Evaluate $\lim_{n\to \infty} \frac{2^{\ln(\ln(n))}}{n\ln(n)}$

Question

I'm having trouble evaluating the limit: $$\lim_{n\to \infty} \frac{2^{\ln(\ln(n))}}{n\ln(n)}$$ (as it looks like, the limit tends to $0$)

This is what I got until now: $$0\le \lim_{n\to \infty} \frac{2^{\ln(\ln(n))}}{n\ln(n)}\le \lim_{n\to \infty} \frac{2^{\ln(n)}}{n\ln(n)}=\lim_{n\to \infty} \frac{e^{\ln{2^{\ln(n)}}}}{n\ln(n)} = \lim_{n\to \infty} \frac{e^{\ln(n)\ln{2}}}{n\ln(n)}$$

Tried L'Hopital from here, but it seems usefulness.

Would appreciate your advice.

Answer 1

You're almost there. $$\frac{e^{\ln(n)\ln(2)}}{n\ln(n)}= \frac{(e^{\ln(n)})^{\ln(2)}}{n\ln(n)}=\frac{n^{\ln2}}{n\ln(n)} = \frac{n^{\ln2 -1}}{\ln(n)}$$ which should tend to zero as $\ln2<1$.

Answer 2

You can transform in the exponential form:

$$2^{\ln(\ln(n))} = e^{\ln(\ln(n))\ln(2)}$$

$$n\ln(n) = e^{\ln(n\ln(n))}$$

Thence your function becomes (using exponential and log properties)

$$e^{\ln(\ln(n))\ln(2) - \ln(n) - \ln(\ln(n))}$$

That is

$$e^{[\ln(2) - 1]\ln(\ln(n)) - \ln(n)}$$

Now, since $\ln(\ln(n))$ grows really slower with respect to $\ln(n)$ you may think that the second terms dominates, thence it remains

$$e^{-\ln(n)} ~~~~~ \to 0 ~~ \text{as} ~~ n\to \infty$$