Convergence of $\{nz^n\}_1^{\infty}.$ 
Discuss completely the convergence and uniform convergence of the sequence $\{nz^n\}_1^{\infty}.$

If $|z|\geq 1$, then $|nz^n|=n|z|^n\geq n$ diverges, so the sequence $nz^n$ also diverges.
If $|z|<1$, it should converge to $0$. So for any $\varepsilon$, we must find $N$ such that $|nz^n|=n|z|^n<\varepsilon$, or in other words $|z|^n<\dfrac{\varepsilon}{n}$ for all $n\geq N$. It should be true since the left hand side converges rapidly to $0$, but how to prove it rigorously?
Then finally, the sequence doesn't converge uniformly in the open disk $|z|<1$, because if it did, for any $\varepsilon$ we must have $N$ such that $|z|^n<\dfrac{\varepsilon}{n}$ for all $|z|<1$ and all $n\geq N$. But we can choose $|z|$ large enough (close enough to $1$) to break this inequality.
So my question is: how to prove that for any $a\in(-1,1)$ and any $\epsilon>0$, there exists $N$ such that $a^n<\dfrac{\varepsilon}{n}$ for all $n\geq N$.

Hint: From real analysis/calculus you may recall the result
$$
\lim_{n\to\infty}\frac{n}{a^n}=0,
$$
whenever the constant $a>1$. An exponential function grows faster than a power function or some catch-phrase like that is sometimes associated with this result.
Fix a constant $a>1$ and consider the numbers $z$ such that $|z|<1/a$.
Remark: You should also prove that the convergence of your sequence is uniform in a closed disk $|z|\le r$, where $r$ is a constant from the interval ...

Reminder: Assume $a>1$, so $a=1+b$ with $b>0$. Then from the binomial theorem you get that
$$
a^n=(1+b)^n=\sum_{k=0}^n{n\choose k}b^k>{n\choose 2}b^2.
$$
Thus
$$
0<\frac{n}{a^n}<\frac{n}{b^2 {n\choose 2}}=\frac{2}{b^2(n-1)}\to0,\ \text{as $n\to\infty.$}.
$$
