Normal convergence implies uniform absolute convergence but not the other way round

Question

How do I show that normal convergence of a series implies uniform and absolute convergence?

So, a series $f_1+f_2+...$ of functions $f_n:D\rightarrow\mathbb{C}, D\subset\mathbb{C}$ is normally convergent if for all $a\in D$ there is a neighborhood $U$ and a sequence $M_n$ of nonnegative real numbers such that $$|f_n(z)|\leq M_n\text{ for all } z\in U\cap D,$$ and $\sum M_n$ converges.
However I have no idea how to use this definition to prove the above.

Answer 1

The absolute convergence of $\sum f_n$ is clear thanks to comparison test.

To prove uniform convergence, it is enough to prove that it is Cauchy uniform. But $$\sup_z\left|\sum_{p}^{q} f_n(z)\right| \leq \sum_{p}^{q} M_n.$$ The series $\sum M_n$ is absolutely convergent by hypothesis, hence $\lim_{p,q\to +\infty} \sum_{p}^{q} M_n =0$.

Answer 2

Absolute Convergence: Let $f_n(z)<M_n$. Then $$0<\sum |f_n(z)|\leq\sum M_n$$ Since the larger sum is bounded on both sides, to prove convergence we need to show that $|f_n(z)|$ goes to zero as $n$ gets large (which prevents the sum from oscillating). But $\sum M_n$ converges, so $M_n\to 0$ and we are done by the squeeze theorem.

Uniform Convergence: $$\left|\sum f_n(z)\right|\leq\sum\left| f_n(z)\right|\leq\sum M_n$$ holds for all $z$ and so it holds for the sup over all $z$ and so the sequence is Cauchy Uniform and we are done.

Not the Converse: I'm not so sure about this direction. It seems like if you are absolutely normally convergent, setting $$M_n=\sup_zf_n(z)$$ should produce a convergent sequence, by using the same $\epsilon$ and $N$ as in the definition of absolute convergence.

Here's a counter example to a partial converse: Consider $f_n(z)=\frac{1}{n^2z}$ This is absolutely convergent since $\sum\frac{1}{n^2}$ converges, but not normally convergent.