# Thread: Help with a limit

1. ## Help with a limit

Hi,

How I show that

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

thanks.

2. Hello,
Originally Posted by Andres Perez
Hi,

How I show that

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

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

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

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

3. Hello, Andres Perez!

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

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

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

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.

4. Originally Posted by Soroban

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

5. Can we obtain that result without Stirling's approximation?

How we show the divergence of the series?

6. 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 $\displaystyle \lim_{n \to \infty} \sqrt[n]{\frac {n!e^n}{n^n}} \le \lim_{n \to \infty} \frac {n!e^n}{n^n}$

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

8. Originally Posted by Andres 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 $\displaystyle \lim_{n \to \infty}\frac{n!e^n}{n^n}=\infty$
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?