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?

  • $\begingroup$ consider using the alternating series test and $\cos(\pi n) = (-1)^n$ $\endgroup$
    – fGDu94
    Commented May 24, 2021 at 19:19
  • $\begingroup$ The title asks to show (conditional) convergence, while the text says you want to show the series is not absolutely convergent. Which question are you actually asking? (Please edit the question and/or title accordingly.) $\endgroup$ Commented May 24, 2021 at 19:22
  • $\begingroup$ Sorry, I see the confusion. It's the one in the text $\endgroup$
    – user926600
    Commented May 24, 2021 at 19:24

2 Answers 2



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

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)

  • $\begingroup$ Note, the body of the question poses a different question than the title. $\endgroup$ Commented May 24, 2021 at 19:25
  • $\begingroup$ @BarryCipra True, in the original post it was that way. It's now been changed to non-absolutely convergent... $\endgroup$
    – DonAntonio
    Commented May 24, 2021 at 19:34

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.


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