Infinite sum involving repeated logarithm For every positive integer $k$, write
$$\ln_k(x)=\log\log\cdots\log(x),$$
where the logarithm has been taken $k$ times. For instance,
$$\ln_1(x) =\log x,\ln_2(x) =\log\log x,\cdots.$$
Define a function $f:\mathbb{N}\rightarrow \mathbb{R}$ by
$$f(n)=n\prod_{m=1}^{k[n]}\ln_m(n),$$
where $k[n]$ is the largest integer, s.t. $\ln_{k[n]}(n)≥ 1$. Does the series $\sum_{n=3}^{\infty}\frac{1}{f(n)}$ converge?
I tried the condensation test (since 1/f(n) presents n in the denominator), so I considered the sum of (2^n)1/f(2^n): here the isolated 2^n are cancelled out. Then the denominator(from a certain n) presents the term loglog(2^n)=nlog(2*((log(2))^1/n)), so I have again an isolated n in the denominator: I do the condensation test again, considering (2^n)*(2^(2^n))*f(2^(2^n)). So, from a certain n, in the denominator I have logloglog(2^(2^n)), from which I can take out n. I proceed this way doing condensation test infinite time, and I end up with an infinite sum that I don't know wether it converges or diverges,  nor if I can write in a compact way, and I don't know if I am on the right way to solve the problem.
 A: Your sum boils down to
$$S = \sum_{n=3}^{15} \frac{1}{n\ln (n)} + \sum_{n=16}^{3\ 814\ 279} \frac{1}{n\ln (n) \ln(\ln (n))} + \cdots ;$$
the limits of this segmentation are found by successively tetrating $e$:
$$\begin{split}
^1e&=e = 2.718... \\
^2e&=e^e = 15.154... \\
^3e&=e^{e^e} = 3\ 814\ 279.1... \\
&\cdots
\end{split}$$
Such a segmentation, considering the monotonicity of the summand, offers natural grounds for trying out an Oresme-like argument, which seems to have been your intuition. But this doesn't work. Setting $E_q := \lfloor ^q e \rfloor $,
$$\begin{split}
S &\geq \underbrace{\frac{1}{15\ln(15)}+\cdots+\frac{1}{15\ln(15)}}_{15-3+1 \text{ times}} + \underbrace{\frac{1}{E_3\ln(E_3) \ln(\ln(E_3))}+\cdots+\frac{1}{E_3\ln(E_3) \ln(\ln(E_3))}}_{3\ 814\ 279-15+1 \text{ times}} + \cdots \\
&\geq \frac{E_2-E_1+1}{ ^2e\ln (^2e)} + \frac{E_3-E_2+1}{ ^3e \ln(^3e) \ln(\ln(^3e))} + \cdots \\ &\geq  \frac{^2e - e}{^2e\ln(^2e)} + \frac{^3e-{^2e}}{^3e\ln(^3e) \ln(\ln(^3e))} + \cdots = \frac{^2e - e}{^2e \cdot e} + \frac{^3e - {^2e}}{^3e\cdot {^2e} \cdot e} + \cdots
\end{split}$$
which converges to $\approx 0.32617...$, whereas we wanted a divergent series.

Another strategy is to check a related integral. We have
$$S = \int_0^\infty f(\lfloor x \rfloor)\ dx \geq \int_0^\infty f(x)\ dx =: I,  $$
where
$$f(x) := \frac{1}{x \ln_1(x) \cdots  \ln_q(x)}\cdot  \mathbf{1}_{ [{^q e},{^{(q+1)}e}) } (x),$$
and $\mathbf 1_{X}$ is the characteristic function of the set $X$. The above inequality is again due to the monotonicity of the summands of $S$. The integral $I$ can be evaluated by splitting its domain by the linearity of the operation, in a similar fashion to what we did for $S$:
$$I = \sum_{q=1}^\infty \int_{^qe}^{^{(q+1)}e} \frac{1}{x \ln_1(x) \cdots \ln_q(x)}dx$$
Now, observing that for all sensible $x$, by the chain rule,
$$\ln'_{q+1}(x) = \frac{1}{x \ln_1(x) \cdots \ln_q(x)}, $$
we may employ the Fundamental Theorem of Calculus to obtain
$$I = \sum_{q=1}^\infty \left[\ln_{q+1}(^{(q+1)}e) - \ln_{q+1}(^qe)\right] = \cdots =  \sum_{q=1}^\infty [ \ln (e) - \ln (1) ] = \sum_{q=1}^\infty 1 = \infty. $$
This allows us to conclude that $S$ also diverges.
