Where do the following sequences converge pointwise and uniformly? Where do the following sequences converge pointwise and uniformly in $\mathbb C$? 
$$1. (nz^n) \hspace{1cm}2.\left(\frac{z^n}{n}\right) \hspace{1cm} 3. \left(\frac{1}{1+nz}\right) \text{where }Re(z) \geq 0$$
I was wondering is there a way to find where do the sequences converge? I only know that we can guess some domain like the unit open disk, then check it by $\epsilon-\delta$ argument, but ie there a method like how to find it not using ''guess''?
Thanks~
 A: You can try to compare with sequences you know. For example the second one can be done using that for $n \in \mathbb{N}^{*}$ and $z=\rho e^{i\theta}$
$$
\left|\frac{z^n}{n}\right| \leq \left|\rho\right|^n=\rho^n
$$
Which is a geometric sequence that you know it converges only if $\rho<1$. For uniform convergence, can you try to find the upper bound of this sequence ?
For the first one, I can give you a simple critera to check the pointwise convergence of sequences that you feel they'll converge to $0$. If a sequence $\left(a_n\right)_{n \in \mathbb{N}}$ is a sequence of strictly positive real number and if 
$$
\frac{a_{n+1}}{a_n}\underset{n \rightarrow +\infty}{\rightarrow}\ell <1
$$
Then the sequence converges to $0$. You can use it for the first one ( you see that it cannot tends to something else than $0$ or $\infty$ ) and for $z\ne0$
$$
\left|\frac{\left(n+1\right)z^{n+1}}{nz^n}\right|=\frac{n+1}{n}\left|z\right| \underset{n \rightarrow +\infty}{\rightarrow}\left|z\right|=\rho
$$
hence if $\rho<1$ it converges to $0$. If $\rho>1$ hence $nz^n>n$ hence it diverges.
A: There isn't much in the way of guessing going on in, say the first one. 
Just note that 
$$
\lim_{n\to \infty}nz^n=\infty
$$
if $|z|\geq 1$ and $0$ if $|z|<1$. So you need to restrict your attention to the disc. Now, as long as you are working on a smaller disc for even just pointwise convergence, say $|z|\leq1-\delta$ for a positive $\delta$ you have
$$
|nz^n|\leq n(1-\delta)^n\to0
$$
and the convergence is uniform. 
If not, you have 
$$
n\sup_{|z|<1}|z|=n\not\to 0
$$
