Question

How would you evaluate such Limit?
\[\lim_{n\to\infty} \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\]

Answer 1

Here is my attempt.\[\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\times \frac{1\cdot3\cdot5\cdots(2n-1)}{1\cdot3\cdot5\cdots(2n-1)}\]\[\implies \frac{1}{(2n)!}\cdot \left(\frac{(2n)!}{2n!}\right)^2\tag1\]
I think Product of Odd numbers can be represented as \(\large\frac{(2n)!}{2n!}\). Please correct me if I am wrong.
From (1), the limit has now become.\[\lim_{n\to \infty}\frac{(2n)!}{\left(2n!\right)^2}\]
So, how do I evaluate it from here? Of course, I can take Log.
\[\lim_{n\to \infty} \Large e^{\ln \frac{(2n)!}{\left(2n!\right)^2}}\]Or, Let \(L_n\) be \(\frac{(2n)!}{\left(2n!\right)^2}\), which makes the limit \(\;\lim_{n\to\infty}e^{L_n}\).
\[\lim_{n\to \infty} L_n = \lim_{n\to \infty} \ln\frac{(2n)!}{(2n!)^2}\]\[\implies \lim_{n\to \infty} \ln(2n)! - 2\ln(2n!)\]But from here onwards I am clueless. :/

Answer 2

When you wrote that the product of odd numbers can be written as \[\large\frac{(2n)!}{2n!}\]I think it should actually be \[\large\frac{(2n)!}{2^n n!}\]Since you need to multiply each term by 2 on the bottom.

Answer 3

And likewise, \[\large 2\cdot4\cdot6\cdot...\cdot2n =2^n n!\]

Answer 4

Ohh yeah. :/ Silly mistake.
\[\ln(2n!) - 2n\ln2 - 2\ln n!\]

Answer 5

Could you please explain this line a bit more to me also? I'm confused as to where the natural log went.

Let \(L_n=\frac{(2n)!}{\left(2n!\right)^2}\) which makes the limit \(\;\lim_{n\to\infty}e^{L_n}\)

Answer 6

\[\Large e^{\ln \frac{(2n)!}{(2^nn!)^2}}\]Sorry for the bad math language.

Answer 7

I don't think I can evaluate the limit with my approach. I got the problem from this link: http://math.stackexchange.com/questions/139494/another-evaluating-limit-question#139498. But I don't understand their solutions at all, one used double factorial while the other uses some inequality with O notation.

Answer 8

The top solution is not too hard to follow. The big O notation is just a side note.

Answer 9

The best way to approach this limit is to use Stirling's approximation of n! for large n
\[ n! \approx \sqrt{2\pi} n^{n+\frac 12 } e^{-n}
\]
If you do this the ratio will behave as
\[\frac {1} {\sqrt n}
\]
Which thend to zero with n going to infinity.

Answer 10

As long as you know that \((n+1)(n-1) \le n^2\) the top proof given at that link is straightforward using a comparison test.

Answer 11

After reading the link, I agree with @KingGeorge.

Answer 12

Ohh, hmm I will try to understand it then. Sorry for troubling you guys.

Answer 13

And Thanks as well. :-)

Answer 14

Going between these two steps is the hardest. \[= \frac{1 \cdot 3}{2 \cdot 2}\cdot\frac{3 \cdot 5}{4 \cdot 4}\cdot\frac{5 \cdot 7}{6 \cdot 6}\cdots\frac{(2n-3) \cdot (2n-1)}{(2n-2) \cdot (2n-2)}\cdot\left(\frac{2n-1}{(2n)^2}\right) \\ \leq \frac{2n-1}{(2n)^2} \]However, note that by the inequality posted at the top, this means that each individual fraction here is\[{(n+1)(n-1)\over n^2} \leq {n^2 \over n^2}\]So each fraction is less than or equal to 1 except for the last one. Thus, you're left with the conclusion \[\leq \frac{2n-1}{(2n)^2}\]