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I am trying to test the convergence of this series from exercise 8.15(j) in Mathematical Analysis by Apostol:

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

I tried every kind of test. I know it should be possible to use the comparison test but I have no idea on how to proceed. Could you just give me a hint?

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2 Answers 2

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Note that, for every $n$ large enough, $$(\log\log n)^{\log\log n}\leqslant(\log n)^{\log\log n}=\exp((\log\log n)^2)\leqslant\exp(\log n)=n,$$ provided, for every $k$ large enough, $$\log k\leqslant\sqrt{k},$$ an inequality you can probably show, used for $k=\log n$. Hence, for every $n$ large enough, $$\frac1{(\log\log n)^{\log\log n}}\geqslant\frac1n,$$ and the series...

...diverges.

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  • $\begingroup$ Great! How did you discover that $\log \log n \leq \log n$ and $(\log \log n)^2\leq \log n$? $\endgroup$
    – Charlie
    Aug 31, 2014 at 14:09
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    $\begingroup$ I "discovered" nothing, simply I know that every power of $\log k$ is negligible with respect to $k$ and I applied this to some suitable $k\to\infty$. $\endgroup$
    – Did
    Aug 31, 2014 at 14:11
  • $\begingroup$ In addition I cannot get the last part of your answer. Why did you write "provided that for $k$ large enough $\log k \leq \sqrt k$ etc... ? $\endgroup$
    – Charlie
    Aug 31, 2014 at 14:13
  • $\begingroup$ Because this is what the second inequality in the chain of equalities-inequalities is based upon. $\endgroup$
    – Did
    Aug 31, 2014 at 14:15
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    $\begingroup$ @Charlie Let $f_{\beta,\gamma}(x)=x^{\beta}(\log(x))^\gamma$, if $\beta\ne 0$ then we have always $$\lim_{x\rightarrow +\infty}f_{\beta,\gamma}(x)=\lim_{x\rightarrow +\infty}x^{\beta}$$ $\endgroup$
    – Krokop
    Aug 31, 2014 at 14:22
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To avoid relying on precise estimates, one can apply Cauchy's condensation test whenever one has to check the convergence of a series which contains several (iterated) logarithms. Usually, this works quite well:

Applying Cauchy's condensation test, we find that convergence of the series

$$\sum_{n \geq 3} \frac{1}{(\log \log n)^{\log \log n}} \tag{1}$$

is equivalent to the convergence of

$$ \sum_{n \geq 3} \frac{2^n}{(\log n)^{\log n}} \tag{2}$$

It is not difficult to see that $$\frac{2^n}{(\log n)^{\log n}}$$

is an increasing (strictly positive) sequence for sufficiently large $n$, e.g. by checking that

$$\frac{d}{dx} \left(\frac{2^x}{(\log x)^{\log x}}\right) > 0.$$

This shows that $(2)$ does not converge; hence, $(1)$ does not converge.

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  • $\begingroup$ Hmmm... But, relying on "precise estimates" for the last limit, are we? (Just kidding, +1.) $\endgroup$
    – Did
    Sep 1, 2014 at 9:11
  • $\begingroup$ @saz Do you mind if I ask you how to prove that the limit is actually infinity? I tried to use l'Hopital's rule with no success... $\endgroup$
    – Charlie
    Sep 1, 2014 at 12:31
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    $\begingroup$ @Charlie Set $$f(x) := \frac{2^x}{(\log x)^{\log x}}$$ and show that $f'(x) \geq 0$. (Hopital isn't very helpful in this case, since we end up with a limit of a similar form.) $\endgroup$
    – saz
    Sep 1, 2014 at 13:54

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