limit of xth root etc etc

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April 25th 2008, 02:45 PM
polymerase

limit of xth root etc etc

Determine if $\lim_{x\to\infty} \frac{\sqrt[x]{x!}}{x}$ exist (and value) and if it doesn't explain why.
(in case thats too small, it reads the xth root of x! all over x)

Potentially VERY easy series questions?!?

Thanks
April 25th 2008, 02:50 PM
Mathstud28

Quote:

Originally Posted by polymerase

Determine if $\lim_{x\to\infty} \frac{\sqrt[x]{x!}}{x}$ exist (and value) and if it doesn't explain why.
(in case thats too small, it reads the xth root of x! all over x)

Potentially VERY easy series questions?!?

Thanks

use the fact taht $n!\sim{\sqrt{2\pi{n}}n^{n}e^{-n}}$

then it should be apparent
April 25th 2008, 03:01 PM
polymerase

Quote:

Originally Posted by Mathstud28

use the fact taht $n!\sim{\sqrt{2\pi{n}}n^{n}e^{-n}}$

then it should be apparent

Firstly, thank you

Secondly, so is the answer $\frac{1}{e}$ ?

Thirdly, is there another way to do this, because we never learned the relationship between n! and the stuff you stated above :)
April 25th 2008, 03:06 PM
Mathstud28

Quote:

Originally Posted by polymerase

Firstly, thank you

Secondly, so is the answer $\frac{1}{e}$ ?

Thirdly, is there another way to do this, because we never learned the relationship between n! and the stuff you stated above :)

Good job! Indeed the answer is $\frac{1}{e}$ (Clapping)

and you could also do this $y=\lim_{x\to\infty}\frac{(x!)^{\frac{1}{x}}}{x}\Ri ghtarrow{\ln(y)=\frac{\ln(x!)}{x^2}}$

and this is painful but $\ln(x!)\approx{x\ln(x)-x+\frac{\ln(x)}{2}+\frac{\ln(2\pi)}{2}}$

Or you could als use that $n!=\int_0^{\infty}x^{n}e^{-x}dx$

The gamma function and go L'hopitals
April 25th 2008, 03:13 PM
polymerase

Quote:

Originally Posted by Mathstud28

Good job! Indeed the answer is $\frac{1}{e}$ (Clapping)

and you could also do this $y=\lim_{x\to\infty}\frac{(x!)^{\frac{1}{x}}}{x}\Ri ghtarrow{\ln(y)=\frac{\ln(x!)}{x^2}}$

and this is painful but $\ln(x!)\approx{x\ln(x)-x+\frac{\ln(x)}{2}+\frac{\ln(2\pi)}{2}}$

Or you could als use that $n!=\int_0^{\infty}x^{n}e^{-x}dx$

The gamma function and go L'hopitals

REALLY?! thats the alternative! how the %$!@ does our prof. expect us to solve that? we never learn ANY of that!

LOL is there ANOTHER way?:D
April 25th 2008, 03:13 PM
Mathstud28

Quote:

Originally Posted by Mathstud28

Good job! Indeed the answer is $\frac{1}{e}$ (Clapping)

and you could also do this $y=\lim_{x\to\infty}\frac{(x!)^{\frac{1}{x}}}{x}\Ri ghtarrow{\ln(y)=\frac{\ln(x!)}{x^2}}$

and this is painful but $\ln(x!)\approx{x\ln(x)-x+\frac{\ln(x)}{2}+\frac{\ln(2\pi)}{2}}$

Or you could als use that $n!=\int_0^{\infty}x^{n}e^{-x}dx$

The gamma function and go L'hopitals

Expanding upon my first thought after simplifying you would have $\lim_{n\to\infty}\frac{\ln(x)}{x}-1+\frac{\ln(x)}{2x^2}+\frac{\ln(2\pi)}{2x^2}=-1$ therefore $\ln(y)=-1\Rightarrow{y=e^{-1}}$

This came from the ln(x!) approximation over x²
April 25th 2008, 03:15 PM
Mathstud28

Quote:

Originally Posted by polymerase

REALLY?! thats the alternative! how the %$!@ does our prof. expect us to solve that? we never learn ANY of that!

LOL is there ANOTHER way?:D

Haha...Uhm..I am surprised I thought of three ways...uhm...I cant think of anymore...maybe if you look at this post above you cna use that...that is not overly complicated? haha
April 25th 2008, 03:23 PM
polymerase

Quote:

Originally Posted by Mathstud28

Expanding upon my first thought after simplifying you would have $\lim_{n\to\infty}\frac{\ln(x)}{x}-\frac{1}{x}+\frac{\ln(x)}{2x^2}+\frac{\ln(2\pi)}{2 x^2}=-1$ therefore $\ln(y)=-1\Rightarrow{y=e^{-1}}$

This came from the ln(x!) approximation over x²

How do you figure out/calculate that $\ln (x!)=$ above stuff?
April 25th 2008, 03:24 PM
Mathstud28

Quote:

Originally Posted by polymerase

How do you figure out/calculate that $\ln (x!)=$ above stuff?

Uhm....well I have the memorized but if you need to see the entire thing...hold on let me find...an ariticle...ah! here we go read here

Natural logarithm - Wikipedia, the free encyclopedia

sorry edit:
Look here
Factorial - Wikipedia, the free encyclopedia
April 25th 2008, 08:32 PM
Isomorphism

Quote:

Originally Posted by Mathstud28

Or you could als use that $n!=\int_0^{\infty}x^{n}e^{-x}dx$

How will you use this? (Wondering)
May 18th 2008, 06:50 AM
Krizalid

Quote:

Originally Posted by polymerase

LOL is there ANOTHER way?:D

$\begin{aligned}<br /> \frac{\sqrt[x]{x!}}{x}&=\exp \left[ \ln \left( \prod\limits_{k\,=\,1}^{x}{\left( \frac{k}{x} \right)^{1/x}} \right) \right] \\ <br /> & =\exp \left[ \frac{1}{x}\sum\limits_{k\,=\,1}^{x}{\ln \frac{k}{x}} \right] \\ <br /> & =\exp \left[ \int_{0}^{1}{\ln y\,dy} \right],\,\text{as }x\to\infty.<br /> \end{aligned}$
May 18th 2008, 05:25 PM
arbolis

Nice limit question. Using my intuition I thought it would tend to positive infinite.