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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 it's true, I just need to show it the shortest way possible.

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    *I've checked that its true*... Oh really? Then show how you did it.2012-10-30

2 Answers 2

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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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    Good job...you "nailed it!" +$1$2013-04-01
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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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    @BabakSorouh: Thank you for your comment. I really appreciate it.2012-10-30