Convergence of series with iterated $\ln$ Let us consider two function $\textrm{pln}_1\colon \mathbb{N}\to\mathbb{R}$ and $\textrm{pln}_2\colon \mathbb{N}\to\mathbb{R}$:
$$\textrm{pln}_1(n) = n\cdot\ln n\cdot\ln\ln n\cdots\underbrace{\ln\circ\ln\circ\cdots\circ\ln n}_{\text{one time less than it is possible}};$$
and
$$\textrm{pln}_2(n) = n\cdot\ln n\cdot\ln\ln n\cdots\underbrace{\ln\circ\ln\circ\cdots\circ\ln n}_{\text{two times less than it is possible}}.$$
For example:
$$\textrm{pln}_1(1) = 1;$$
$$\textrm{pln}_1(5) = 5\cdot\ln 5\cdot\ln\ln5;$$
$$\textrm{pln}_2(5) = 5\cdot\ln 5;$$
$$\textrm{pln}_1(15) = 15\cdot \ln 15\cdot\ln\ln 15;$$
$$\textrm{pln}_1(16) = 16\cdot \ln 16\cdot\ln\ln 16\cdot\ln\ln\ln 16;$$
$$\textrm{pln}_2(16) = 16\cdot \ln 16\cdot\ln\ln 16.$$
What can we say about convergence of the following series:
$$S_1=\sum_{n=1}^{+\infty}\frac{1}{\textrm{pln}_1(n)}?$$
$$S_2=\sum_{n=3}^{+\infty}\frac{1}{\textrm{pln}_2(n)}?$$
 A: It turns out that the first sum diverges!  Define the iterated exponential sequence $a_0=1, a_{n+1} = e^{a_n}$.  I'm not sure that it's known whether $a_k = \exp(\exp(\exp(\exp(\cdots\exp(1)\cdots)$ is always a non-integer (for $k>0$), but I'm going to assume it is, and this argument only requires minor adjustments otherwise.
Let $b_k = \lceil a_k \rceil$ for $k>0$ and let's examine the value of $\text{pln}_1(b_k)$, which by assumption will be exactly
$$b_k \ln b_k \ln\ln b_k \cdots \underbrace{\ln \circ \ln \cdots \circ \ln}_{\text{$k+1$ times}}\; b_k.$$
Clearly $b_k > a_k$, $\ln b_k > \ln a_k$, $\ln \ln b_k > \ln \ln a_k$ and so on for each iterated logarithm.  So let's get an upper bound on the differences $\ln b_k - \ln a_k$ and so forth:
$$\begin{align}
b_k &< a_k + 1, \\
\ln b_k &< \ln(a_k + 1) = \ln a_k + \ln(1 + a_k^{-1}) < a_{k-1} + a_k^{-1}, \\
\ln \ln b_k &< \ln(a_{k-1} + a_k^{-1}) = \ln a_{k-1} + \ln(1 + (a_{k-1}a_k)^{-1}) < a_{k-2} + (a_{k-1} a_k)^{-1},
\end{align}$$
You can see an inductive pattern forming, which eventually leads to:
$$
\begin{align}
\underbrace{\ln \circ \ln \cdots \circ \ln}_{\text{$k$ times}}\; b_k &< a_0 + (a_1 a_2 \cdots a_k)^{-1} = 1 + (a_1 a_2 \cdots a_k)^{-1}, \\
\underbrace{\ln \circ \ln \cdots \circ \ln}_{\text{$k+1$ times}}\; b_k &< (a_1 a_2 \cdots a_k)^{-1}.\\
\end{align}
$$
So then $$\text{pln}_1(b_k) < \frac{b_k \ln b_k \ln \ln b_k \cdots \overbrace{\ln \circ \ln \cdots \circ \ln}^{\text{$k$ times}}\; b_k}{a_k \ln a_k \ln \ln a_k \cdots \underbrace{\ln \circ \ln \cdots \circ \ln}_{\text{$k$ times}}\; a_k},$$
which we can prove from the above bounds is very close to $1$.  Indeed, the above calculations let us bound the absolute difference of logs of the numerator and denominator by:
$$a_k^{-1} + (a_k a_{k-1})^{-1} + \cdots + (a_k a_{k-1} \cdots a_1)^{-1} < k/a_k,$$
which converges to $0$ very quickly as $k \to\infty$.  It follows that $\text{pln}_1(b_k)$ is bounded below by a positive constant close to $1$, so that $\sum_n  1/\text{pln}_1(n)$ diverges.
In fact, heuristically we expect the difference $\lceil a_k \rceil - a_k$ to be randomly distributed in $(0,1)$, in particular it should get arbitrarily close to $0$.  So it's extremely plausible (albeit hopeless to prove) that $\text{pln}_1(n)$ gets arbitrarily close to $0$, meaning $\limsup_{n\to\infty} 1/\text{pln}_1(n) = +\infty$.
We can also show the second sum diverges by a similar calculation.  Reusing the above sequence $\{b_k\}$ we can write the sum as:
$$ \sum_{n=b_1}^{b_2-1} \frac{1}{n \ln n} + \sum_{n=b_2}^{b_3-1} \frac{1}{n \ln n \ln \ln n} + \cdots,$$
where the $k$th summation is
$$S_k := \sum_{n=b_k}^{b_{k+1}-1} \frac{1}{n \ln n \cdots \underbrace{\ln \circ \ln \cdots \circ \ln}_{\text{$k$ times}}\; n}.$$
Let's try to get a lower bound for $S_k$.  By comparison to the integral,
$$S_k > \int_{b_k}^{b_{k+1}} \frac{dx}{x \ln x \cdots \underbrace{\ln \circ \ln \cdots \circ \ln}_{\text{$k$ times}}\; x} = \left.\underbrace{\ln \circ \ln \cdots \circ \ln}_{\text{$k+1$ times}}\; x\,\right|_{x=b_k}^{b_{k+1}} \\
> \underbrace{\ln \circ \ln \cdots \circ \ln}_{\text{$k+1$ times}} \; a_{k+1} - \underbrace{\ln \circ \ln \cdots \circ \ln}_{\text{$k+1$ times}} \; b_k > 1 - (a_k a_{k-1} \cdots a_1)^{-1}.$$
Since $S_k > \tfrac12$, it follows that $\sum_k S_k = \sum_n 1/\text{pln}_2(n)$ diverges.
