I want to prove the radius of convergence for the following is$~{1\over 3}~$
$${3\over\sqrt{1-9x^2}}\tag{1}$$
By the advice from@Kavi Rama Murthy, I thought the following.
$$f(y):={1\over\sqrt{1+y}}=\sum_{i=0}^{\infty}{f^{(i)}(0)\over i!}y^{i}\tag{2}$$
$$a_{i}:={f^{(i)}(0)\over i!}={(-1)^{i}\over 2^i(i!)}\prod_{j=1}^{i}\left(2j-1\right)\tag{3}$$
Using d'alembert's principle with above coefficient, the radius of convergence of$~(1+y)^{-1/2}~$can be determined as$~1~$which is proven by following.
$$\lim_{i\to\infty}\left|{a_{i+1}\over a_{i}}\right|\tag{4}$$
$$=\lim_{i\to\infty}\left|{(-1)^{i+1}\displaystyle\prod_{j=1}^{i+1}(2j-1)\over 2^{i+1}(i+1)!}{2^{i}i!\over(-1)^{i}\displaystyle\prod_{j=1}^{i}(2j-1)}\right|\tag{5}$$
$$=\lim_{i\to\infty}\left|{(-1)(2(i+1)-1)\over 2(i+1)}\right|\tag{6}$$
$$=\lim_{i\to\infty}\left|{-(2i+2-1)\over 2(i+1)}\right|\tag{7}$$
$$=\lim_{i\to\infty}\left|{-(2i+1)\over 2(i+1)}\right|\tag{8}$$
$$=\lim_{i\to\infty}{(2i+1)\over 2(i+1)}\tag{9}$$
$$=\lim_{i\to\infty}{2\over 2}=1\tag{10}$$
How can I proceed from here?
I made the path for the solution in my own though I have no confidence for it so far.
$$g(x):={3\over\sqrt{1-9x^2}}\tag{1}$$
$$f(y):={1\over\sqrt{1+y}}=\sum_{i=0}^{\infty}{f^{(i)}(0)\over i!}y^{i}\tag{2}$$
$$a_{i}:={f^{(i)}(0)\over i!}={(-1)^{i}\over 2^i(i!)}\prod_{j=1}^{i}\left(2j-1\right)\tag{3}$$
$${1\over\sqrt{1+y}}=\sum_{i=0}^{\infty}\left({(-1)^{i}\over 2^i(i!)}\left(\prod_{j=1}^{i}\left(2j-1\right)\right)y^i\right)\tag{4}$$
$$\left|y\right|<1~\text{must be held to make summation finite}\tag{5}$$
$$\text{This}~1~\text{is a radius of convergence}\tag{6}$$
This statement is proven using D'Alembert's Ratio Test.
$$\underbrace{y=-9x^2}_{\text{substitution}}\tag{7}$$
$${1\over\sqrt{1-9x^2}}=\sum_{i=0}^{\infty}\left({(-1)^{i}\over 2^i(i!)}\left(\prod_{j=1}^{i}\left(2j-1\right)\right)\left(-9x^2\right)^i\right)\tag{8}$$
$$\left|-9x^2\right|<1\tag{9}$$
$$\left|9x^2\right|<1\tag{10}$$
$$9x^2<1\tag{11}$$
$$x^2<{1\over 9}~~\Leftrightarrow~~x<\left|{1\over 3}\right|\tag{12}$$
$$\therefore~~{1\over 3}~~\text{is a radius of convergence of}~{1\over\sqrt{1-9x^2}}\tag{13}$$
$$\therefore~~{1\over 3}~~\text{is also a radius of convergence of}~{3\over\sqrt{1-9x^2}}\tag{14}$$