Determine convergence of the series $\sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!}$

Recall that

$$(2n-1)!!=\frac{(2n)!}{2^n\,n!}$$

and

$$(2n)!!=2^n\,n!$$

Thus, the ratio is

$$\frac{(2n)!}{4^n\,(n!)^2}$$

Now use Stirling's formula

$$n!=\sqrt{2\pi n}(n/e)^n\left(1+O\left(\frac{1}{n}\right)\right)$$

to find that

$$\frac{(2n-1)!!}{(2n)!!}=\sqrt{\frac{1}{\pi n}}+O\left(n^{-3/2}\right)$$

which shows that the series diverges.

Hint: $$ \frac{(2n-1)!!}{(2n)!!}\ge\frac{(2n-2)!!}{(2n)!!}=\frac1{2n} $$

Another approach can be this. We have, by Taylor series, $$\arcsin\left(x\right)=\sum_{n\geq0}\frac{\left(2n\right)!}{4^{n}\left(n!\right)^{2}\left(2n+1\right)}x^{2n+1} $$ with $\left|x\right|<1 $, then if we take the derivative $$\frac{1}{\sqrt{1-x^{2}}}=\sum_{n\geq0}\frac{\left(2n\right)!}{4^{n}\left(n!\right)^{2}}x^{2n} $$ and now it is clear that if you take the limit as $x\rightarrow1 $ the LHS goes to infinity, so the series diverges.