# Proving the limit of $\frac{\log(n)^{\log(n)}}{1.01^{n}}$

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!

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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
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. :) – Babak S. Oct 30 '12 at 11:21
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
I've checked that its true... Oh really? Then show how you did it. – Did Oct 30 '12 at 11:57

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 $n\rightarrow\infty$.

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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. – Babak S. Oct 30 '12 at 12:57
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.

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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. – Babak S. Oct 30 '12 at 12:58
@barto: Thanks for comment. Offcourse $\ln=\log_{e}$. – Mhenni Benghorbal Oct 30 '12 at 12:59
@BabakSorouh: Thank you for your comment. I really appreciate it. – Mhenni Benghorbal Oct 30 '12 at 13:00