Is $n^{1/\log \log n} = O(1)$?

Question

Is $n^{1/\log \log n} = O(1)$ ?

Suppose that $n^{1/\log \log n} = c$ where $c$ is constant.

Taking logs of both sides,

$$\frac{1}{\log \log n}\log n = \log c.$$

I am not able to spot an error. Please help

Answer 1

The function $n^{1/\log \log n}$ tends to infinity, since $$ n^{1/\log\log n} = e^{\log n/\log\log n}, $$ and $\log n/\log \log n \longrightarrow \infty$.

Answer 2

In your last step, a contradiction is reached since (by L'Hopital) $$\lim_{n\to\infty}\frac{\log n}{\log\log n}=\lim_{n\to\infty}\frac{1/n}{1/(n\log n)}=\lim_{n\to\infty}\log n$$ and $\log$ is unbounded.

Answer 3

There isn't an error. You supposed that $n^{1/\log\log n}$ is constant and you derived the contradiction that $\log n/\log\log n$ is constant. Therefore, your supposition must be false.

It would have been better to have written that $n^{1/\log\log n}\leq c$ for all $n$, since you're really trying to show that the function is bounded above by a constant, not that is constant.

Answer 4

By way of contradiction, suppose $n^{1/ \log \log n} = O(1)$. Then, by definition of the big-O notation, there exist positive constants $c$ and $n_0$ such that $n^{1/\log \log n} \le c$ for all $n \ge n_0$. Taking log of both sides, we obtain that $\frac{\log n}{ \log \log n}$ is at most some constant $c' = \log c$ for all sufficiently large $n$. But this is impossible because it can be shown (for eg by L'Hopital's rule) that $\lim_{n \rightarrow \infty} \frac{\log n}{\log \log n} = \infty$.