Showing that $\frac{z^n}{n^2+z^n}$ converges uniformly I am trying to prove that $f_n(z)=\frac{z^n}{n^2+z^n}$ converges uniformly for $|z|<1$. This would mean that:
$$\forall \varepsilon>0, \exists p \in \mathbb{N}:\left|\frac{z^n}{n^2+z^n}-f(z)\right|<\varepsilon\ \ \ \text{if } n\geq p$$
Where $f(z) = \lim_n \frac{z^n}{n^2+z^n}$. I already calculated this limit and got 0. This means that in order to prove that this converges uniformly we just need to show that:
$$\forall \varepsilon>0, \exists p \in \mathbb{N}:\left|\frac{z^n}{n^2+z^n}\right|<\varepsilon\ \ \ \text{if } n\geq p$$
The thing is that I don't know how to continue the proof from this point on. How can I prove this?

If $n>1$ and $|z|<1$, then$$\left|\frac{z^n}{n^2+z^n}\right|\leqslant\frac1{n^2-1},$$since $|z^n|<1$ and $|n^2+z^n|\geqslant n^2-|z^n|>n^2-1$. So, for each $\varepsilon>0$, take $N\in\Bbb N$ such that $n\geqslant N\implies\frac1{n^2-1}<\varepsilon$; such a $N$ exists, since $\lim_{n\to\infty}\frac1{n^2-1}=0$. So, if $n\geqslant N$ and $|z|<1$, you have$$\left|\frac{z^n}{n^2+z^n}\right|<\varepsilon.$$
