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

shamieh
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A little confused here..

what does $$\frac{(2n-1)!}{2n!}$$ as n---> $$\infty$$ = 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
 
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shamieh said:
A little confused here..

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

$\chi$ $\sigma$
 
Last edited:
So you're saying 2n! = 2n(n-1)?
 
I don;t know how to solve: as n ---> $$\infty$$ $$\frac{(2n - 1)!}{2n!}$$ I don't get it at all...
 
shamieh said:
I don;t know how to solve: as n ---> $$\infty$$ $$\frac{(2n - 1)!}{2n!}$$ 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 $\chi$ $\sigma$
 
Thanks... Took me a second to see what was going on. I get it now.so then if i had $$\frac{(2n-1)!}{(2n + 1)!}$$ then I would get as n-> infinity $$(2n+1)(2n)$$ right?
 
shamieh said:
if i had $$\frac{(2n-1)!}{(2n + 1)!}$$ then I would get as n-> infinity $$(2n+1)(2n)$$ right?
What do you mean by saying you would get $$(2n+1)(2n)$$?
 
If i took n --> infinity I would get that result.
 
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.
 
  • #10
Evgeny.Makarov said:
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
 
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