I want to find an example of a sequence $a_n >0$ such that its series $\sum a_n$ diverges, but when we consider $$\sum_{i=2}^n \frac{a_n}{\ln(\ln(n))}$$ then this series is in fact convergent.
Any idea what to start with, what motivation would you use to find such a sequence?
Take $$a_n=\frac{1}{(n\, \ln n) \ln( \ln n)}$$ Then $$\sum_{n=3}^\infty a_n= \sum_{n=3}^\infty \frac{1}{(n \,\ln n) \ln( \ln n)} $$ diverges , but $$\sum_{n=3}^\infty \frac{a_n}{\ln(\ln n)}= \sum_{n=3}^\infty\frac{1}{(n\,\ln n) \ln( \ln n)}. \frac{1}{\ln(\ln n)}=\sum_{n=3}^\infty \frac{1}{n \,\ln n [\ln (\ln n)]^2}< \infty$$
The convergence and divergence are follows from:
Theorem 1: Suppose $a_1 \geq a_2 \geq \cdots \geq 0. $ Then $\sum a_n < \infty \iff \sum 2^k a_{2^k} < \infty$
Theorem 2: If $p>1$, then $$\sum \frac{1}{n\, (\ln n)^p} $$ converges, and diverges otherwise