Limit of $\frac{\log(R)^p}{ R^n }$ as $ R \rightarrow \infty$

Question

I have been reading this old post:

Showing that an entire function with the following inequality is constant.

In the solution to the problem, we claim the following for $C$ constant, $n$ is a positive integer and $R$ is a positive, real number and $p >1$:

$$ \frac{n! C \; \log(R)^p}{R^n} \rightarrow 0 $$ for $R \rightarrow \infty $.

I have tried out L'Hôpitals rule as well as substituting R for an exponential function but with no useful results. I've done loads of problems like this in the past but I'm totally stuck on this one for some reason - a hint or two would be much appreciated!

Answer 1

Remark that it suffices to prove that for any $p>0$: $$\frac{(\log R)^p}{R} \to 0$$ when $R \to \infty.$ What happens when you apply L'Hopital's rule? Can you apply it again?

Answer 2

First of all, we can prove that $\;\color{brown}{z+1\leqslant e^z\,\text{ for all }\,z\in\Bbb Z}\,.\;\;\color{blue}{(1)}$

If $\;z\in\Bbb Z\;$ and $\;z\leqslant-1\;,\;$it results that $\;z+1\leqslant0<e^z\,.$
If $\;z\in\Bbb Z\;$ and $\;z\geqslant0\;,\;$ we proceed by induction on $\,z:$
$\quad$for $\;z=0\;,\;$ it results that $\;z+1=1=e^z\,;$
$\quad$we suppose that $\,(1)\,$ is true for $\;z=z_0\in\Bbb Z^+_0\,$ and prove
$\quad\text{that }\,(1)\,$ is also true for $\;z=z_0+1\,.$
$\quad z+1=(z_0\!+\!1)+1\leqslant e^{z_0}\!+1\leqslant2e^{z_0}\!<e\!\cdot\!e^{z_0}\!=e^{z_0+1}\!=e^z.$

Now, we will prove that $\;\color{brown}{\log x<x\;\text{ for any }x>0}\,.\quad\color{blue}{(2)}$

For any $\,x>0\,,\,$ there exists $\,z\in\Bbb Z\,$ such that $\,e^z\leqslant x<e^{z+1}.$
Hence, $\;\log x<\log e^{z+1}=z+1\underset{(1)}{\leqslant}e^z\leqslant x\,.$

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Let $\,C\,$ be a constant, let $\,n\,$ be a positive integer and $\,p>1\,.$

For any $\,R>1\,,\;$ it results that

$\left|\dfrac{n!C\big(\!\log R\big)^p}{R^n}\right|=\dfrac{n!|C|\big(\!\log R\big)^p}{R^n}=n!|C|\left(\!\dfrac{\log R}{R^{\frac np}}\!\right)^p=$

$=n!|C|\left(\!\dfrac{2p\log R^{\frac n{2p}}}{nR^{\frac np}}\!\right)^p\underset{\overbrace{\;(2)\,\text{ for }\,x=R^{\frac n{2p}}\;}}{\leqslant}n!|C|\left(\!\dfrac{2pR^{\frac n{2p}}}{nR^{\frac np}}\!\right)^p=$

$=n!|C|\left(\!\dfrac{2p}{nR^{\frac n{2p}}}\!\right)^p=\dfrac{n!|C|2^pp^p}{n^pR^{\frac n2}}=\dfrac{2^pp^pn!|C|}{n^p\sqrt{R^n}}\,.$

Since $\;\left|\dfrac{n!C\big(\!\log R\big)^p}{R^n}\right|\leqslant\dfrac{2^pp^pn!|C|}{n^p\sqrt{R^n}}\;$ for any $\;R>1\;$ and $\;\lim\limits_{R\to+\infty}\dfrac{2^pp^pn!|C|}{n^p\sqrt{R^n}}=0\;,\;$ by applying the Squeeze Theorem,
it follows that $\;\lim\limits_{R\to+\infty}\dfrac{n!C\big(\!\log R\big)^p}{R^n}=0\,.$