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Study the convergence of the next series: $$\sum_{n=1}^\infty \frac{(2n)!!}{(2n+1)!!} $$

My solution: since $$\frac{(2n)!!}{(2n+2)!!} \leq \frac{(2n)!!}{(2n+1)!!}$$ forall $n \in \mathbb{N}$ and since $$\sum_{n=1}^\infty \frac{(2n)!!}{(2n+2)!!} = \sum_{n=1}^\infty \frac{1}{2n+2} = \infty$$ then the first series diverges. Is there anything wrong? Thanks in advance.

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Looks fine to me. – Brian M. Scott Jan 5 '13 at 18:56
That's all correct! I've voted to close rather than posting an answer because I don't think the nature of the question will give rise to many useful answers. (Not that it's a bad question, of course!) – Clive Newstead Jan 5 '13 at 18:56
@Clive Newstead: why close? To me, this site also serves as a reference manual for solutions. It is unlikely someone would google for this precise question successfully, but you never know. – gnometorule Jan 5 '13 at 19:11
It looks nice and straightforward: good job. – DonAntonio Jan 5 '13 at 19:16
@gnometorule: It will still exist and be searchable on Google (or even the search function here). Closure just means new answers can't be posted. – Clive Newstead Jan 5 '13 at 19:41
up vote 1 down vote accepted

Yes. What you have done is correct. You can attempt to quantify the divergence a bit better. Note that $(2n)!! = 2^n n!$ and $(2n+1)!! = \dfrac{(2n+2)!}{2^{n+1} (n+1)!}$. Hence, $$\dfrac{(2n)!!}{(2n+1)!!} = 2^{2n+1} \dfrac{(n+1)! n!}{(2n+2)!} = \dfrac{2^{2n+1}}{\dbinom{2n+2}{n+1} \cdot (n+1)}$$ We have that $\dbinom{2n+2}{n+1} \sim \dfrac{4^{n+1}}{ \sqrt{\pi(n+1)}}$. Hence, $$\dfrac{2^{2n+1}}{\dbinom{2n+2}{n+1} \cdot (n+1)} \sim \dfrac{2^{2n+1} \sqrt{\pi}}{4^{n+1} \sqrt{n+1}} = \dfrac{\sqrt{\pi}}2 \cdot \dfrac1{\sqrt{n+1}}$$ Hence, $$\sum_{n=1}^{\infty}\dfrac{(2n)!!}{(2n+1)!!} \,\,\,\,\,\,\,\,\text{diverges as} \,\,\,\,\,\, \dfrac{\sqrt{\pi}}2 \cdot \sum_{n=1}^{\infty}\dfrac1{\sqrt{n+1}}$$

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