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 ?

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    $\begingroup$ Have you tried $\lim_{n\to\infty}n\cdot u_n$? $\endgroup$ – Mikasa Mar 5 '13 at 19:32
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    $\begingroup$ can this be done using cauchy's condensation test? $\endgroup$ – still_learning Mar 5 '13 at 19:52
  • $\begingroup$ André Nicolas has already answered this question, but if you're interested in seeing some motivation for a slighly different approach, see this post. $\endgroup$ – Dave L. Renfro Mar 6 '13 at 16:24

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.

  • $\begingroup$ Interesting philosophy in the remark, are there any specific schools of thought using similar views? $\endgroup$ – jimjim Mar 5 '13 at 20:17
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    $\begingroup$ 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. $\endgroup$ – André Nicolas Mar 5 '13 at 20:23

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