if  $$z_n\to 0$$, then $$\frac{1}{n}(z_1+\cdots + z_n)\to 0$$  I'm sorry for asking this one, I've seen it in may places before but I cannot think of a way to solve it, and I don't know how to search it here. Also, I've seen the solution before for real numbers.
I'm self-studying a bit of complex analysis, and I'm attempting to figure out the following.   Suppose $\lim_{n\to\infty}z_n=A$. How can I show that $$ \lim_{n\to\infty}\frac{1}{n}(z_1+\cdots+z_n)=A. $$  Is there a clever way to write the limit to make it more approachable? Thank you.
$\sum\limits_{n=1}^{\infty}\frac{1}{1+z^n}$, $|z|&gt;1$.  There are two facts that my professor uses that I am confused about.  The first is: $|1+z^n| \geq ||z|^n-1|$, I believe this is true for any $|z|$.  The other is: $\frac{1}{|z|^n-1} \leq \frac{2}{|z|^n}$, I believe this is also true for any $|z|$.  Can anyone prove these statements for me?