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 $n$.



Convergence of a general Bertrand series
The sum diverges. This is [Putnam Problem A4, 2008](http://kskedlaya.org/putnam-archive/). 


