Hi,
How I show that
$\displaystyle \lim_{n\to \infty} \frac{n!e^n}{n^n}=\infty $
thanks.
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
Hello, Andres Perez!
Ratio test . . .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.
you already asked this and i already gave you a hint. not very polite of you to re-post Andres.probably.
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.How we show the divergence of the series?
a variation on the root test may work though.
note that $\displaystyle \lim_{n \to \infty} \sqrt[n]{\frac {n!e^n}{n^n}} \le \lim_{n \to \infty} \frac {n!e^n}{n^n}$