Does $\sum_{k=1}^∞ \frac1{\prod_{i=0}^{l(k)} (\ln^i k) · {(\ln^{l(k)} k)}^{l(k)}}$ converge, where $l(x) = \min \{ c\in\Bbb{N} : \ln^c x < e \}$? 
Define $\def\lnc{\operatorname{lnc}}$$\lnc x$ is the minimum natural $c$ such that $\ln^c x < e$, for each real $x ≥ 1$.
Prove that $\displaystyle\sum_{k=1}^∞ \frac1{\prod_{i=0}^{\lnc k} (\ln^i k) · {(\ln^{\lnc k} k)}^{\lnc k}}$ diverges.
Prove that $\displaystyle\sum_{k=1}^∞ \frac1{\prod_{i=0}^{\lnc k} (\ln^i k) · {(\ln^{\lnc k} k)}^{(\lnc k)^r}}$ converges for every $r > 1$.

I came up with this for fun, just to try to squeeze the gap between the well-known boundary:

*

*$\displaystyle\sum_{k=1}^∞ \frac1{\prod_{i=0}^p (\ln^i k)}$ diverges for every natural $p$.


*$\displaystyle\sum_{k=1}^∞ \frac1{\prod_{i=0}^p (\ln^i k) · (\ln^p k)}$ converges for every natural $p$.
My proofs are below. Any comments are welcome! Downvoters should note that such questions are explicitly encouraged.
 A: Let $\def\lnc{\operatorname{lnc}}$$M(c)$ as the minimum natural $k ≥ 1$ such that $\lnc k = c$.
Then $\exp^c(1) ≤ M(c) ≤ \exp^c(1) + 1 ≤ \exp^c(2)$ for any natural $c$.
So 
$$
\begin{split}
\sum_{c=2}^\infty \sum_{k=M(c)}^{<M(c+1)}
         \frac1{\prod_{i=0}^{\lnc k} (\ln^i k) · {(\ln^{\lnc k} k)}^{\lnc k}}
  &= \sum_{c=2}^\infty \sum_{k=M(c)}^{<M(c+1)}
         \frac1{\prod_{i=0}^c (\ln^i k) · {(\ln^c k)}^c} \\
  &\ge \sum_{c=2}^\infty \int_{M(c)}^{M(c+1)}
         \frac1{\prod_{i=0}^c (\ln^i x) · {(\ln^c x)}^c}\ dx \\
  &= \sum_{c=2}^\infty \Big[\! -\frac1{c·{(\ln^c x)}^c} \Big]_{x=M(c)}^{M(c+1)} \\
  &\ge \sum_{c=2}^\infty \frac1{c} \Big( \frac1{(1+1/M(c))^c} - \frac1{e^c} \Big) \\
  &\to \infty,
\end{split}
$$
 where we obtained $\ln^c M(c) ≤ 1+1/M(c)$ by using $\ln(x+1) ≤ 1+1/x$ for every real $x > 0$.
But 
$$
\begin{split}
\sum_{c=1}^∞ \sum_{k=M(c)}^{<M(c+1)} \frac1{\prod_{i=0}^{\lnc k} (\ln^i k) · {(\ln^{\lnc k} k)}^{(\lnc k)^r}}
 &= \sum_{c=1}^∞ \sum_{k=M(c)}^{<M(c+1)} \frac1{\prod_{i=0}^c (\ln^i k) · {(\ln^c k)}^{c^r}} \\
 &\le \sum_{c=1}^∞ \int_{M(c)-1}^{M(c+1)-1} \frac1{\prod_{i=0}^c (\ln^i x) · {(\ln^c x)}^{c^r}}\ dx \\
 &= \sum_{c=1}^∞ \Big[\! -\frac1{c^r·{(\ln^c x)}^{c^r}} \Big]_{x=M(c)-1}^{M(c+1)-1} \\
 &\le \sum_{c=1}^∞ \frac1{c^r·{(\ln^c (\exp^c(1)-1))}^{c^r}} \\
 &\le \sum_{c=1}^∞ \frac1{c^r·{(1-1/(\exp^c(1)-1))}^{c^r}}\\
 &< \infty,  
\end{split}
$$
for every real $r > 1$,
 where we obtained $\ln^c (\exp^c(1)-1) ≥ 1-1/(\exp^c(1)-1)$ by using $\ln(x-1) ≥ \ln(x)-\frac1{x-1}$ for every real $x ≥ 2$.
