Find the domain of convergence of the series :
$(1)$ $$\sum\limits_{n=1}^{\infty}\frac{1\cdot3\cdot5\cdots(2n-1)}{n!}\Big(\frac{1-z}{z}\Big)^n$$
$(2)$ $$\frac{1}{2}z+\frac{1\cdot3}{2\cdot5}z^2+\frac{1\cdot3\cdot5}{2\cdot5\cdot8}z^3 +\dots$$
I was reading Rudin and I know the following theorems
$\star $ For any sequence $\{c_n\}$ of positive numbers, $$\lim_{n\to\infty} \inf \frac{c_{n+1}}{c_n} \leq \lim_{n\to\infty} \inf \sqrt[n]{c_n},$$ $$ \lim_{n\to\infty} \sup \sqrt[n]{c_n} \leq \lim_{n\to\infty} \sup \frac{c_{n+1}}{c_n}.$$
$\star $ Cauchy-Hadamard Theorem
My Attempt :
$(1)$ If $c_n = \frac{1\cdot3\cdot5\cdots(2n-1)}{n!},$ then $\lim_{n\to \infty}\frac{c_{n+1}}{c_n}=\lim_{n\to \infty}\frac{2n+1}{n+1}=2$
Hence the series converges if $|\frac{1-z}{z}|<1/2$ or if $|1-z|<|z|/2$ i.e., $|z|<3/2.$
$(2)$ Here similarly $\lim_{n\to \infty}\frac{c_{n+1}}{c_n}=\lim_{n\to \infty}\frac{2n+1}{3n-1}=2/3$
Hence series converges if $|z|<3/2.$
Is this correct? $
Let us first write
$$\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{n!}=\frac{1\cdot2\cdot3\cdot4\cdot\ldots\cdot(2n)}{2^nn!n!}=\frac{(2n)!}{2^n(n!)^2}$$
and now apply the quotient test:
$$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{(2n+2)!}{2^{n+1}\left((n+1)!\right)^2}\frac{(1-z)^{n+1}}{z^{n+1}}\cdot\frac{2^n(n!)^2}{(2n)!}\frac{z^n}{(1-z)^n}\right|=$$
$$=\frac{(2n+1)(2n+2)}{2(n+1)^2}\left|\frac{1-z}z\right|\xrightarrow[n\to\infty]{}2\left|\frac{1-z}z\right|\stackrel?<1\iff\left|\frac z{1-z}\right|>2$$
and putting $\;z=x+iy\;$ , the last inequality is the same as
$$x^2+y^2>4\left[(x-1)^2+y^2\right]\implies 3x^2-8x+4+3y^2<0\implies$$
$$3\left(x-\frac43\right)^2+3y^2<\frac43\iff \left|z-\frac43\right|<\frac23\;,\;\;z\neq1\; $$
which is almost an open disk of radius $\;\frac23\;$ around $\;\left(\frac43,0\right)\;$ on the complex plane.