Help with a limit

**Andres Perez** · April 9th 2009, 10:33 AM

Hi,

How I show that

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

thanks.

**Moo** · April 9th 2009, 10:44 AM

Hello,

Originally Posted by Andres Perez

Hi,

How I show that

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

thanks.

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

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

See here : http://en.wikipedia.org/wiki/Stirling%27s_approximation

**Soroban** · April 9th 2009, 02:31 PM

Hello, Andres Perez!

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

Ratio test . . .

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

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

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

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

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

**NonCommAlg** · April 9th 2009, 02:48 PM

Originally Posted by Soroban

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

we can't say anything about convergence or divergence of the series because $\lim_{n\to\infty} \frac{a_{n+1}}{a_n}=1.$ even if it diverges, which it does but for another reason, we can't deduce that $\lim_{n\to\infty}a_n=\infty.$

**Andres Perez** · April 9th 2009, 05:58 PM

Can we obtain that result without Stirling's approximation?

How we show the divergence of the series?

**Jhevon** · April 17th 2009, 08:11 AM

you already asked this and i already gave you a hint. not very polite of you to re-post Andres.

Originally Posted by Andres Perez

Can we obtain that result without Stirling's approximation?

probably.

How we show the divergence of the series?

we don't have a series here, we have a limit. doing the ratio test was a beat around the bush way to attack this problem. of course, it yields no fruit since the limit was exactly 1, it gives us no information.

a variation on the root test may work though.

note that $\lim_{n \to \infty} \sqrt[n]{\frac {n!e^n}{n^n}} \le \lim_{n \to \infty} \frac {n!e^n}{n^n}$

**Andres Perez** · April 17th 2009, 08:58 AM

on the Jevon's inequality we get that $\lim_{n \to \infty}\frac{n!e^n}{n^n}\geq \lim_{n \to \infty}\sqrt[n]{\frac{n!e^n}{n^n}}=1$ , so it doesn't says anything about $\lim_{n \to \infty}\frac{n!e^n}{n^n}=\infty$

**Jhevon** · April 17th 2009, 09:08 AM

Originally Posted by Andres Perez

on the Jevon's inequality we get that $\lim_{n \to \infty}\frac{n!e^n}{n^n}\geq \lim_{n \to \infty}\sqrt[n]{\frac{n!e^n}{n^n}}=1$ , so it doesn't says anything about $\lim_{n \to \infty}\frac{n!e^n}{n^n}=\infty$

ah yes, the ratio and root tests are equivalent when both work

without Stirling's, this is a beast.

did you try using power series as i suggested before?

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