I want to prove the following:
The power series $\sum_{m=0}^{ \infty}{n (z^n)}$ does not converge on any point of the unit circle.
The power series $\sum_{m=0}^{ \infty} {z^n/(n^2)}$ converges at every point of the unit circle.
The power series $\sum_{m=0}^{ \infty} {z^n/ n}$ converges at every point of the unit circle except z=1.
Shall I use radius of convergence to prove those statements?
I know that for a holomorphic function $f$ whose power series has coefficient $a_n$ is given as
$$\frac{1}{R}= \lim_{x \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$$
But if I use that then for the first power series we get R=1 so the disk of convergence is $|z| \le 1$. So how can I prove that it does not converge on any point of circle? Here circle means I guess boundary points of disk i.e $|z|=1$.Please help
Best Answer
1. $\sum nz^n$
Since $nz^n\to \infty$ if $|z|=1$, then $\sum nz^n$ does not converges in any point of the unit circle.
2. $\sum z^n/n^2$
Since $|z^n/n^2|=1/n^2$ for all $|z|=1$, then $\sum z^n/n^2$ converges at every point in the unit circle as $\sum 1/n^2$ does ($p$-series $p=2$.)
3. $\sum z^n/n$. You can see Martin Argerami's answer here. I reproduce it here.
If $z=1$ then $\sum z^n/n=\sum 1/n$ is divergent (harmonic series). If $|z|=1$ and $z \neq 1,$ write $z=e^{2\pi i t}$ with $t \in (0, 1)$ and apply Dirichlet's test: if $ \{a_{n}\}$ is a sequence of real numbers and $\{b_{n}\}$ a sequence of complex numbers satisfying
then $\sum a_nb_n$ converges. Let $a_n=1/n$, so $a_n$ satisfies $a_{n+1} \leq a_n$ and $\lim_{n\to\infty}a_n=0$. Let $b_n=e^{2\pi i nt}$, then $$\left|\sum_{n=1}^Nb_n\right|= \left|\sum_{n=1}^Ne^{2\pi i n t}\right|=\left|\frac{e^{2\pi i t}-e^{2\pi i (N+1)t}}{1-e^{2\pi i t}}\right|\leq\frac{2}{|1-e^{2\pi i t}|}=M \hspace{0.4cm} \text{for all} \hspace{0.4cm} N\in\mathbb{N}.$$ Thus $\sum a_nb_n=\sum z^n/n$ converges for every point in the unit circle except $z=1.$