Does the series converge

Question

We know that all series of the following form diverge: \begin{equation} S_k = \sum_{n=\left\lceil \mathrm{e}^k \right\rceil}^\infty \frac{1}{n (\ln n) (\ln \ln n)\dots(\ln^k n)} \end{equation} where the notation $\ln^k$ is function composition. Additionally, this series is interesting because for large $k$, it diverges very slowly.

Then does the following series converge? \begin{equation} S = \sum_{n=1}^\infty \frac{1}{f(n)} \end{equation} where $f(n)$ is defined recursively by the monotonic function \begin{equation} f(n) = \begin{cases} n & \text{if }n \le \mathrm{e} \\ n \cdot f{\left(\ln n\right)} & \text{otherwise} \end{cases} \end{equation}

The first few terms of this series are \begin{equation} 1 + \frac{1}{2} + \frac{1}{3 \ln 3} + \frac{1}{4 \ln 4} + \dots + \frac{1}{16 (\ln 16) (\ln \ln 16)} + \dots \end{equation}

and the maximum power of $\ln$ in the denominator gradually increases.

This series should eventually grow slower than all $S_k$ (it is in some kind of sense a limit) because the terms will eventually drop below those of $S_k$ for any particular $k$. But at the same time it seems to grow faster than all series of the form \begin{equation} T_k = \sum_{n=\left\lceil \mathrm{e}^k \right\rceil}^\infty \frac{1}{n (\ln n) (\ln \ln n)\dots(\ln^{k-1} n){(\ln^k n)}^2} \end{equation} which are known to converge, albeit also very slowly.

Answer 1

Since $f$ is monotonic, we can use the integral comparison. For any fixed $u \geqslant e$, we have

$$\int_u^{e^u} \frac{dx}{f(x)} = \int_u^{e^u} \frac{1}{f(\ln x)} \frac{dx}{x} = \int_{\ln u}^u \frac{dt}{f(t)}.$$

Since $\int_1^e \frac{dx}{f(x)} > 0$, we have

$$\int_1^{\infty} \frac{dx}{f(x)} = \sum_{k = 0}^{\infty} \int_{x_k}^{x_{k+1}} \frac{dx}{f(x)} = \sum_{k = 0}^{\infty} \int_1^e \frac{dx}{f(x)} = +\infty,$$

where $x_0 = 1$ and $x_{k+1} = e^{x_k}$ for all $k$. Thus the integral, and hence the series diverges. Very very slowly.