Proving the limit of log⁡(n)log⁡(n)1.01nCan anyone please show me a simple way (if there is one) to show thatlimn→∞log⁡(n)log⁡(n)1.01n=0And thatlimn→∞1.01nn!=0I've checked that it's true, I just need to show it the shortest way possible.

For the second limit use this fact that1.01nn!&lt;2nn!and the fact that 2nn!→0 when n tends to infinity. In fact,0&lt;2nn!=21⋅22⋅23⋅⋅⋅2n≤21⋅22⋅23⋅⋅⋅23=21⋅22×(23)n−2and you know that since 23&lt;1 then (23)n−2→0 when n→∞

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ropowiec2gkc

Answered question

2022-03-30

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

Answer & Explanation

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} \leq \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 {(\frac{2}{3})}^{n - 2}

and you know that since

\frac{2}{3} < 1

then

{(\frac{2}{3})}^{n - 2} \to 0

when

n \to \infty

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