Find the region of convergence

Question

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:

$$a_p = \left\{ \begin{array}{ll} \frac{(-1)^p}{p} & \mbox{if } p=n(n+1) \\ 0 & \mbox{in other case} \end{array} \right.$$

$$\overline{\lim}_{n \to \infty} \sqrt[p(p+1)]{ \left\lvert \frac{(-1)^p}{p} \right\rvert }= \overline{\lim}_{n \to \infty} \sqrt[p(p+1)]{ \frac{1}{p} }= \overline{\lim}_{n \to \infty} \frac {1}{\sqrt[p(p+1)]{ p }}=0$$

Answer 1

Note that we have directly

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

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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

$$a_p=\begin{cases} \frac{(-1)^p}{p}&p=n(n+1)\\\\ 0&,\text{otherwise} \end{cases}$$

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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 $|z|>1$.

Answer 2

1. You are right.
2. Yes, the radius of convergence is $0$. This means that it converges if and only if $z=0$.
3. 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 $1$.