It is known that the following series diverges:

$$\sum_{n=1}^\infty\frac{1}{n\log n\log\log n\ldots\log^{(k)}n}$$

where $\log^{(k)}$ is natural logarithm iterated $k$ times, and $k=L(n)$ is the smallest nonnegative integer such that $1<\log^{(k)}n\le e$. This series is greater than

$$\int_1^e\frac{\mathrm{d}x}{x}+\int_e^{e^e}\frac{\mathrm{d}x}{x\log x}+\int_{e^e}^{e^{e^e}}\frac{\mathrm{d}x}{x\log x\log\log x}+\cdots=1+1+1+\cdots,$$

so it increases at the level of $L(x)=\log^*\!x$ (iterated logarithm).

We may obtain a series that diverges more slowly by multiplying a similar function of $k$, instead of $n$, to the denominator:

$$\sum_{n=1}^\infty\frac{1}{\big(\prod_{j=0}^k\log^{(j)}n\big)k\log k\log\log k\ldots\log^{(L(k))}k}.$$

This series should increase at the level of $k_2=L(k)$. We can similarly iterate this process to get a series:

$$\sum_{n=1}^\infty\frac{1}{\big(\prod_{j=0}^{L(n)}\log^{(j)}n\big) \big(\prod_{j=0}^{L(L(n))}\log^{(j)}L(n)\big)\ldots\big(\prod_{j=0}^{L^{(m)}(n)}\log^{(j)}L^{(m-1)}(n)\big)}$$

where $m$ is the smallest nonnegative integer such that the smallest term in the product in the denominator is in $(1,e]$. This series should diverge at the level of “iterated $\log^*$.”

We may obtain a series that diverges even more slowly by multiplying a similar product of $m$, instead of $n$, to the denominator. Furthermore, we can repeat this iteration process (from $n$ to $k$, then from $k$ to $m$, etc) indefinitely like how we construct ordinals. After that, each term in the resulting series would still be a finite product because $n$ is finite.

The question is: does this resulting series, when the iteration process is repeated indefinitely, converge? If it converges, can its value be bounded? If it diverges, can its rate of increasing be described?

Intuitively, I think it might converge because if it diverges it will be increasing more slowly than any function I can imagine. But this is not a rigorous reason to show that it converges.

  • $\begingroup$ The series shouldn't start at $n=1$ since $\log 1=0$, but anyway you may just exploit the Cauchy condensation test. If $a_n$ is decreasing to zero, $$\sum a_n\qquad\text{and}\qquad \sum 2^n a_{2^n}$$ have the same behaviour. $\endgroup$ – Jack D'Aurizio Sep 11 '19 at 16:00
  • 1
    $\begingroup$ @jack-daurizio It seems that Cauchy condensation test would only transform the series to something of similar difficulty (with a form very similar to the original series). $\endgroup$ – Yizhen Chen Sep 12 '19 at 8:16
  • 1
    $\begingroup$ I don't think ordinals tag is right for this question. $\endgroup$ – L. McDonald Sep 29 '19 at 4:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.