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For any positive integer $n$, show that $x\gg 1 \rightarrow \log{x} < \sqrt[n]{x}$

I approached this problem by showing that $\lim_{x \to \infty} \frac{\log{x}}{\sqrt[n]{x}}= 0$

I tried to show that the limit converges to zero by showing that the integral test, $$\int_{1}^{\infty}{\log{x}\over\sqrt[n]{x}} dx $$ converges, which would in turn mean that the limit converges to zero (due to the $n^{\text{th}}$-term test)

But I can't get how to solve this improper integral, and also don't know whether my approach is right in the first place.

Hanul Jeon
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the24
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2 Answers2

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By L'Hopital's rule $\lim \frac {\log \, x} {\sqrt [n] {x}}=\lim \frac {1/x} {\frac 1 n x^{\frac 1 n -1}}=\lim \frac n {x^{\frac 1 n}}=0$.

Kavi Rama Murthy
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  • oh my I missed that one. What if I just went with the integral test? How would I solve that improper integral? – the24 Jul 22 '19 at 07:58
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    Integral cannot be used here. Mere existence of $\int_1^{\infty} f(x)dx$ does not guarantee that $f(x) \to 0$ as $x \to \infty$. – Kavi Rama Murthy Jul 22 '19 at 08:00
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Using the mother of all inequalities abouzt the exponential, $$ e^t\ge 1+t,$$ and multiplicativity, we find that for $y>(n+1)^{n+1}$, $$ e^y=\left(e^{\frac y{n+1}}\right)^{n+1}\ge \left(1+\frac y{n+1}\right)^{n+1}>\frac{y^{n+1}}{(n+1)^{n+1}}>y^n,$$ or: $$ \sqrt[n]{e^x}>y.$$ Hence for $x>e^{(n+1)^{n+1}}$, we have (setting $y=\ln x$) $$\sqrt[n]x>\ln x.$$

Hagen von Eitzen
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