What Does \(\frac{(2n-1)!}{2n!}\) Approach as \(n\) Approaches Infinity?

[shamieh]

A little confused here..

what does \(\displaystyle \frac{(2n-1)!}{2n!}\) as n---> \(\displaystyle \infty\) = to?

How would I look at that by inspection and figure it out because I am confused..

isn't 2! = 1 * 2
and 3! = 1 * 2 * 3?

but what is 2n! factorial... Also What is 2n-1! factorial equal to

 

[chisigma]

A little confused here..

what does \(\displaystyle \frac{(2n-1)!}{2n!}\) as n---> \(\displaystyle \infty\) = to?

How would I look at that by inspection and figure it out because I am confused..

isn't 2! = 1 * 2
and 3! = 1 * 2 * 3?

but what is 2n! factorial... Also What is 2n-1! factorial equal to

I suppose that Your 'confusion' is due to a problem of notation. The denominator is...

$\displaystyle 2\ n! = 2\ n\ (n-1)\ ...\ 2\ \cdot 1 \ne (2\ n)! = 2\ n\ (2\ n -1)\ (2\ n - 2)...\ 2\ \cdot 1$

Kind regards

$\chi$ $\sigma$

 

[shamieh]

So you're saying 2n! = 2n(n-1)?

 

[shamieh]

I don;t know how to solve: as n ---> \(\displaystyle \infty\) \(\displaystyle \frac{(2n - 1)!}{2n!}\) I don't get it at all...

 

[chisigma]

I don;t know how to solve: as n ---> \(\displaystyle \infty\) \(\displaystyle \frac{(2n - 1)!}{2n!}\) I don't get it at all...

Is...

$\displaystyle \frac{(2\ n - 1)!}{2\ n!} = \frac{(2\ n - 1)\ (2\ n - 2)\ ...\ 2\ \cdot 1}{2\ n\ (n-1)\ (n-2)\ ... \ 2\ \cdot 1}= \frac{1}{2}\ (2\ n -1)\ (2\ n -2)\ ...\ (n+1)$

... so that... Kind regards $\chi$ $\sigma$

 

[shamieh]

Thanks... Took me a second to see what was going on. I get it now.so then if i had \(\displaystyle \frac{(2n-1)!}{(2n + 1)!}\) then I would get as n-> infinity \(\displaystyle (2n+1)(2n)\) right?

 

[Evgeny.Makarov]

if i had \(\displaystyle \frac{(2n-1)!}{(2n + 1)!}\) then I would get as n-> infinity \(\displaystyle (2n+1)(2n)\) right?

What do you mean by saying you would get \(\displaystyle (2n+1)(2n)\)?

 

[shamieh]

If i took n --> infinity I would get that result.

 

[Evgeny.Makarov]

First,
\[
\frac{(2n-1)!}{(2n+1)!}=\frac{(2n-1)!}{(2n-1)!(2n)(2n+1)}=\frac{1}{2n(2n+1)}
\]
and not $2n(2n+1)$. Second, how can you write that taking the limit as $n\to\infty$ gives an expression that still contains $n$? For example, if $f(n)=n$, then which is correct: $\lim_{n\to\infty}f(n)=n$ or $\lim_{n\to\infty}f(n)=\infty$?

In post #3, you wrote $2n! = 2n(n-1)$. How did a factorial, which is a product of many factors, turn into a product of just three factors?

If you expect helpers to give you careful and correct answers, you should also double-check what you write.

 

[shamieh]

First,
\[
\frac{(2n-1)!}{(2n+1)!}=\frac{(2n-1)!}{(2n-1)!(2n)(2n+1)}=\frac{1}{2n(2n+1)}
\]
and not $2n(2n+1)$. Second, how can you write that taking the limit as $n\to\infty$ gives an expression that still contains $n$? For example, if $f(n)=n$, then which is correct: $\lim_{n\to\infty}f(n)=n$ or $\lim_{n\to\infty}f(n)=\infty$?

In post #3, you wrote $2n! = 2n(n-1)$. How did a factorial, which is a product of many factors, turn into a product of just three factors?

If you expect helpers to give you careful and correct answers, you should also double-check what you write.

You are correct