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Is the sum $$ S= \sum_{n=2}^\infty \frac{1}{ \log^1n \log^2n \log^3n \cdots\log^{TL(n)}n} $$ convergent?
Here $\log^i n$ denotes the $i$'th iterate of $\log$ (in base 2) of $n$, so $\log^2n$ means $\log\log n$, etc., and $T(n)$ is the tower of $n$ (stack of $n$ 2's) defined by $T(1)=2$ $T(n+1)=2^{T(n)}$ for $n\ge1$. We then set $$TL(n) := \text{the "towerian log"} = \sup \bigl\{ k : T(k) \le n < T(k+1) \bigr\} .$$

MOTIVATION : Generalizing the following that are called Bertrand series (I think):The harmonic series $\sum_{n\ge1} 1/n$ is divergent, as are all of the series $$ \sum_{n\ge2} \frac{1}{n\log n},\quad \sum_{n\ge2} \frac{1}{n\log n\log^2 n},\quad \sum_{n\ge2} \frac{1}{n\log n\log^2n\log^3n},\ldots\,. $$

Here the product of iterated logs is pushed as far as possible, and its size depends on the parameter $n$.

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The sum diverges. This is Putnam Problem A4, 2008.

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  • $\begingroup$ Thank you for this fast and apparently accurate answer. I had made up this problem ten years ago didn't ask the right people. $\endgroup$ – Jérôme JEAN-CHARLES Jan 5 '10 at 2:44
  • $\begingroup$ Do you agree at a glance that the result (divergence) remains the same if we use floor function to round everything inside integer. Meaning replacing any the Log function by floor(log). Floor(x) being the highest integer less than x or equal to it. $\endgroup$ – Jérôme JEAN-CHARLES Sep 25 '10 at 15:10

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