# Order of $f(n) = \sqrt[\log n]{n} \cdot n^{\sqrt[\log n]{n}}$

This is as far as i've gone: $$f(n) = \sqrt[\log n]{n} \cdot n^{\sqrt[\log n]{n}} \iff f(n)^{1 / \sqrt[\log n]{n}} = (\sqrt[\log n]{n})^{1 / \sqrt[\log n]{n} } \cdot n \iff f(n) = n^{\sqrt[\log n]{n}}$$

since $\lim \limits_{x\to \infty} \sqrt[x]{x} = 1$

Now i haven't been able to proceed from there. If i divide $f(n)$ by $\sqrt[\log n]{n}$ i will prove that it's $\mathcal{O}\left(\sqrt[\log n]{n}\right)$ but i think i need to do better than that.. Help?

$$\sqrt[\log n]{n} = n^{1/\log(n)} = \left(e^{\log(n)}\right)^{1/\log(n)} = e$$ So overall you have $f(n) = e \cdot n^e$
Observe that $$\log\left(\sqrt[\log n]{n}\right) = \log\left(n^{1 / \log n}\right) = 1$$
One may write, as $n \to \infty$, $$f(n) = \sqrt[\log n]{n} \cdot n^{\sqrt[\log n]{n}}=n^{\large \frac1{\log n}}\cdot n^{\large n^{\frac1{\log n}}}=e^{\large \frac{\log n}{\log n}}n^{\large e^{\frac{\log n}{\log n}}}=e \cdot n^{\large e^1}\sim e \cdot n^{e}.$$
• $f(n) = e \cdot n^e$ is true for all $n$ right? – rookie Mar 28 '17 at 10:19
• @stud_iisc $f(n)$ is meaningless for $n=0$ and for $n=1$. As the OP asked for an order of $f(n)$, I've then considered that $n\to \infty$. – Olivier Oloa Mar 28 '17 at 12:25
• Yes I understand that the order of $f(n)$ is what we need to find. And yeah $f(n)$ in it's original form is meaningless for $n=0$ or $1$. I just wanted to say, shouldn't the fifth similarity be just an equality? Otherwise the answer is sound. – rookie Mar 28 '17 at 14:44