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.

Can anyone please show me a simple way (if there is one) to show that $$\lim_{n\to \infty}\frac{\log(n)^{\log(n)}}{1.01^{n}}=0$$ And that $$\lim_{n\to \infty}\frac{1.01^{n}}{n!}=0$$

I've checked that its true, I just need to show it the the shortest way possible.

Thanks!

share|improve this question
1  
Define "the shortest way possible". also, how can we prove that its "the shortest way possible" if you did not say what it is ? –  Belgi Oct 30 '12 at 11:19
1  
For the first one: take the logarithm of the limit and prove that $$\lim_{n\to\infty}\log\left(\frac{\log(n)^{\log(n)}}{1.01^n}\right)=\infty$$ –  barto Oct 30 '12 at 11:20
    
Fix the title. :) –  B. S. Oct 30 '12 at 11:21
1  
The second one is a known limit. For values of $n$ that are large enough, $n!$ is multiplied with a large value, while $1.01^n$ is multiplied with the constant $1.01$. –  barto Oct 30 '12 at 11:26
1  
I've checked that its true... Oh really? Then show how you did it. –  Did Oct 30 '12 at 11:57

2 Answers 2

up vote 6 down vote accepted

For the second limit use this fact that $$\frac{1.01^n}{n!}<\frac{2^n}{n!}$$ and the fact that $\frac{2^n}{n!}\longrightarrow 0$ when $n$ tends to infinity. In fact, $$0<\frac{2^n}{n!}=\frac{2}{1}\cdot\frac{2}{2}\cdot\frac{2}{3}\cdot\cdot\cdot\frac{2}{n}\leqslant\frac{2}{1}\cdot\frac{2}{2}\cdot\frac{2}{3}\cdot\cdot\cdot\frac{2}{3}=\frac{2}{1}\cdot\frac{2}{2}\times\bigg(\frac{2}{3}\bigg)^{n-2}$$ and you know that since $\frac{2}{3}<1$ then $\big(\frac{2}{3}\big)^{n-2}\longrightarrow 0$ when $n\rightarrow\infty$.

share|improve this answer
    
Doesn't this argument work just as well with $1.01$s in the numerator as $2$s? –  Jason DeVito Oct 30 '12 at 12:46
    
@JasonDeVito: Yes Jason, it works indeed. I thought maybe the squeeze theorem could illustrate the OP how the limit gets zero. You are absolutely right. Thanks. –  B. S. Oct 30 '12 at 12:57
1  
It was more to check my own understanding. I also think that, aesthetically, a $2$ looks better than a $1.01$ in the numerator ;-) +1 from me. –  Jason DeVito Oct 30 '12 at 13:54
    
Good job...you "nailed it!" +1 –  amWhy Apr 1 '13 at 0:11

I already solved a problem for you related to this problem. You are right. The first limit is $0$. Here how to prove it. Making the change of variables $m=\ln(n)$ yields

$$\lim_{n\to \infty}\frac{\log(n)^{\log(n)}}{1.01^{n}}= \lim_{m\to \infty}\frac{m^m}{e^{\ln(1.01)e^m}}=y\,.$$

Taking the $\ln$ (the logarithmic function) to both sides of the last equation gives

$$\implies \ln(y)=m\ln(m)-\ln(1.01)e^m \,,$$

which follows from the properties of the logarithmic function. Taking the limit of the last equation gives

$$ \implies \lim_{m\to \infty}\ln(y)= \ln(\lim_{m\to \infty}y) = \lim _{m\to \infty} (m\ln(m)-\ln(1.01)e^m)\rightarrow -\infty $$

$$ \implies \lim_{m\to \infty} y = e^{-\infty}=0 \,.$$

Interchanging the order of the limit is justified by the continuity of the logarithmic function.

share|improve this answer
    
Maybe the step $\displaystyle\lim_{n\to\infty}(m\cdot\ln(m)-\ln(1.01)\cdot e^m)=-\infty$ is not very clear. But note that $m\ln(m)<m^2$ for $m$ large enough, so we have $\displaystyle\lim_{n\to\infty}(m\cdot\ln(m)-\ln(1.01)\cdot e^m)\leq\lim_{n\to\infty}(m^2-c\cdot e^m)$ for some positive constant $c$. The last limit is clearly $-\infty$. ($\ln(x)$ means the natural logarithm of $x$, I'm not sure whether it is the same as $\log(x)$.) –  barto Oct 30 '12 at 12:53
    
+1 you did the first one in a clear way. –  B. S. Oct 30 '12 at 12:58
    
@barto: Thanks for comment. Offcourse $\ln=\log_{e}$. –  Mhenni Benghorbal Oct 30 '12 at 12:59
1  
@BabakSorouh: Thank you for your comment. I really appreciate it. –  Mhenni Benghorbal Oct 30 '12 at 13:00

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.