Generalized $p$-Series and a Related Series The main question of this post is motivated by the following few paragraphs.
Motivation: Generalized $p$-Series
We define a family of sequences indexed by $k$ by $$a^{p}_{n,k}=\prod_{j = 0}^{k}\left(\log^{j} n\right)^{-p_{j}},$$ where $p = \left(p_{j}\right)$ is a sequence of real numbers. Here, we're taking $\log^{j}n$ to be the $j$-iterated logarithm ($\log \cdots \log n$ with $j$ iterations of $\log$), where $\log^{0} n = n$. This can be seen as a generalization of the $p$-series.
For example, with $k = 0,$ we have the typical $p$-series studied in calculus, $\sum_{n} n^{-p},$ which converges iff $p>1$.
On the other hand, we get an infinite family of divergent series by taking $p = \mathrm{\bf 1}$, the sequence of all $1$'s. For the first few $k$: $$\sum_{n} \frac{1}{n} = \infty,$$ $$\sum_{n} \frac{1}{n \log n} = \infty,$$ $$\sum_{n} \frac{1}{n \log n \log \log n} = \infty,$$ and so on.
Note that $a^{\mathrm{\bf 1}}_{n,k} \leq a^{\mathrm{\bf 1}}_{n, k'}$ for all $k\geq k'$ and $n$ where both terms are defined, so this is family of sequences gets (pointwise) smaller as $k\to\infty$, while their sums remain divergent.
This generalized $p$-series comes with the following convergence test (which is equivalent to the typical $p$-series test for $k = 0$):
Theorem. [Generalized $p$-Series Test] Let $a^{p}_{n,k}$ be defined as above. Then, $\sum_{n} a^{p}_{n,k}$ converges if and only if there exists $\ell \leq k$ such that $p_{\ell}>1$ and $p_{j}=1$ for all $j<\ell$.
Proof.
First, suppose that there exists $\ell \leq k$ such that $p_{\ell}>1$ and $p_{j}=1$ for all $j<\ell$.
Using the integral test, we test for convergence by comparing with $$\int_{n_{0}}^{\infty}\prod_{j = 0}^{k}\left(\log^{j} x\right)^{-p_{j}}dx,$$ where $n_{0}$ is chosen sufficiently large so that the highest iterated logarithm is defined and so that it is large enough anywhere else we need large values of $n_{0}$.
Suppose that $\ell$ is the first index exhibiting $p_{\ell} = 1 + \varepsilon$ for some $\varepsilon>0$. Substituting $u = \log^{\ell}x$, we get $du = \prod_{j=0}^{\ell-1}\left(\log^{j} x\right)^{-1}dx$. Hence, with $n_{0}'$ being the properly adjusted lower bound on the integral, our integral becomes $$\int_{n_{0}'}^{\infty}u^{-\left(1+\varepsilon\right)}\prod_{j = 1}^{k - \ell}\left(\log^{j}u\right)^{-p_{j + \ell}}du.$$
Since any polynomial dominates any logarithm asymptotically, and since our $n_{0}$ was luckily chosen large enough, we can sacrifice half an $\varepsilon$ from the exponent of $u$ to get $$\int_{n_{0}'}^{\infty}u^{-\left(1+\varepsilon\right)}\prod_{j = 1}^{k - \ell}\left(\log^{j}u\right)^{-p_{j + \ell}}du \leq \int_{n_{0}'}^{\infty}u^{-\left(1+\varepsilon/2\right)} du < \infty.$$
I leave the other direction of the proof, that convergence implies that there exists $\ell \leq k$ such that $p_{\ell}>1$ and $p_{j}=1$ for all $j<\ell$, to the reader.
QED
I also leave it as a (very tedious) exercise to the reader to prove this using the Cauchy condensation test.
Main Question
In the above discussion, we were considering sequences with a fixed number of logarithmic factors. Suppose that we allow for the number of logarithmic factors in the denominator to grow with $n$.
Define the sequence $$a^{p}_{n,*} = \prod_{j=0}^{\log^{*}n}\left(\log^{j}n\right)^{-p_{j}},$$ where $p=\left(p_{j}\right)$ is an infinite sequence of real numbers and $\ell = \log^{*}n$ (the iterated logarithm) is defined as the smallest positive integer so that $\log^{\ell}n\leq1$, with $\ell = 0$ for $n = 1$.
When does $\sum_{n}a^{p}_{n,*}$ converge?
