Q) The radius of convergence of the power series$$\sum_{m=0}^{\infty} \frac{(3m)!}{(m!)^3}x^{3m}$$ is: _____________
My Approach :- Power series $$f(x) = \sum_{n=0}^{\infty }c_{n}(x-a)^{n}$$ converges if :
$$ \lim_{n\rightarrow \infty }\left | \frac{c_{n+1}*(x-a)^{n+1}}{c_{n}*x^n} \right | < 1$$
Here, power series is defined as $$\sum_{m=0}^{\infty }\frac{(3m)!}{(m!)^3}\;x^{3m}$$
So, given power series converges if :
$$\lim_{m\rightarrow \infty } \left | \frac{\frac{((3(m+1))!)*x^{(3(m+1))}}{((m+1)!)^3}}{\frac{((3m)!)*x^{3m}}{(m!)^3}} \right | < 1$$
$$\Rightarrow \lim_{m\rightarrow \infty } \left | \frac{\frac{((3m+3)!)*x^{3m}x^3}{((m+1)(m!))^3}}{\frac{((3m)!)*x^{3m}}{(m!)^3}} \right | < 1$$
$$\Rightarrow \lim_{m\rightarrow \infty } \left | \frac{(3m+3)(3m+2)(3m+1)((3m)!)x^{3m}x^3 (m!)^3}{(m+1)^3(m!)^3 ((3m)!)x^{3m} } \right | < 1$$
$$\Rightarrow \lim_{m\rightarrow \infty } \left | \frac{(3m+3)(3m+2)(3m+1)x^3 }{(m+1)^3 } \right | < 1$$
On Dividing by $m^3$ in both numerator and denominator :
$$\Rightarrow \lim_{m\rightarrow \infty } \left | \frac{(3 + \frac{3}{m})(3+ \frac{2}{m})(3+\frac{1}{m})x^3 }{(1+\frac{1}{m})(1+\frac{1}{m})(1+\frac{1}{m})} \right | < 1$$
$$\Rightarrow | (3x)^3| < 1 \\ \Rightarrow -1 < (3x)^3 < 1$$
$$\Rightarrow (-1)^{1/3} < 3x < 1^{1/3}$$
$$\Rightarrow \frac{(-1)^{1/3}}{3} < x < \frac{1^{1/3}}{3}$$
So, Radius of Convergence will be $\frac{1^{1/3}}{3}$
Please tell me whether it is correct or not and if it is wrong then please correct it.
An alternate way:
By applying Stirling's Formula
$$ n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^n $$ we have:
\begin{align} \left(\frac{(3m)!}{(m!)^3}\right)^{1/3m}&\sim\left(\frac{\sqrt{3m}\ 3^{3m}}{\sqrt{m}}\right)^{1/3m}\sim 3 \end{align} Hence the radius of convergence is $\frac{1}{3}$.