Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Does the following series converge:

$$\sum_{n=3}^\infty \frac {1}{(\log n)^{\log(\log(n)}}$$

I've tried all test I know... Any ideas ?

share|cite|improve this question
Have you tried $\lim_{n\to\infty}n\cdot u_n$? – Babak S. Mar 5 '13 at 19:32
can this be done using cauchy's condensation test? – MSEoris Mar 5 '13 at 19:52
André Nicolas has already answered this question, but if you're interested in seeing some motivation for a slighly different approach, see this post. – Dave L. Renfro Mar 6 '13 at 16:24
up vote 8 down vote accepted

Note that $(\log n)^{-\operatorname{loglog} n}=e^{-(\operatorname{loglog} n)^2}$, since $\log n=e^{\operatorname{loglog} n}$.

For large $n$, we have $(\operatorname{loglog} n)^2\lt \log n$, so for large $n$ the $n$-th term is greater than $\frac{1}{n}$.

The fact that $(\operatorname{loglog} n)^2$ is eventually dominated by $\log n$ is just the familiar fact that $e^x\gt x^2$ for large enough $x$.

Remark: In dealing with convergence of series, it is often better to ask oneself first: How fast are the terms approaching $0$? Looking instead for a test to use tends to distance us from the concrete reality of the series.

share|cite|improve this answer
Interesting philosophy in the remark, are there any specific schools of thought using similar views? – Arjang Mar 5 '13 at 20:17
The remark was not really of a philosophical nature, though it could be interpreted as a tilt towards Platonism as opposed to Formalism. However, what was meant is that in problem-solving, concrete confrontation of the problem can be more effective than searching through the toolchest for a suitable tool. – André Nicolas Mar 5 '13 at 20:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.