Uniform convergence of $f_n(z)=nz^n$ in the set $|z|<\frac{1}{2}$

Question

In an exercise I have to prove that $f_n(z)=nz^n$ converges uniformly for $|z|<\frac{1}{2}$.

So I have to prove that:

$$\forall \varepsilon>0, \exists N \in \mathbb{N}:|nz^n-f(z)|<\varepsilon\ \ \ \text{if } n\geq N$$

My question is, how can I find that $f(z)$?

I've tried calculating the $\lim_n nz^n$ but I got stuck. How can I evaluate this limit?

Answer 1

Since you're taking the limit wrt $n$, treat $z$ as a constant. To make things easier while taking powers, consider the polar form: $$\lim_{n\rightarrow\infty}nz^n=\lim_{n\rightarrow\infty}n\underbrace{\left|z\right|^n}_{\leq\frac{1}{2^n}}e^{in\arg(z)}$$

So clearly the magnitude of the terms just goes to zero. There's only one complex number with magnitude zero.

Answer 2

Observation: The point-wise limit is zero as discussed in the comments. [edit: Strictly speaking, we don't need to know this in order to prove that the uniform limit is zero, but it's kind of a helpful clue.]

Now, we will show that there exists a sequence of real numbers $(a_n)$ converging to zero such that $|a_n| \geq |f_n(z)| = |f_n(z)-f(z)|$ for all $n$ and all $|z|<1/2$. It is $a_n := n2^{-n}$ $(n \geq 1)$. Thus, for a given $\varepsilon>0$, for some $N$, we have $\varepsilon > |a_n|$ for all $n \geq N$. So, $\varepsilon > |a_n| \geq |f_n(z)-f(z)|$ for all $|z|<1/2$ and all $n \geq N$. In particular, $\varepsilon > |f_n(z)-f(z)|$ for all $|z|<1/2$ and all $n \geq N$. QE-deedly-D