Clearly, for all $k\geq0$, there exists $N>0$ such that $a^{p}_{n,*}\leq a^{p}_{n,k}$ for all $n\geq N$, since (eventually) there are more $\log$ factors in the denominator of $a^{p}_{n,*}$, so any $p$ that works for $\sum_{n} a^{p}_{n,k}$ also works for $\sum_{n} a^{p}_{n,*}$ by the comparison test.
On the other hand, if we define $\mathrm{\bf 1}_{k}$ to be the sequence of all $1$'s, except at index $k$ where it equals $2$, then $\sum_{n} a^{\mathrm{\bf 1}_{k}}_{n,k}$ converges for all $k\geq 0$, and $\mathrm{\bf 1}_{k}\to \mathrm{\bf 1}$ as $k\to\infty$.
Would the unbounded number of (further-iterated) $\log$ factors that eventually show up in the denominator allow for $\sum_{n}a^{\mathrm{\bf 1}}_{n,*}$ to converge, since $a^{p}_{n,*}$ feels, in some sense, like a limiting case of the $a^{p}_{n,k}$ (like how $\mathrm{\bf 1}_{k}\to\mathrm{\bf 1}$)?
 A: It turns out that my main question is very similar to Problem A4 of the 2008 Putnam exam.
A4. Define $f:\mathbb{R}\to\mathbb{R}$ by $$f(x) = \begin{cases} x & \mbox{if $x \leq e$} \\ x f(\log x) &
\mbox{if $x > e$.} \end{cases}$$ Does $\displaystyle{\sum_{n=1}^\infty \frac{1}{f(n)}}$ converge?
In the notation of my problem, if we let $\ell_{n} = \log^{*}n$, then $$\frac{1}{f\left(n\right)} = \prod_{j=0}^{\ell_{n} - 1}\left(\log^{j}n\right)^{-1},$$ which is $a^{\mathrm{\bf 1}}_{n,*}$ but without the final $\log^{\ell_{n}}n$ factor in the denominator. However, this is easily resolved by noting that $\log^{\ell_{n}}n\leq 1,$ by definition, and so $\frac{1}{f\left(n\right)} = a^{\mathrm{\bf 1}}_{n,*} \log^{\ell_{n}}n \leq a^{\mathrm{\bf 1}}_{n,*}$. So if we can show that the sum in A4 diverges, then $\sum_{n}a^{\mathrm{\bf 1}}_{n,*}$ also diverges by the comparison test.
Since it's clear that $\frac{1}{f\left(x\right)}$ is positive and decreasing with $x$ on $\left[1,\infty\right)$, we will use the integral test again!
To determine if $$\int_{1}^{\infty}\frac{dx}{f\left(x\right)}$$ diverges or not, let's partition $\left[1,\infty\right)$ into $\left[1,e\right)$, $\left[e,e^{e}\right)$, etc. In other words, using $\exp^{j}\left(x\right)$ to denote the $j$-iterated exponential (taking $\exp^{0}\left(x\right)=x$), we have the partition $$\left[1,\infty\right)=\bigcup_{j=0}^{\infty}\left[\exp^{j}\left(1\right),\exp^{j+1}\left(1\right)\right).$$ Then, we have $$\int_{1}^{\infty}\frac{dx}{f\left(x\right)} = \sum_{j=0}^{\infty}\int_{\exp^{j}\left(1\right)}^{\exp^{j+1}\left(1\right)}\frac{dx}{f\left(x\right)}.$$
Fix $\ell\in\mathbb{N}$. Then, make the substitution $u = \log^{\ell}x$, $du = \prod_{j=0}^{\ell-1}\left(\log^{j}x\right)^{-1}dx$ to get $$\int_{\exp^{\ell}\left(1\right)}^{\exp^{\ell+1}\left(1\right)}\frac{dx}{f\left(x\right)} = \int_{1}^{e}u^{-1} du = 1.$$
$$\text{Therefore, }\int_{1}^{\infty}\frac{dx}{f\left(x\right)},\text{ } \sum_{n=1}^{\infty}\frac{1}{f\left(n\right)},\text{ and, hence, }\sum_{n}a^{\mathrm{\bf 1}}_{n,*}\text{ all diverge.}$$
Since I wanted to know what conditions on the sequence $p$ of real numbers were required for convergence, we can state it very concisely: $$\sum_{n}a^{p}_{n,*}\text{ converges iff }\sum_{n}a^{p}_{n,k}\text{ converges for some }k\geq 0.$$
