# Limit (7)

• Dec 12th 2009, 04:38 PM
NonCommAlg
Limit (7)
Evaluate $\lim_{n\to\infty} \sqrt[n]{\binom{n}{\lfloor na \rfloor}},$ where $0 < a < 1$ is given.
• Dec 13th 2009, 02:49 PM
redsoxfan325
Quote:

Originally Posted by NonCommAlg
Evaluate $\lim_{n\to\infty} \sqrt[n]{\binom{n}{\lfloor na \rfloor}},$ where $0 < a < 1$ is given.

Spoiler:
$\frac{1}{a^a(1-a)^{1-a}}$
but as of right now I have no proof of this conjecture.
• Dec 13th 2009, 03:13 PM
redsoxfan325
Ignore this. I accidentally clicked "Submit" halfway through my proof.
• Dec 13th 2009, 03:17 PM
redsoxfan325
Quote:

Originally Posted by NonCommAlg
Evaluate $\lim_{n\to\infty} \sqrt[n]{\binom{n}{\lfloor na \rfloor}},$ where $0 < a < 1$ is given.

And here's my proof. Because I don't want to repeatedly type this, and I don't do anything with it for a while, let $k=\lfloor na\rfloor$. Also, assume there is a $\lim_{n\to\infty}$ in front of each of these.

Spoiler:
Rewrite this expression as

$\exp\left[\frac{1}{n}\ln\left(\frac{n!}{(n-k)!k!}\right)\right]=\exp\left[\frac{1}{n}\left(\ln(n!)-\ln((n-k)!)-ln(k!)\right)\right]$

Using Stirling's approximation $\ln n!\approx n\ln n-n$, the above equals

$\exp\left[\frac{1}{n}\big(n\ln(n)-n-(n-k)\ln(n-k)+(n-k)-k\ln(k)+k\big)\right]=$ $\exp\left[\frac{1}{n}\big(n\ln(n)-n\ln(n-k)+k\ln(n-k)-k\ln(k)\big)\right]=$

$\exp\left[\frac{1}{n}\left(n\ln\left(\frac{n}{n-k}\right)+k\ln\left(\frac{n-k}{k}\right)\right)\right]=$ $\exp\left[\ln\left(\frac{n}{n-k}\right)+\frac{k}{n}\ln\left(\frac{n}{k}-1\right)\right]=$

$\frac{n}{n-k}\cdot\left(\frac{n}{k}-1\right)^{k/n}=\frac{1}{1-a}\left(\frac{1}{a}-1\right)^a=\frac{(1-a)^a}{(1-a)a^a}=\boxed{\frac{1}{a^a(1-a)^{1-a}}}$
• Dec 14th 2009, 01:51 AM
simplependulum
Can i use Riemann Sum ?
• Dec 14th 2009, 08:09 PM
NonCommAlg
Quote:

Originally Posted by redsoxfan325
And here's my proof. Because I don't want to repeatedly type this, and I don't do anything with it for a while, let $k=\lfloor na\rfloor$. Also, assume there is a $\lim_{n\to\infty}$ in front of each of these.

Spoiler:
Rewrite this expression as

$\exp\left[\frac{1}{n}\ln\left(\frac{n!}{(n-k)!k!}\right)\right]=\exp\left[\frac{1}{n}\left(\ln(n!)-\ln((n-k)!)-ln(k!)\right)\right]$

Using Stirling's approximation $\ln n!\approx n\ln n-n$, the above equals

$\exp\left[\frac{1}{n}\big(n\ln(n)-n-(n-k)\ln(n-k)+(n-k)-k\ln(k)+k\big)\right]=$ $\exp\left[\frac{1}{n}\big(n\ln(n)-n\ln(n-k)+k\ln(n-k)-k\ln(k)\big)\right]=$

$\exp\left[\frac{1}{n}\left(n\ln\left(\frac{n}{n-k}\right)+k\ln\left(\frac{n-k}{k}\right)\right)\right]=$ $\exp\left[\ln\left(\frac{n}{n-k}\right)+\frac{k}{n}\ln\left(\frac{n}{k}-1\right)\right]=$

$\frac{n}{n-k}\cdot\left(\frac{n}{k}-1\right)^{k/n}=\frac{1}{1-a}\left(\frac{1}{a}-1\right)^a=\frac{(1-a)^a}{(1-a)a^a}=\boxed{\frac{1}{a^a(1-a)^{1-a}}}$

your final answer is correct but you need to be more careful with using the approximation $\ln n!\approx n\ln n-n.$

simplependulum, you can use whatever you want! (Wink)
• Dec 14th 2009, 08:10 PM
redsoxfan325
How could I make it more rigorous?
• Dec 14th 2009, 08:22 PM
NonCommAlg
Quote:

Originally Posted by redsoxfan325
How could I make it more rigorous?

by using $n! = \sqrt{2 \pi n} \left(\frac{n}{e} \right)^n (1 + O(1/n)).$ see Stirling's approximation - Wikipedia, the free encyclopedia.
• Dec 14th 2009, 11:11 PM
simplependulum
I got the approximation of $\ln(n!)$ when $n$ tends to infinty , using Riemann Sum

$\ln(n!) = \ln(1) + \ln(2) + ... + \ln(n)$

$= n \frac{ \ln(1/n) + \ln(2/n) + ... + \ln(n/n) }{n} + n\ln(n)$

$= n \int_0^1 \ln(x) ~dx + n\ln(n)$

$= n \left [ x\ln(x) - x \right ]_0^1 + n\ln(n)$

$= n\ln(n) - n$ which equals to what redsoxfan325 mentioned .

I believe the $n\ln(n) - n$ is correct but i'm not sure if the above is a rigorous proof .
• Dec 15th 2009, 09:17 PM
Drexel28
Quote:

Originally Posted by simplependulum
I got the approximation of $\ln(n!)$ when $n$ tends to infinty , using Riemann Sum

$\ln(n!) = \ln(1) + \ln(2) + ... + \ln(n)$

$= n \frac{ \ln(1/n) + \ln(2/n) + ... + \ln(n/n) }{n} + n\ln(n)$

$= n \int_0^1 \ln(x) ~dx + n\ln(n)$

$= n \left [ x\ln(x) - x \right ]_0^1 + n\ln(n)$

$= n\ln(n) - n$ which equals to what redsoxfan325 mentioned .

I believe the $n\ln(n) - n$ is correct but i'm not sure if the above is a rigorous proof .

Well the equals sign is pushing it. In your Riemann sum step what you really mean as that as $n$ increases without bound $\frac{ \ln(1/n) + \ln(2/n) + ... + \ln(n/n) }{n}\to\int_0^1\ln(x)\text{ }dx$. You of course knew that, but in case anyone misinterpreted it.