ropowiec2gkc

2022-03-30

Proving the limit of $\frac{{\mathrm{log}\left(n\right)}^{\mathrm{log}\left(n\right)}}{{1.01}^{n}}$
Can anyone please show me a simple way (if there is one) to show that
$\underset{n\to \mathrm{\infty }}{lim}\frac{{\mathrm{log}\left(n\right)}^{\mathrm{log}\left(n\right)}}{{1.01}^{n}}=0$
And that
$\underset{n\to \mathrm{\infty }}{lim}\frac{{1.01}^{n}}{n!}=0$
I've checked that it's true, I just need to show it the shortest way possible.

Malia Booth

For the second limit use this fact that
$\frac{{1.01}^{n}}{n!}<\frac{{2}^{n}}{n!}$
and the fact that $\frac{{2}^{n}}{n!}\to 0$ when n tends to infinity. In fact,
$0<\frac{{2}^{n}}{n!}=\frac{2}{1}\cdot \frac{2}{2}\cdot \frac{2}{3}\cdot \cdot \cdot \frac{2}{n}\le \frac{2}{1}\cdot \frac{2}{2}\cdot \frac{2}{3}\cdot \cdot \cdot \frac{2}{3}=\frac{2}{1}\cdot \frac{2}{2}×{\left(\frac{2}{3}\right)}^{n-2}$
and you know that since $\frac{2}{3}<1$ then ${\left(\frac{2}{3}\right)}^{n-2}\to 0$ when $n\to \mathrm{\infty }$

Do you have a similar question?