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

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Discussion Overview

The discussion centers around the limit of the expression \(\frac{(2n-1)!}{2n!}\) as \(n\) approaches infinity. Participants express confusion regarding the factorial notation and the behavior of the expression in the limit, exploring various interpretations and calculations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants express confusion about the factorial notation, particularly distinguishing between \(2n!\) and \(2n(n-1)\).
  • One participant attempts to clarify that \(2n!\) represents the product of all integers from \(1\) to \(2n\), while \(2n(n-1)\) is not a correct representation of \(2n!\).
  • Another participant proposes that \(\frac{(2n-1)!}{2n!}\) simplifies to \(\frac{1}{2}(2n-1)(2n-2)\ldots(n+1)\), indicating a potential approach to evaluate the limit.
  • There is a suggestion that if \(\frac{(2n-1)!}{(2n+1)!}\) is considered, it would lead to a different expression, specifically \(\frac{1}{2n(2n+1)}\), which raises questions about the limit as \(n\) approaches infinity.
  • Some participants challenge the correctness of earlier statements, particularly regarding the simplification of factorial expressions and the implications of taking limits.

Areas of Agreement / Disagreement

Participants generally do not reach a consensus on the correct interpretation of the factorial expressions or the limits involved. Multiple competing views remain, particularly regarding the simplifications and the behavior of the expressions as \(n\) approaches infinity.

Contextual Notes

There are limitations in the discussion regarding the clarity of factorial notation and the assumptions made in the simplifications. Some participants express uncertainty about the correct approach to evaluating the limits.

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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