Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Why is $\lim\limits_{n \to \infty}\frac{1}{n} \sqrt[n] {n^n}=1$ where $\lim\limits_{n \to \infty} \sqrt[n] {n!}=\infty$ ?

We all know that $n^n > n! \ : \forall n$ so how come the factorial "beats" the exponent when it's nth rooted and going to infinity ?

The factorial is behaving as if it's bigger than $n^n$.

share|improve this question
1  
Are you sure about the question [n^n]^(1/n) ? –  Claude Leibovici Dec 8 '13 at 18:50
    
Answers are not matching with the question. –  Dutta Dec 9 '13 at 17:24

3 Answers 3

up vote 26 down vote accepted

Your first limit does not make sense. That is, it doesn't make sense to have the limiting variable outside the limit.

Perhaps you mean $$ \lim_{n\to\infty}\frac1n\sqrt[n]{n^n}=1 $$ In that case, then you get, in a similar fashion, for the second limit $$ \lim_{n\to\infty}\frac1n\sqrt[n]{n!}=\frac1e $$

share|improve this answer
    
Yeah I did it from memory and messed up... It doesn't makes sense now that I think about it. Thanks. If you can please delete this question. –  GinKin Dec 8 '13 at 19:03
4  
Would the downvoter care to comment? –  robjohn Dec 8 '13 at 21:00
    
@robjohn Ping the moderators and you will get the following reply: "We are watching what we can to see if there is anything we can see that the script has not. Keep us informed.". But no action will be taken and they will keep watching till the entire site collapses. –  user17762 Dec 8 '13 at 21:12
    
@user17762 "till the entire site collapses." wat –  GinKin Dec 8 '13 at 21:14
1  
@Pleasedeleteaccount what happened ? why are you so mad ? not all users are thankless just so you know... –  GinKin Dec 9 '13 at 19:11

Because $$\lim_{n\to\infty}\sqrt[n]{n^n}=\lim_{n\to\infty}n=\infty.$$

share|improve this answer

$\lim \frac{1}{n}(n^n)^{\frac{1}{n}} = \lim \frac{n}{n} = 1$

For $(n!)^{\frac{1}{n}}$ take $u_n = n!$. $\lim \frac{u_{n+1}}{u_n} = \lim (n+1) = \infty$ that gives $\lim (n!)^{\frac{1}{n}} = \infty$, as $\lim \inf \frac{u_{n+1}}{u_n} \le \lim \inf (u_n)^{\frac{1}{n}} \le \lim \sup (u_n)^{\frac{1}{n}} \le \lim \sup \frac{u_{n+1}}{u_n}$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.