MHB Is the Double Factorial Series Convergent with Stirling's Asymptotic Formula?

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Hi, question is - show that the following series is convergent: $ \sum_{s}^{} \frac{(2s-1)!}{(2s)!(2s+1)}$

Hint: Stirlings asymptotic formula - which I find is : $n! = \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n $

I can see how this formula would simplify - but can't see how it relates to the double factorial !
 
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Um, not sure why no replies, always feel free to tell me bluntly if I must do something else or different...

I know of course that (2s-1)! = (2s-1)(2s-3)(2s-5) ...3.1 but cannot see how Stirling's formula helps or even relates...

Liebnitz' criteria requires L'Hospital and I don't know how to differentiate a dbl factorial. Tried with Wolfram and it returns a complex series that isn't going to help. I also don't know how to integrate a dbl factorial, so the integral test won't help.

I tried the ratio test (the dbl factorials simplified nicely) but got L=1, i.e. inconclusive.

I tried expanding a few terms, got -7/8, 11/16, -225/64, 133/256 ... and don't see a partial sum formula emerging from that.

Any suggestions?
 
Hint:

Convert the double factorial to a regular factorial.
 
ognik said:
Hi, question is - show that the following series is convergent: $ \sum_{s}^{} \frac{(2s-1)!}{(2s)!(2s+1)}$

Hint: Stirlings asymptotic formula - which I find is : $n! = \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n $

I can see how this formula would simplify - but can't see how it relates to the double factorial !
As well as the Stirling approximation formula for $n!$, there is a less well known Stirling formula for the double factorial, namely $$n! \approx \Bigl(\frac2\pi\Bigr)^{\frac14(1-\cos(n\pi))}\sqrt\pi n^{(n+1)/2}e^{-n/2}$$ (see Double factorial: Introduction to the factorials and binomials). That might perhaps be helpful here.
 
If the hint didn't help you might want to see Example 4.5 in https://zaidalyafeai.files.wordpress.com/2015/09/advanced-integration-techniques.pdf. There I do a conversion between double factorial and a regular factorial.
 
ognik said:
Hi folks, thanks for all help. Once I had that suggestion, I found some useful identities here - Double Factorial -- from Wolfram MathWorld - that sorted me out.

Please post your solution if you have time. This will help future thread readers.
 

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