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Find the limit

Alexmahone

Active member
Jan 26, 2012
268
Find $\displaystyle\lim_{n\to\infty}\frac{(2n)!!}{(2n-1)!!}$.

Is the answer $\displaystyle\lim_{n\to\infty} \frac{2}{1}\cdot\frac{4}{3}\cdot\frac{6}{5}\cdots \frac{2n}{2n-1}=\infty$?
 
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chisigma

Well-known member
Feb 13, 2012
1,704
Find $\displaystyle\lim_{n\to\infty}\frac{(2n)!!}{(2n-1)!!}$.
Is...

$\displaystyle \frac{(2n)!!}{(2n-1)!!}= \prod_{k=0}^{n} (1+ \frac{1}{2k+1})$ (1)

... and because the series $\displaystyle \sum_{k=0}^{\infty} \frac{1}{2 k+1}$ diverges, also the 'infinite product diverges and is...

$\displaystyle \lim_{n \rightarrow \infty} \frac{(2n)!!}{(2n-1)!!}= \prod_{k=0}^{\infty} (1+ \frac{1}{2k+1})= + \infty $ (2)

Kind regards

$\chi$ $\sigma$
 
Jan 31, 2012
54
Find $\displaystyle\lim_{n\to\infty}\frac{(2n)!!}{(2n-1)!!}$.

Is the answer $\displaystyle\lim_{n\to\infty} \frac{2}{1}\cdot\frac{4}{3}\cdot\frac{6}{5}\cdots \frac{2n}{2n-1}=\infty$?

Maybe that will help you:


$$\frac{(2n)!!}{(2n-1)!!}=\frac{((2n)!!)^2}{(2n)!}=\frac{2^{2n}(n!)^2}{(2n)!}$$


And use Stirling's formula(with $2n$ instead $n$)
 
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