Finding radius of convergence of following series $$f(x)=\sum^{\infty}_{n=0}\frac{4^n\cdot n!}{(2n+1)}x^{2n+1}$$
What i try : let $\displaystyle a_{n}=\frac{4^n\cdot n!}{(2n+1)}x^{2n+1}$. Then $\displaystyle a_{n+1}=\frac{4^{n+1}\cdot (n+1)!}{(2n+3)}x^{2n+3}.$
So $$\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_{n}}\bigg|<1$$
$$\lim_{n\rightarrow \infty}4|x^2|\bigg|\frac{(n+1)(2n+1)}{(2n+3)}\bigg|<1$$
Here the left side quantity is $\infty$
I did not understand How can i find radius of convergence. Help me please
