Convergence of the alternating series $\sum_{n=1}^\infty \frac{(-1)^n(2n)!!}{(2n+1)!!}$

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.

1 Answer

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

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

• I see, however, is there any more elementary way to solve this? – Peter Liu Mar 18 '16 at 1:29
• 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 $-1\leq x\leq1$. – Marco Cantarini Mar 18 '16 at 8:29
• Thanks. Do you have some links for this equality? – Peter Liu Mar 19 '16 at 14:31