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 $n\geq N$.

One way is to look at the ratio of terms
$$\left|\frac{a_{n+1}}{a_n} \right|= \frac{n+1}n |z|,$$
which goes to $|z|$ as $n\to \infty$. Eventually $|\frac{a_{n+1}}{a_n}|$ will be $<|z|+\epsilon<1$, whence continued multiplication by a number $<1$ will get the terms as small as you like:
$$a_{n+M} < a_nr^M,$$
where $r = |z|+\epsilon <1$
