It is required to find the values of x for which the series whose terms are given by $$ a_n=\frac{1.3.5...(2n-1)}{2.4.6...(2n)}x^n$$
I’m new to this type of series( I don’t see how radio test or root test can be used here). Any help will be highly appreciated.
Hint:
$$\frac{1.3.5....(2n-1)}{2.4.6 . . .2n}=\frac{n!}{(2^n.n!)^2}=\frac{ 1}{2^{2n}.n!}$$
So if $x<1$ the series converges to 0.