TITLE: Determine convergence of the series $\sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!}$
QUESTION [9 upvotes]: I tried to use D'Alambert theorem to determine convergence of the series $\sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!}$.
$$\lim_{n \to \infty} \frac{a_{n+1}}{a_{n}} = \lim_{n \to \infty} \frac{(2n-1)!!(2n+1)(2n)!!}{(2n-1)!!(2n)!!(2n+2)} = \lim_{n \to \infty} \frac{2n+1}{2n+2} = 1$$
but this test is inconclusive.
I think a comparison test might give a result, but with which series should I compare it to?

REPLY [5 votes]: 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.