# Convergence of the series $\sum\limits_{n=3}^\infty (\log\log n)^{-\log\log n}$

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?

## 2 Answers

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.

• Great! How did you discover that $\log \log n \leq \log n$ and $(\log \log n)^2\leq \log n$? – Charlie Aug 31 '14 at 14:09
• 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 $k\to\infty$. – Did Aug 31 '14 at 14:11
• 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... ? – Charlie Aug 31 '14 at 14:13
• Because this is what the second inequality in the chain of equalities-inequalities is based upon. – Did Aug 31 '14 at 14:15
• @Charlie Let $f_{\beta,\gamma}(x)=x^{\beta}(\log(x))^\gamma$, if $\beta\ne 0$ then we have always $$\lim_{x\rightarrow +\infty}f_{\beta,\gamma}(x)=\lim_{x\rightarrow +\infty}x^{\beta}$$ – Krokop Aug 31 '14 at 14:22

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.

• Hmmm... But, relying on "precise estimates" for the last limit, are we? (Just kidding, +1.) – Did Sep 1 '14 at 9:11
• @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... – Charlie Sep 1 '14 at 12:31
• @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.) – saz Sep 1 '14 at 13:54