How to estimate $n/e^{\sqrt{\log n}}$?

Question

Suppose $f(n)= \frac{n}{e^{\sqrt{\log n}}}.$ So my question is: how can I simplify $f(n).$ Is it possible to write it down like $n^{1-c}$ where is $c$ is a constant?

Answer 1

You can write this as

$$f(n) = \frac{e^{\log n}}{e^{\sqrt{\log n}}} = \exp (\log n - \sqrt{\log n})$$

Now for any $\epsilon > 0$, we eventually have $\log(n) - \sqrt{\log n} > (1 - \epsilon) \log n$, and so

$$f(n) > \exp( (1 - \epsilon) \log n) = n^{1 - \epsilon}$$ eventually. So although $f(n) / n \to 0$, $f(n)$ cannot be bounded above by any power function below $n$.

Answer 2

If you let $n=e^{x^2}$ (where $x$ is large, assuming you're interested in the asymptotics as $n\to\infty$), then $\sqrt{\log n}=\sqrt{x^2}=x$ and thus

$$f(n)=n/e^{\sqrt{\log n}}=e^{x^2}/e^x=e^{x^2-x}\gt e^{rx^2}=n^r$$

when $x$ is sufficient large, for any $r\lt1$. Write $r=1-c$ with $0\lt c$ and you're done.