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How to solve $$\lim_{n\rightarrow \infty}\frac{(\log n)^p}{n}$$

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up vote 2 down vote accepted

Note that $\displaystyle\frac{(\log n)^p}{n}=p^p\cdot\left(\frac{\log k}k\right)^p$, where $k=n^{1/p}\to\infty$ when $n\to\infty$.

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Let $n = e^t$ and let $r = \lceil p \rceil+1$. Then $e^t > \dfrac{t^{r}}{r!}$, since $\displaystyle e^t = \sum_{k=0}^{\infty} \dfrac{t^k}{k!}$. Hence, $$0 \leq \lim_{n \to \infty} \dfrac{\log^p n}{n} = \lim_{t \to \infty} \dfrac{t^p}{e^t} < \lim_{t \to \infty} \dfrac{r!}{t^{r-p}} \leq \lim_{t \to \infty}\dfrac{r!}{t} = 0$$

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Replace $n$ by $2^n$ and take the limit to $\infty$.

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That, assuming the sequence in $n$ converges. – Pedro Tamaroff Jun 28 '12 at 18:29

Let $\displaystyle a(n) = \frac{\log^p n}{n}$. Notice that $\displaystyle \frac{a(n^2)}{a(n)} = \frac{2^p}{n}$.

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hi. My first thought was to replace $n$ by $2^n$. Then you're done. – user 1618033 Jun 28 '12 at 18:12
    
@Chris: There are many ways to go here! Another choice would be $n\to e^n$. – user26872 Jun 28 '12 at 18:18
    
well, that way is already used by Marvis. :-) – user 1618033 Jun 28 '12 at 18:20
    
Oops. Sometimes I am not a careful reader! – user26872 Jun 28 '12 at 18:24
    
hehe. Neither do i all the time. – user 1618033 Jun 28 '12 at 18:26

Apply $\,[p]+1\,$ times L'Hospital's rule to$\,\displaystyle{f(x):=\frac{\log^px}{x}}$: $$\lim_{x\to\infty}\frac{\log^px}{x}\stackrel{L'H}=\lim_{x\to\infty}p\frac{\log^{p-1}(x)}{x}\stackrel{L'H}=\lim_{x\to\infty}p(p-1)\frac{\log^{p-2}(x)}{x}\stackrel{L'H}=...\stackrel{L'H}=$$ $$\stackrel{L'H}=\lim_{x\to\infty}p(p-1)(p-2)...(p-[p])\frac{\log^{p-[p]-1}(x)}{x}=0$$ since $\,p-[p]-1<0\,$

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Let $A=\lim_{n\to\infty} \frac{(\log n)^p}{n}$.Taking $log$ both sides gives $\log A=\lim_{n\to\infty} (p\log \log n -\log n)=\lim_{\log n\to\infty} (p\log \log n -\log n) $.Take $y=1/\log n\implies \log A=\lim_{y\to0^+} (-p\log y -1/y) \implies \log A=\lim_{y\to0^+} \frac{-py\log y -1}{y}$.This limit surely goes to $-\infty$ and hence $\log A=-\infty\implies A=0$.Thus the limit is $0$.

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