Tight bound for $e^{\sqrt{\log n}}$

Question

I want to get upper and lower bounds of $e^{\sqrt{\log n}}$. It is trivial that $e^{\sqrt{\log n}}\leq e^{\log n}=n$. But is it a tight bound? i.e., $e^{\sqrt{\log n}}=\Theta(n)$ or $e^{\sqrt{\log n}}=\Theta(n^{1-c})$ for some $c$?

Answer 1

Say you look for $\alpha$ such that $e^{\sqrt{\ln n}} = \Theta(n^\alpha)$. This means the following is finite:

$$\lim_{n \to \infty} \frac{e^{\sqrt{\ln n}}}{n^\alpha}$$

Simplify the fraction by taking logarithms:

$$\sqrt{\ln n} - \alpha \ln n$$

It is seen that this tends to $-\infty$ as $n \to \infty$, whatever $\alpha > 0$ may be.

Would need more details to know if these (sloppy) bounds are enough for you...

Answer 2

Since $\log(n) <n^c/c$ for any $c>0$, $e^{\sqrt{\log(n)} } <e^{n^{c/2}/\sqrt{c}}$.