Find the region of convergence Find the region of convergence
$$i) \sum_{n=1}^{\infty} \frac{(-1)^nz^{2n-1}}{(2n-1)!} $$
$$ii) \sum_{n=1}^{\infty} n!z^n $$
$$iii)\sum_{n=1}^{\infty} \frac{(-1)^nz^{n(n+1)}}{n} $$
I got the following:
For (i) I obtained that the convergence region is all $\mathbb{C}$.
For (ii) I obtained that the radius of convergence is $0$, does this mean that it only converges around the point $a = 0$?.
For (iii) I'm not sure of the following result:
$$a_p =
\left\{
 \begin{array}{ll}
  \frac{(-1)^p}{p}  & \mbox{if } p=n(n+1) \\
  0 & \mbox{in other case} 
 \end{array}
\right.$$
$$\overline{\lim}_{n \to \infty} \sqrt[p(p+1)]{ \left\lvert \frac{(-1)^p}{p} \right\rvert }= \overline{\lim}_{n \to \infty} \sqrt[p(p+1)]{ \frac{1}{p}  }= \overline{\lim}_{n \to \infty} \frac {1}{\sqrt[p(p+1)]{ p  }}=0$$
 A: *

*You are right.

*Yes, the radius of convergence is $0$. This means that it converges if and only if $z=0$.

*If $z\ne0$, then\begin{align}\lim_{n\to\infty}\frac{\left|\frac{(-1)^{n+1}z^{(n+1)(n+2)}}{n+1}\right|}{\left|\frac{(-1)^nz^{n(n+1)}}n\right|}&=\lim_{n\to\infty}\frac n{n+1}|z|^{2n+2}\\&=\begin{cases}\infty&\text{ if }|z|>1\\1&\text{ if }|z|=1\\0&\text{ if }|z|<1\end{cases}\end{align}and therefore the radius of convergence of your series is $1$.

A: Note that we have directly
$$\limsup_{n\to\infty} \sqrt[n]{\left|\frac{(-1)^n z^{n(n+1)}}{n} \right|}=\begin{cases}\infty &,|z|>1\\\\1&,|z|=1\\\\
0&,|z|<1\end{cases}$$


Another way to proceed is to note that the series of interest is indeed a power series given by
$$\begin{align}
\sum_{n=1}^\infty \frac{(-1)^nz^{n(n+1)}}{n}&=0z^0+0z^1+\frac{(-1)^1}{1}z^2+0z^3+0z^4+0z^5+\frac{(-1)^2}{2}z^6+\cdots\\\\
&=\sum_{p=1}^\infty a_p z^p\tag1
\end{align}$$
where the coefficient $a_p$ is given by
$$a_p=\begin{cases}
\frac{(-1)^p}{p}&p=n(n+1)\\\\
0&,\text{otherwise}
\end{cases}$$

Then, the series in $(1)$ has a radius of convergence, $R$
$$R=\frac1{\limsup_{p\to\infty}\sqrt[p]{\left|a_p\right|}}=1$$
Hence, the series converges for $|z|<1$ and diverges for $|z|>1$.
