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

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?

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.

$$\sum_n \frac{1}{n}$$ 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.

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

• Oh, ok. Thanks a lot. Do you have any advice on how to show $\sum_n P(Z>\sqrt{2log log n}(1+\epsilon)) < \infty$ for $Z \sim N(0,1)$? Dec 1, 2018 at 6:21
• @Xiaomi Sorry, I suck at probability theory.
– xbh
Dec 1, 2018 at 7:15

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.

• Thanks a lot. Do you have any advice on how to show $\sum_n P(Z>\sqrt{2log log n}(1+\epsilon)) < \infty$ for $Z \sim N(0,1)$? The common probability inequalities do not seem tight enough to show this.. Dec 1, 2018 at 6:24
• You may use the integral for $N(0,1)$ for each member of the series and estimate it: $$P(Z>\sqrt{2\log \log n}(1+\epsilon)) \sim \int_{\sqrt{2\log \log n}(1+\epsilon)}^{\infty} e^{\frac{-x^2}{2}}dx$$ Dec 1, 2018 at 6:48

Credits to user MoonKnight.

$$\log 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.