Questions about the powerseries $\sum_{n=0}^{\infty}\frac{1}{n^d}z^n,d\in\mathbb{R},z\in\mathbb{C}$

Question

The convergenceradius is $1$ that means if $|z|= 1$ then the convergencebehaviour depends on $d$.

I have several questions about this issue:

1.Why is $\sum_{n=0}^{\infty}\frac{1}{n^d}$ absolutely convergent for $d>1$ and $|z|=1$

2.Why is the implication true:

series absolutely divergent $\Rightarrow$ series is divergent

3.Why is $\sum_{n=0}^{\infty}\frac{1}{n^d}z^n$ divergent for $d\leq 0$ and $|z|=1$

4.Why does the convergencebehaviour depends on $z\in\{a\in\mathbb{C}||a|=1\}$ for $0<d\leq 1$?

I know that if $d$ is for example $1$ then $\sum_{n=0}^{\infty}\frac{1}{n^d}z^n$ converges absolutely for z = $-1$ and diverges for z = $1$. But the statement above says we can go with $d$ as close to 0 as we want and we will still find a $z$ with $|z|=1$ for which $\sum_{n=0}^{\infty}\frac{1}{n^d}z^n$ converges can somebody please show me the proof for all those questions I have?

Thank you!

Answer 1

1. Apply integral test. (Thanks to T. Bongers for pointing out an error I made earlier).
2. False. $\sum \frac {(-1)^{n}} n$ is absolutely divergent but not divergent.
3. The general term does not tend to $0$ so the series is divergent.
4. For $0<d<1$ the series diverges absolutely whenever $|z|=1$. [Compare with $\sum \frac 1 n$]. The series converges for $z=-1$ and diverges for $z=1$. For $z=-1$ use Alternating Series Test.