I'm having trouble with the sequence: $\displaystyle a_n=\frac{1*3*5*...*(2n-1)}{n!}$

It seems that the limit must exists since each factor$\displaystyle a_n=\frac{1}{1}*\frac{3}{2}*\frac{5}{3}*\frac{2n-1}{n}$ has a finite limit.

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- Jun 30th 2010, 11:26 AMadkinsjrLimit of a Sequence
I'm having trouble with the sequence: $\displaystyle a_n=\frac{1*3*5*...*(2n-1)}{n!}$

It seems that the limit must exists since each factor$\displaystyle a_n=\frac{1}{1}*\frac{3}{2}*\frac{5}{3}*\frac{2n-1}{n}$ has a finite limit. - Jun 30th 2010, 11:41 AMHallsofIvy
Write 1*3*5*...(2n-1) as $\displaystyle \frac{1*2*3*4*5*...*(2n-1)(2n)}{2*4*6*...**(2n)}= \frac{(2n)!}{((1)(2))(2(2))*(3(2))*...*(n(2))$$\displaystyle =\frac{(2n)!}{2^n n!}$.

Now, $\displaystyle a_n= \frac{(2n)!}{2^n(n!)^2}$. - Jun 30th 2010, 12:05 PMadkinsjr
Thanks, but you have a latex error.

- Jun 30th 2010, 02:35 PMchisigma
It is easy to see that is...

$\displaystyle a_{n+1} = a_{n} \frac{2n+1}{n+1}$ (1)

... so that is...

$\displaystyle \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}} = 2$ (2)

... and that means that is...

$\displaystyle \lim_{n \rightarrow \infty} a_{n} = \infty$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$