# Convergence of a general Bertrand series

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 $n$.

Your sum is larger than $$\sum_{n=2}^\infty \frac{1}{ n \log^1n \log^2n \log^3n \cdots\log^{TL(n)}n}$$ which is divergent by Cauchy condensation test.
On a second thought, this is clearly wrong as the summand depends on $$n$$. I will delete this answer soon.