Determine if $\displaystyle \lim_{x\to\infty} \frac{\sqrt[x]{x!}}{x}$ 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

Printable View

- Apr 25th 2008, 02:45 PMpolymeraselimit of xth root etc etc
Determine if $\displaystyle \lim_{x\to\infty} \frac{\sqrt[x]{x!}}{x}$ 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 - Apr 25th 2008, 02:50 PMMathstud28
- Apr 25th 2008, 03:01 PMpolymerase
- Apr 25th 2008, 03:06 PMMathstud28
Good job! Indeed the answer is $\displaystyle \frac{1}{e}$ (Clapping)

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

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

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

The gamma function and go L'hopitals - Apr 25th 2008, 03:13 PMpolymerase
- Apr 25th 2008, 03:13 PMMathstud28
Expanding upon my first thought after simplifying you would have $\displaystyle \lim_{n\to\infty}\frac{\ln(x)}{x}-1+\frac{\ln(x)}{2x^2}+\frac{\ln(2\pi)}{2x^2}=-1$ therefore $\displaystyle \ln(y)=-1\Rightarrow{y=e^{-1}}$

This came from the ln(x!) approximation over x² - Apr 25th 2008, 03:15 PMMathstud28
- Apr 25th 2008, 03:23 PMpolymerase
- Apr 25th 2008, 03:24 PMMathstud28
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 - Apr 25th 2008, 08:32 PMIsomorphism
- May 18th 2008, 06:50 AMKrizalid
$\displaystyle \begin{aligned}

\frac{\sqrt[x]{x!}}{x}&=\exp \left[ \ln \left( \prod\limits_{k\,=\,1}^{x}{\left( \frac{k}{x} \right)^{1/x}} \right) \right] \\

& =\exp \left[ \frac{1}{x}\sum\limits_{k\,=\,1}^{x}{\ln \frac{k}{x}} \right] \\

& =\exp \left[ \int_{0}^{1}{\ln y\,dy} \right],\,\text{as }x\to\infty.

\end{aligned}$ - May 18th 2008, 05:25 PMarbolis
Nice limit question. Using my intuition I thought it would tend to positive infinite.