How fast does $\exp(\sqrt{\ln(x)})$ grow?

Question

At this Wikipedia article I found that there is a multiplication algorithm which can multiply two $n$ digit numbers with a complexity of $$O\left(n2^{\sqrt{2\ln(n)}}\ln(n)\right).$$

I want to get a better grasp of this. More specifically, how fast does $$2^{\sqrt{2\ln(n)}}$$ grow?

Since we have $2^\sqrt{2}\approx e-0.05$, I am already happy with upper bounds on $$e^{\sqrt{\ln(n)}}.$$

For example, do we have $e^{\sqrt{\ln(n)}}<\sqrt{n}$ at some point? What about $e^{\sqrt{\ln(n)}}<\ln(n)$ for $n$ large enough?

Answer 1

Note that:

$$e^{\sqrt{\ln(x)}}=x^{1/\sqrt{\ln(x)}}$$

and thus this grows slower than $x^{1/r}$ for any $r>0$. Likewise, we also have

$$e^{\sqrt{\ln(x)}}=[\ln(x)]^{\sqrt{\ln(x)}/\ln(\ln(x))}$$

and thus this grows faster than $[\ln(x)]^r$ for any $r>0$. The same asymptotics hold for $2^{\sqrt{\ln(x)}}$ as well.