Proving that $f_n(z)=nz^n$ converges uniformly for $|z|<\frac{1}{2}$.

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:

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

I calculated $\lim_n nz^n=0$, so we basically need to prove that:

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

My approach:

we can write $z=re^{i\varphi}$

then we have that $|nz^n|<\varepsilon$ is equivalent to $nr^n<\varepsilon$ and we have that $r<\frac{1}{2}$ but I got stuck here. How can I procede?

Answer 1

Hint : show for any $r\in\mathbb{R}$ s.t. $0\leq r\leq 1/2$, for any $\epsilon>0$, there exists $N\in\mathbb{N}$ s.t. $\forall n\geq N$, $nr^n\leq \epsilon$. Then apply the result to the complex case, with $|e^{i\varphi}|=1$.

Edit : If you don't know how to show $\lim_{n\rightarrow \infty}nr^n=0$, then there are several ways to show. Most convenient way in this case will be the ratio test : $$\lim_{n\rightarrow\infty}\bigg|\frac{(n+1)r^{n+1}}{nr^n}\bigg|=\lim_{n\rightarrow\infty}\frac{n+1}{n}r=r<1,$$ therefore the series is absolutely convergent. You don't need to find explicit $N$ (of course you can though) to show the result. I am sure if you learned introduction to analysis courses then you learned some convergence tests. It is useful tool even later on, so you better be able to use some of them!

Answer 2

We have the estimation for the uniform norm between $f_n$ and the (to be uniform) limit for $n>2$: $$ \begin{aligned} \|f_n-0\| &=\max_{|z|\le 1/2}|nz^n|\\ &=\frac n{2^n}=\frac n{(1+1)^n}=\frac n{1+n+\frac 12n(n-1)+\dots}\\ &<\frac n{\frac 12n(n-1)}=\frac 2{n-1}\ . \end{aligned} $$ So $\|f_n-0\|\to 0$.