I had to graph it to see it. How can I prove it?
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What is your exact question? Do you want a proof of $\log^2(n)/\sqrt{n}\to 0$ or that of an inequality $\log^2 n < \sqrt{n}$ for large $n$? – Hanul Jeon Dec 18 '16 at 13:24
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I want to show that $logn*logn$ is of Ordo($\sqrt{n}$) – Samu Dec 18 '16 at 13:25
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It is the former statement I mentioned. I am going to give a proof. – Hanul Jeon Dec 18 '16 at 13:27
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@Hanul Jeon Thank you :) – Samu Dec 18 '16 at 13:27
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Before proving it let us examine the following inequality: $$\log x \le x.$$ It holds for any $x$. Now substitute $x$ to $x^{1/8}$ then we have $$\log x \le 8 x^{1/8}.$$
Now the proof is direct: you can see that $\log^2{n} / \sqrt{n}\le 64/n^{1/4}$ for any $n$.
Hanul Jeon
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When you make the substitution it is confusing to reuse the variable. It would be better to define $y=x^{1/8}$ (or do you want $y=x^8?)$ and rewrite in terms of $y$ – Ross Millikan Dec 18 '16 at 13:46
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Suppose we establish that $\log(x)<x$. This is quite trivial. Now let $x=y^{1/n}$,
$$\frac1n\log(y)=\log(y^{1/n})<y^{1/n}$$
$$\log(y)<ny^{1/n}$$
Square both sides.
$$\log^2(y)<n^2y^{2/n}=\mathcal O(y^{2/n})$$
Let $n>4$ and we are done.
Simply Beautiful Art
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What is true is more general: for any $\alpha,\beta>0$, $\;\ln^{\alpha} x=o(x^\beta)\;$ near $\;\infty$.
This is easily deduced from the basic case $\alpha=\beta=1$, or directly proved showing that, if $x$ is large enough, $$\ln x < x^{\tfrac{\beta}{2\alpha}}$$ (compare the derivatives of both sides of the inequality, and use a corollary of the mean-value theorem).
Bernard
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