How do I determine the divergence/convergence of $\sum_n \frac{1}{\log(\log(n))}$?

Question

I am working through some problems in Durrett's probability book, and one of them involves a variant of the law of iterated logarithm.

I've managed to reduce the result to showing that

$$\sum_n \frac{1}{\log \log (n)}\exp(-\log \log(n)) < \infty$$

Using upperbounds for tail probabilities of standard normal. But, as I'm really not good with this stuff, I'm unsure how I am supposed to show this (if it's indeed true?).

Clearly without the exponential, the series would diverge since $\log (n) \leq n$. But, with the exponential it seems as though this could converge.

Could someone advise me how to complete this last step?

Answer 1

We can simplify $$\exp(- \log \log n) = \frac{1}{e^{\log \log n}} = \frac{1}{\log n}$$ Since $(\log \log n) \log n \leq n$, this diverges.

Answer 2

$$ \frac 1{(\log \log n )\exp(\log \log n)} = \frac 1 {\log \log n \cdot \log n} \geqslant \frac 1{\log ^2 n} \geqslant \frac 1{n^{1/2 \times 2}} = \frac 1n, $$ so it still diverges.

Answer 3

$\sum_n \frac{1}{n}$ is divergent. So is $\sum_n \frac{1}{\log(\log(n))}$

But I feel $\sum_n \frac{\exp(-\log(\log(n))}{\log(\log(n))}$ might still be convergent, you just need to find another way to prove it.

Answer 4

Cauchy condensation shows $$\sum_n \frac{1}{\log(\log(n))} \sim \sum_n \frac{2^n}{\log(\log(2^n))} = \sum_n \frac{2^n}{\log(n\log(2))} = \sum_n \frac{2^n}{\log(n) + \log(\log(2))} $$ So, it is divergent.

Answer 5

Credits to user MoonKnight.

$\log n <n ; $

And once more:

$\log (\log n) \lt \log n \lt n.$

$\dfrac {1}{n} \lt \dfrac{1}{\log n} \lt \dfrac{1}{\log (\log n)}.$

Comparison test.