Prove divergence of series

Question

The question is to prove that the series $$\sum_{n=1}^\infty\frac1{4\log n\log\log n}$$ diverges.

I tried using the Cauchy condensation test but could not prove it.

Answer 1

By comparison: $$\log n \log \log n \leq n \log n$$ for all sufficiently big $n$, so $ \sum_n \frac{1}{\log n \log \log n} $ diverges by comparison with $ \sum_n \frac{1}{n \log n} $. (Which itself diverges by e.g. the integral test, noting that $\int \frac{dx}{x \log x } = \log\log x$.)

Answer 2

$\log{\log{2^n}}=\log{n}+\log{\log{2}}$

$\log{2^n}=n\log{2}$

You can use Cauchy's condensation theorem:

$$\sum_{n=2}^\infty\frac{2^n}{4n\log{n}\log{2}+n\log{2}\log{\log{2}}}$$

This series clrearly diverges.

Note that:

$$\sum_{n=2}^\infty\frac{2^n}{4n\log{n}\log{2}+n\log{2}\log{\log{2}}}\geq \sum_{n=2}^\infty\frac{2^n}{8n\log{n}\log{2}}\geq \sum_{n=1}^\infty\frac{2^n}{8n^2 \log{2}}$$

Answer 3

For large $n,$

$$\ln n\cdot \ln \ln n < (\ln n)^2 < (\sqrt n)^2 =n.$$

Taking reciprocals and comparing to the harmonic series then shows our series diverges.