# Evaluating limit

• Mar 6th 2013, 01:04 PM
Stormey
Evaluating limit
Hi.

Ho can i show that $\displaystyle \lim_{n\rightarrow \infty}\frac{n\ln n}{\ln n!}=1$?

I tried to use the sandwich rule to say that:

$\displaystyle \frac{n\ln n}{\ln n!}\geq \frac{n\ln n}{\ln n^n}=\frac{n\ln n}{n\ln n}=1$

but couldn't find anything bigger to squeeze it between.
what other ways are there to evaluate the limit here?

any guidance would greatly appreciated.
• Mar 6th 2013, 01:40 PM
ILikeSerena
Re: Evaluating limit
Quote:

Originally Posted by Stormey
Hi.

Ho can i show that $\displaystyle \lim_{n\rightarrow \infty}\frac{n\ln n}{\ln n!}=1$?

I tried to use the sandwich rule to say that:

$\displaystyle \frac{n\ln n}{\ln n!}\geq \frac{n\ln n}{\ln n^n}=\frac{n\ln n}{n\ln n}=1$

but couldn't find anything bigger to squeeze it between.
what other ways are there to evaluate the limit here?

any guidance would greatly appreciated.

Hi Stormey! :)

You can find an upper limit here including the 3 line derivation for it.
• Mar 7th 2013, 12:41 AM
Stormey
Re: Evaluating limit
Hi ILikeSerena! :)
Thanks for the help.

it says there:

$\displaystyle n\ln (\frac{n}{e})+1\leq \ln n!$

and i assume it's concluded by $\displaystyle \ln n$'s integral, and we didn't learn integrals yet, so i suppose i need to address it in some other way.
Actually, my assignment requires to show that $\displaystyle n\ln n=\Theta (\ln n!)$, so i just need to show that $\displaystyle n\ln n$ is bounded by $\displaystyle \ln n!$, and that $\displaystyle \ln n!$ is bounded by $\displaystyle n\ln n$.

Sorry for the ignorance, as much as i love math, calculus is my greatest weakness. (Crying)