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I know the series $$\sum_{N \leq n}{\frac{1}{n \ln(n) \ln(\ln(n)) ... (\ln^k)(n)}}$$ ($\ln^k$ is $\ln$ iterated $k$ times, not $\ln$ to the power of $k$) diverges for all constant $k$, as long as $N$ is big enough for it to be defined. However, I was wondering what happens if we extend that to infinity.

We can define a function $$f(n) = \begin{cases} 1 & n\leq 1 \\ n\cdot f(\ln(n)) & 1 < n \end{cases} $$ Or equivalently $f(n) = n \ln(n) \ln(\ln(n)) ...$ as long as the $\ln$ is greater than $1$.

Does the series $$\sum_{n=1}^{\infty} {\frac{1}{f(n)}}$$ converge? I thought about using the integral test and partitioning it based on the tetrations of $e$, but I can't think of any way to estimate $$\int_{^{k}e}^{^{k+1}e}{\frac{dx}{x \ln(x) \ln(\ln(x)) ... (\ln^k)(x)}}$$

Command Master
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    Note that, by the chain rule, $$\int_{{}^k e}^{{}^{k+1}e} \frac{\mathrm{d}x}{x (\log x)(\log^{\circ 2}x)\cdots(\log^{\circ k}x)}=\left[\log^{\circ(k+1)}x\right]_{{}^k e}^{{}^{k+1}e}=1.$$ (Alternatively, apply the substitution $u=1/x$ to find that the integral for $k+1$ is equal to the integral for $k$.) – Sangchul Lee Mar 08 '23 at 07:09
  • @SangchulLee thanks! So the infinite integral diverges and so does the series? – Command Master Mar 08 '23 at 07:14
  • You might look at Schlömilch's Test, a generalization of the Cauchy Condensation Test that is often useful in such situations when you are composing multiple functions. See here, for instance. – Integrand Mar 08 '23 at 15:27
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    Duplicate of this question. – metamorphy Mar 13 '23 at 08:03

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