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I tried finding a few inequalities but all didn't seem to get me anywhere.
My try:
By $\log(x)<x \quad \forall x\ge 0$ we can see pretty easily that
$\log_2(n)^a < n^a$
or in another way - since $\log_2(n)=2\log_2(\sqrt{n}) < 2\sqrt{n}$ so
$ \log_2(n)^a = 2^a\log_2(\sqrt{n})^a < 2^a\sqrt{n}^a$
Am I missing something?
How can I reduce the power of $a$? I tried using the limits definition but also got stuck

Thank you in advance

Giorgos Giapitzakis
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Roach87
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1 Answers1

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Let set $f(x)=\dfrac{\log_2(x)^a}{\sqrt{x}}$

Notice that $f'(x)=\dfrac{\overbrace{\log_2(x)^a}^{>0}\overbrace{(2a-\ln(x))}^{<0}}{\underbrace{2x\sqrt{x}\ln(x)}_{>0}}<0\ $ at infinity (i.e. when considering $x\gg 1$).

Since $f$ is continuous on $[1,+\infty)$ and $\searrow\ $ at infinity and $f(x)\ge 0$ , then $f$ is bounded by some constant $M$ on this interval.

Now exploit the additive property of the logarithm:

$0\le f(x^2)=\dfrac{\log_2(x^2)^a}{\sqrt{x^2}}=\dfrac{2^a\log_2(x)^a}{\sqrt{x}\sqrt{x}}=\dfrac{2^af(x)}{\sqrt{x}}\le \dfrac{\overbrace{2^aM}^\text{bounded}}{\sqrt{x}}\to 0\ $ at infinity.

Apply to $x=\sqrt{n}$ to get $f(n)\to 0\iff \log_2(n)^a=o(\sqrt{n})$

zwim
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