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~
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.