1. ## 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.

2. ## Re: Evaluating limit

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.

3. ## 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.