When is $\sum_{n \ge 0} g_n(z)$ analytic? Let $D$ be an open subset of $\mathbb{C}$ where $g_n(z)$, $n \in\mathbb{N}$ are analytic. Then $$f(z)=\sum_{n \ge 0}g_n(z)$$
is analytic on $D$ iff $\sum_{n\ge 0}g_n(z)$ is locally uniformly convergent. 
Firstly, what is the proof of this, and is there a way of intuiting why pointwise convergence is not strong enough to imply $f$'s analyticity, and why (non-local) uniform convergence is not required?
 A: I'm sorry for the discussions in the comments. They are not helpful, so let me try to make my point clear. You asked for intuition on
a) why uniform convergence is needed,
b) why it is sufficient, and
c) why it is only needed locally.
The hardest point here is b), and I'm not sure if I can explain this entirely intuitionally, so let's do the easy stuff first.
Ad a), pointwise convergence only is never good enough to preserve continuity; there are always series of arbitrarily nice (non constant) functions converging pointwise to a discontinuous function. On the other hand, uniform convergence will always results in a continuous function.
Ad c), as I tried to point out in my comment, analyticity is a local property by definition. A function is analytic if it is analytic in any point, i.e. if any point has an open neighbourhood on which the function is a convergent power series.
Therefore, any condition that is good enough to make sure a series converges to an analytic function will work locally too.
For a), finally, we can observe the wonderful miracle of function theory: holomorphic functions are analytic, roughly because their derivatives are again holomorphic, which is pretty strong, but the conditions imposed on their differentiability are not to strong. We only need a continuous function that fulfills a certain condition, where the occurring terms are compatible with uniform convergence.
I'm thinking of Morera's theorem, of course, that is, a continuous function, defined on some open domain in $\mathbb{C}$, is holomorphic on this domain if its contour integral over any closed curve in this domain vanishes.
Now, one is encouraged to gain an intuitive understanding of what it means for a contour integral to vanish. If you know some physics, you could first convince yourself of the relationship between the (converse of the) above statement and Green's theorem and try to combine it with the physical idea behind the latter.
Unfortunately, I don't know this stuff good enough to give good further explanation of this.
(Anyone more familiar with this kind of interpretation may very well add extra information or references where this is explained. Therefore, I made this post community wiki.)
A: Can't help you much on the intuition, but I know one proof that $f$ is analytic using Cauchy's theorem. 
Let $f_n = \sum_n g_n$ and pick a point $\alpha \in {\rm Int}(\Gamma)$ where $\Gamma$ is a compact region where $f_n$ converges uniformly. Since $f_n$ is analytic 
$$f_n(\alpha) = \frac{1}{2\pi i}\int_\gamma \frac{f_n(z) dz}{z-\alpha}$$
where $\gamma$ is the boundary of $\Gamma$. To show that $f$ is analytic in $\alpha$ we only need to show that $f(\alpha) = I$ where
$$I =  \frac{1}{2\pi i}\int_\gamma \frac{f(z) dz}{z-\alpha}$$
But this is easy as
$$|I - f(\alpha)| \leq |I - f_n(\alpha)| + |f_n(\alpha) - f(\alpha)|\\=\frac{1}{2\pi i}\int_\gamma \frac{|f_n(z)-f(\alpha)| dz}{z-\alpha} + |f_n(\alpha) - f(\alpha)| \leq  2|f_n - f|_\infty \to 0 {\rm ~~for~~} n\to \infty$$
by the uniform convergence of $f_n$.
