In what set of complex numbers does these series $(a_n)_{n=1}^{\infty}$ convergence/divergence? If series convergences is the convergence steady (uniform?) within the convergence area.
$(a)$ $a_n=\frac{(-1)^n z}{n}$
\begin{align}
r&=\lim_{n\to\infty}\Bigg| \frac{\frac{(-1)^{n+1}}{n+1}}{\frac{(-1)^n}{n}}\Bigg| \\
&=\lim_{n\to\infty}\Bigg| \frac{(-1)^n (-1)^1}{n+1} \frac{n}{(-1)^n}\Bigg| \\
&=\lim_{n\to\infty}\Bigg|\frac{-n}{n+1}\Bigg|=|-1|=1
\end{align}
So it converges for $\forall z \in \mathbb{C}$ when $|z|<1$. Doesn't this mean it diverges if $|z|>1$?
$(b)$ $a_n=nz^n$
\begin{align}
r=&\lim_{n\to\infty}\Bigg| \frac{(n+1)z^{n+1}}{nz^n} \Bigg| \\
&=\lim_{n\to\infty}\Bigg| \frac{(n+1)z^nz}{nz^n} \Bigg| \\
&=\lim_{n\to\infty}\Bigg| \frac{nz+z}{n} \Bigg| \\
&=\lim_{n\to\infty}\Bigg| z+\frac zn \Bigg| \\
&=|z|
\end{align}
Converges when $|z|<1$ for all $z\in\mathbb{C}$.
$(c)$ $a_n=\frac{z^n}{n}$
\begin{align}
r&=\lim_{n\to\infty}\frac{\sqrt[n]{|z^n|}}{\sqrt[n]{n}} \\
&=\frac{|z|}{1}=|z|
\end{align}
Converges when $|z|<1$ for all $z\in\mathbb{C}$.
Best Answer
$$\lim_{n\to\infty }\left|\frac{a_{n+1}}{a_n}\right|=1$$ where $a_n=\frac{(-1)^n}{n}$. Then the serie converge for all $z\in\mathbb C$ such that $|z|<1$.