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I want to find out whether the series $\sum_{n=1}^\infty \frac{(-1)^n(2n)!!}{(2n+1)!!}$ convergent and I know the alternating series test. However, I don't know whether the absolute term converges to 0 or not. I already show that it is not absolutely convergent. Thanks.

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One may observe that, as $n \to \infty$, by the use of the Stirling formula, $$ \frac{(2n)!!}{(2n+1)!!}=\frac{2^n\cdot n!}{\frac{(2n+1)!}{2^n n!}}=\frac{2^{2n} (n!)^2}{(2n+1)!}\sim \frac{\sqrt{\pi}}2\frac1{\sqrt{n}} $$ and the initial series is not absolutely convergent.

In fact, as $n \to \infty$, by using the asymptotic expansion, $$ n! = \sqrt{2 \pi}n^{n+1/2} e^{-n} \left( 1 + O\left(\frac1n\right)\right) $$ we get

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

the initial series is convergent being the sum of two convergent series.

One may prove that

$$ \sum_{n\geq1}(-1)^n\frac{(2n)!!}{(2n+1)!!}=\frac{\sqrt{2}}2\log \left(1+\sqrt{2}\right)-1. $$

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  • $\begingroup$ I see, however, is there any more elementary way to solve this? $\endgroup$ – Peter Liu Mar 18 '16 at 1:29
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    $\begingroup$ Maybe it's interesting to note that holds $$\frac{\sinh^{-1}\left(x\right)}{\sqrt{1+x^{2}}}=\sum_{n\geq0}(-1)^{n}\frac{(2n)!!}{(2n+1)!!}x^{2n+1}$$ for $-1\leq x\leq1$. $\endgroup$ – Marco Cantarini Mar 18 '16 at 8:29
  • $\begingroup$ Thanks. Do you have some links for this equality? $\endgroup$ – Peter Liu Mar 19 '16 at 14:31

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