Can we say that $ 2^\frac{n}{\log(n)} \sim 2^\frac{\log(n)}{\log(\log(n))}$?

Question

Can we assert and proove that :

$$ 2^\frac{n}{\log(n)} \sim 2^\frac{\log(n)}{\log(\log(n))}$$ And What inequality relating two parts can be proved ?

Answer 1

No, it's false since

$$ \frac{2^{n/\log n}}{2^{\log n / \log\log n}} = 2^{\frac{n}{\log n} - \frac{\log n}{\log\log n}} \not\to 1 $$

as $n \to \infty$. In fact the ratio tends to $\infty$ since

$$ \frac{n}{\log n} - \frac{\log n}{\log\log n} \to \infty $$

as $n \to \infty$, so for any $C > 0$ there will be an $N > 0$ such that

$$ 2^\frac{n}{\log n} > C 2^\frac{\log n}{\log\log n} $$

for all $n > N$.