Can anyone help me show that $ \sum_{n=2}^{\infty} \frac{\cos(\pi n)}{\log(\log(n))}$ isn't absolutely convergent?

Question

I have to show that $\displaystyle \sum_{n=2}^{\infty} \frac{\cos(\pi n)}{\log(\log(n))}$ isn't absolutely convergent.

I thought about doing the following:

$\begin{align}\displaystyle\sum_{n=2}^{\infty}\left|\frac{\cos(\pi n)}{\log(\log(n))}\right|&=\sum_{n=2}^{\infty} \frac{1}{|\log(\log(n))|} \geq \sum_{n=2}^{\infty} \frac{1}{|\log(n-1)|} \geq\\&\geq\sum_{n=2}^{\infty} \frac{1}{n}.\end{align}$

My problem is, that I don't know if

$\big|\log(\log(n))\big|\leq\big|\log(n-1)\big|\leq n$

is a true statement. Can anybody help me out?

Answer 1

Hint:

$$\cos\pi n=(-1)^n\;,\;\;\text{and then something about Leibniz...}$$

The above refers to the original question in the body of the question. About absolute convergence, as it appears some 15 minutes later: you can easily prove it doesn't happen with Cauchy's Condensation Test (work this out)

Answer 2

Things are actually simpler than you've made them out to be. The general inequality $\log x\lt x$ holds for all $x\gt0$. This tells us that

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

for all $n\gt1$. If $n\gt2$ all three quantities are positive (i.e., $3\gt e$, so $\log3\gt1$, so $\log(\log3)\gt0$), so one has

$${1\over|\log(\log n)|}\gt{1\over n}\quad\text{for }n\ge3$$

which is enough to show that your series is not absolutely convergent.

Remark: the inequality $|\log(\log n)|\lt n$ actually holds for $n=2$ as well, even though $\log(\log2)$ is negative. but we don't need to know this for the question of determining convergence/divergence of the infinite series.