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.

Can anyone please show me a simple way (if there is one) to show that

And that

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

Malia Booth

Beginner2022-03-31Added 16 answers

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,

$0<\frac{{2}^{n}}{n!}=\frac{2}{1}\cdot \frac{2}{2}\cdot \frac{2}{3}\cdot \cdot \cdot \frac{2}{n}\le \frac{2}{1}\cdot \frac{2}{2}\cdot \frac{2}{3}\cdot \cdot \cdot \frac{2}{3}=\frac{2}{1}\cdot \frac{2}{2}\times {\left(\frac{2}{3}\right)}^{n-2}$

and you know that since$\frac{2}{3}<1$ then ${\left(\frac{2}{3}\right)}^{n-2}\to 0$ when $n\to \mathrm{\infty}$

and the fact that

and you know that since

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