Hi,

How I show that

$\displaystyle \lim_{n\to \infty} \frac{n!e^n}{n^n}=\infty $

thanks.

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- Apr 9th 2009, 10:33 AMAndres PerezHelp with a limit
Hi,

How I show that

$\displaystyle \lim_{n\to \infty} \frac{n!e^n}{n^n}=\infty $

thanks. - Apr 9th 2009, 10:44 AMMoo
Hello,

Using Stirling's approximation for n!, it is quite straightforward.

For a large n, $\displaystyle n! \sim \sqrt{2\pi n} \cdot \frac{n^n}{e^n}$

See here : http://en.wikipedia.org/wiki/Stirling%27s_approximation - Apr 9th 2009, 02:31 PMSoroban
Hello, Andres Perez!

Quote:

How do I show that: .$\displaystyle \lim_{n\to \infty} \frac{n!e^n}{n^n}\:=\:\infty $

$\displaystyle R \;=\;\frac{a_{n+1}}{a_n} \;=\;\frac{(n+1)!e^{n+1}}{(n+1)^{n+1}}\cdot\frac{n ^n}{n!e^n} \;=\;\frac{e^{n+1}}{e^n}\cdot\frac{(n+1)!}{n!}\cdo t\frac{n^n}{(n+1)^{n+1}}$

. . $\displaystyle = \;\frac{e}{1}\cdot\frac{n+1}{1}\cdot\frac{n^n}{(n+ 1)(n+1)^n} \;=\;e\cdot\frac{n^n}{(n+1)^n} \;=\;e\cdot\left(\frac{n}{n+1}\right)^n $

Divide top and botom by $\displaystyle n\!:\;\;e\cdot\left( \frac{1}{1+\frac{1}{n}}\right)^n \;=\;e\cdot\frac{1}{(1 + \frac{1}{n})^n}$

Then: .$\displaystyle \lim_{n\to\infty} R \;=\;\lim_{n\to\infty}\bigg[e\cdot\frac{1}{(1+\frac{1}{n})^n}\bigg] \;=\;e\cdot\frac{1}{\lim(1+\frac{1}{n})} \;=\;e\cdot\frac{1}{e} \;=\;1$

. . . . . . . . . . . . . . . . . . . . . . . . . . .$\displaystyle ^{^{n\to\infty}}$

Since $\displaystyle \lim_{n\to\infty}|R| \:\not<\:1$, the series diverges.

- Apr 9th 2009, 02:48 PMNonCommAlg
- Apr 9th 2009, 05:58 PMAndres Perez
Can we obtain that result without Stirling's approximation?

How we show the divergence of the series? - Apr 17th 2009, 08:11 AMJhevon
you already asked this and i already gave you a hint. not very polite of you to re-post Andres.probably.

Quote:

How we show the divergence of the series?

a variation on the root test may work though.

note that $\displaystyle \lim_{n \to \infty} \sqrt[n]{\frac {n!e^n}{n^n}} \le \lim_{n \to \infty} \frac {n!e^n}{n^n}$ - Apr 17th 2009, 08:58 AMAndres Perez
on the Jevon's inequality we get that $\displaystyle \lim_{n \to \infty}\frac{n!e^n}{n^n}\geq \lim_{n \to \infty}\sqrt[n]{\frac{n!e^n}{n^n}}=1$, so it doesn't says anything about $\displaystyle \lim_{n \to \infty}\frac{n!e^n}{n^n}=\infty$

- Apr 17th 2009, 09:08 AMJhevon