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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?

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    $\begingroup$ I might just find the divergent series I could compare series $\sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!}$ to. $\frac{1}{2n-3}=\frac{(2n-1)!!}{(2n-1)!!(2n-3)}=\frac{(2n-1)!!}{(2n-3)!!}<\frac{(2n-1)!!}{(2n-3)!!}$ and series $\sum_{n=1}^{\infty}\frac{1}{2n-3}$ is divergent. $\endgroup$
    – purgerica
    Jun 22 '15 at 0:33
  • $\begingroup$ Stirling's approximation will show you that $(2n-1)!!/(2n)!! = (\pi n)^{-1/2}+O(n^{-3/2})$ as $n \to \infty$, so you could use that to show divergence. $\endgroup$
    – Chappers
    Jun 22 '15 at 0:36
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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.

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    $\begingroup$ You probably want to write your last statement differently, e.g. change $O$ to $\Theta$ or just omit the $O$ and change the $=$ sign to $\sim$. (The problem is that $2^{-n} = O\left(n^{-1/2}\right)$ as well, but the sum of $2^{-n}$ converges.) $\endgroup$
    – Jim Belk
    Jun 22 '15 at 0:53
  • $\begingroup$ @jimbelk Thanks! Edited. +1 for your useful comment $\endgroup$
    – Mark Viola
    Jun 22 '15 at 0:56
  • $\begingroup$ It should be $\frac{1}{\sqrt{\pi n}}$ instead of $\sqrt{\frac{\pi}{n}}$ $\endgroup$
    – user3154270
    Jun 13 '19 at 16:50
  • $\begingroup$ @user3154270 Indeed and I have edited accordingly. The conclusion was not impacted, but thank you for the comment. $\endgroup$
    – Mark Viola
    Jun 13 '19 at 23:17
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Hint: $$ \frac{(2n-1)!!}{(2n)!!}\ge\frac{(2n-2)!!}{(2n)!!}=\frac1{2n} $$

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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.

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