\left| \sum_{r=1}^{K} \frac{\sin (r \theta)}{r} \right| \leq \sum_{r=1}^{K} \frac{r \theta}{r} = K \theta.$$
For the second part of the sum, we start by noting
$$\sin \frac{\theta}{2} \cdot \sum_{r=K+1}^M \frac{\sin (r \theta)}{r} = $$
$$\frac{\cos ((K+1/2) \theta)}{K+1} + \sum_{r=K+2}^M \cos ((r-1/2) \theta)\left( \frac{1}{r-1} - \frac{1}{r} \right)+ \frac{\cos((M+1/2) \theta)}{M}$$
so
$$\left| \sum_{r=K+1}^M \frac{\sin (r \theta)}{r} \right| \leq \frac{1}{\sin (\theta/2)} \left( \frac{1}{K+1} + \sum_{r=K+2}^M \left( \frac{1}{r-1} - \frac{1}{r} \right)+ \frac{1}{M}\right) $$
$$ = \frac{2}{\sin (\theta/2) (K+1)} \leq \frac{2}{(K+1)(\theta/\pi)} = \frac{2 \pi}{(K+1)\theta}.$$
We have used the bound $\sin (\theta/2) \geq \theta/\pi$ for $\theta \in (0 , \pi)$, which is true because $\sin$ is concave.
In short,
$$\left| \sum_{r=1}^M \frac{\sin (r \theta)}{r} \right| \leq K \theta + \frac{2 \pi}{(K+1) \theta}.$$
Choose $K$ such that $K \theta$ is neither near $0$ nor $\infty$, and this quantity will be bounded, so we have proved the lemma. $\square$.
Why can't the radius of convergence $\sqrt{5}$ be increased to include the origin?
The real part of logarithm (no matter what branch you take) is $\log |z|$. This is unbounded as $z\to 0$. But a power series is bounded on compact subsets of its disk of convergence. Therefore, no power series representing the logarithm (whatever branch) can have $0$ inside of its disk of convergence.
Can the Taylor series converge to a function that is not a branch of $\log$?
By definition, anything you get from Taylor series of $f$ is some branch of $f$, because it's related to $f$ by analytic continuation.
how do I actually prove that there is a branch of log that the Taylor series about $z_0=−2+i$ converges to?
There is nothing magical about negative real axis: you can cut the plane along any other half-line from $0$ to $\infty$ and define a branch of $\log$ in the slit plane (by interpreting $\arg z$ there). For example, the half-line $\{(2-i)t:t\ge 0\}$ would do. Or simply define $$\operatorname{Log}z =\log( z/(-2+i))+\log (-2+i)$$ where $\log$ is the principal branch.
Best Answer
Power series don't converge uniformly on their disc of convergence. But they do so locally, meaning that they converge uniformly on every compact subset of their disc of convergence. The Weierstraß-M-test can be used to prove this.
First notice that every compact subset of a disc centered at $z_0$ is contained in a compact disc also centered at $z_0$. So it's sufficient to show uniform convergence on compact discs $K_r(z_0)$ with radius $r$ smaller than the radius of convergence $R$. On such a disc we have
$$\vert a_k(z-z_0)^k\vert=\vert a_k\vert\vert z-z_0\vert^k\leq\vert a_k\vert r^k,$$
and the series with these terms is known to converge because the original power series converges absolutely. Now we can apply the test to show uniform convergence on the compact disc. But it won't generally converge on the entire disc of convergence. For instance, the series expansion of the exponential function doesn't. Neither does the series expansion of $\frac{1}{1-x}$ at $0$.