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Can someone explain me the strategy for calculating this limit and other limit like this?$$\lim_{n\to\infty}\frac{n}{(\ln n)^{-p}}.$$

I have tried with the squeeze rule, l'Hopital and other simple strategies without success.

Thank you.

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  • $\begingroup$ What is your $p$? $\endgroup$ – ploosu2 Jan 24 '14 at 15:13
  • $\begingroup$ @ploosu $p$ is a fixed real number. Thank for your help. $\endgroup$ – Charlie Jan 24 '14 at 15:15
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L'Hopital's rule will get you there.

$$ \begin{align} \lim_{n\to \infty}{\frac{n}{(\ln{n})^{-p}}} &= \lim_{n\to \infty}{\frac{f(n)}{g(n)}} \\ &= \lim_{n\to \infty}{\frac{f'(n)}{g'(n)}} \\ \end{align} $$

For $p\lt 0$, the power in the denominator is positive. This means we get: $$ f'(n) = 1\\ g'(n) = (-p)*(\ln{n})^{-p-1}*\frac{1}{n} $$ Simplifying our equation: $$ \lim_{n\to \infty}{\frac{n}{(\ln{n})^{-p}}} = \lim_{n\to \infty}{\frac{n}{-p(\ln{n})^{-p-1}}} \\ $$ This can be continued until the power in the denominator is zero, leaving only a constant value $C$. Therefore: $$ \lim_{n\to \infty}{\frac{n}{(\ln{n})^{-p}}} = \lim_{n\to \infty}{\frac{n}{C}} = \infty \\ $$

For $p>0$, the power in the denominator will be negative. This simplifies our equation to: $$ \lim_{n\to \infty}{\frac{n}{(\ln{n})^{-p}}} = \lim_{n\to \infty}{n(\ln{n})^{p}} = \infty \\ $$

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$$\lim_{n\to\infty}\frac n{\log^{-p}n}=\infty\;,\;\;\forall\,p\in\Bbb R$$

You can show the above by cases: for $\;p\ge 0\;$ it's trivial since the product of sequences diverging to infinity diverges to infinity, and for $\;p<0\;$ you can prove it by means of l'Hospital's rule ($\;[k]+1\;$ times) and the function

$$\frac x{\log^kx}\;,\;\;k>0$$

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  • $\begingroup$ @AlexR No, technically the above covers all the cases. l'H's rule is only for the case $\;p<0\;$ , the other case being trivial, as pointed out there. $\endgroup$ – DonAntonio Jan 24 '14 at 15:27
  • $\begingroup$ Okay I see that you have to use $$\frac n{\log^{-p} n} = \frac n{\log^{\lceil p \rceil} n} \cdot \frac n{\log^{-p-\lceil p \rceil} n}$$ To complete it for any $p$. I suggest you add that as a remark. $\endgroup$ – AlexR Jan 24 '14 at 15:29
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    $\begingroup$ @AlexR, I think you're really confussing things here: I don't need any $\;p\;$ in my last line. I just used $\;k>0\;$ ...who cares how we call constants/variables? Oh, I see...because of the k times thing? Well, that's a little...well, I shall edit. Thanks. $\endgroup$ – DonAntonio Jan 24 '14 at 15:31
  • $\begingroup$ That was exactly my point. (+1) $\endgroup$ – AlexR Jan 24 '14 at 15:33
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Ok, the $p$ doesn't even matter. If it's positive you have $n(\ln n)^p$ which clearly goes to infinity as $n \to \infty$. If it's negative you still have the $n$ upstairs and $n$ grows much faster than $(\ln n)^{-p}$.

Make the substitution $s = \ln n$. It is equivalent to consider this subsituted version when $s \to \infty$.

You have $n = e^s$ and the function you are considering becomes

$\frac{e^s}{s^{-p}}$

Are you familiar with this sort of function as $s \to \infty$?

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Case I. $p\ge 0$. Then $$ \frac{n}{(\ln n)^{-p}}=n(\ln n)^{p}>n, $$ and as $n\to\infty$, so does $\frac{n}{(\ln n)^{-p}}$.

Case II. $p<0.$ Then for $q=-p>0$, let $$ f(x)=x(\ln x)^q. $$ Clearly, $$ f'(x)=(\ln x)^q+qx(\ln x)^{p-1}\ge q\mathrm{e}, \quad \text{for}\quad x\ge \mathrm{e}, $$ and thus $$ f(x)=f(\mathrm{e})+\int_{\mathrm{e}}^x f'(t)\,dt\ge \,f(\mathrm{e})+(x-\mathrm{e})q\mathrm{e} \to \infty, $$ as $x\to\infty$.

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