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Denote $\log^{\circ j}$ to be $\log$ composed with itself $j$ times, i.e. $$\log^{\circ 0}(n) = n,\ \ \log^{\circ 1}(n) = \log(n),\ \ \log^{\circ 2}(n) = \log(\log(n))$$ and so on.

As stated in Julien Clancy's answer to this question, the following series diverges for every $k$: $$\sum_{i=2}^\infty \frac{1}{\prod_{j=0}^{k}\log^{\circ j}(n)} = \infty$$

and the following series converges for every $k$: $$\sum_{i=2}^\infty \frac{1}{\log^{\circ k}(n)\prod_{j=0}^{k}\log^{\circ j}(n)} < \infty$$

For example, when $k=2$ we have: $$\sum_{i=2}^\infty \frac{1}{n\cdot \log(n)\cdot \log(\log(n))} = \infty$$ but $$\sum_{i=2}^\infty \frac{1}{n\cdot \log(n)\cdot \log(\log(n))^2} < \infty$$

Intuitively, by increasing $k$ the first series becomes very close to converging, but it still diverges for all $k$.

This naturally raises the question - Is the (first) series still diverging when $k$ is increasing with $i$? More rigorously, is the following series diverges: $$\sum_{i=2}^\infty \frac{1}{\prod_{j=0}^{i}\log^{\circ j}(n)}$$

The usual condensation test approach seems hopeless here. Maybe comparing the series to an integral could work. I don't even have any intuition whether this should diverge or converge!

Rei Henigman
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  • You can also denote $F_j$ as $\log^{\circ j}$ (\log^{\circ j}). By $\log$, do you mean $\ln$? – J.G. Mar 19 '23 at 20:09
  • Yes, By $log$ I mean $ln$. And for the notation of $F_j$, I followed the first answer to this question: https://math.stackexchange.com/questions/926247/notation-for-repeated-composition-of-functions – Rei Henigman Mar 19 '23 at 20:13
  • We can compare the series to an integral $\int_2^\infty \frac{1}{\prod_{j=0}^{i}F_j(x)} dx$. If the integral converges, then the series must also converge, but $F_j(n)$ has a very slow growth, so it's not clear... – rumathe Mar 19 '23 at 20:26
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    You might get inspired by this post? https://math.stackexchange.com/q/1764465/1104384 – Bruno B Mar 19 '23 at 20:29
  • Do you mean $\sum_{n=2}^{\infty},$ rather than $\sum_{i=2}^\infty?$ – Thomas Andrews Mar 19 '23 at 20:34
  • @BrunoB Well, that's a bit awkward - I completely missed the fact that when $n < \left\lceil \mathrm{e}^k \right\rceil$, my summands are not necessarily even positive! I think this question should be closed, as the linked question makes much more sense, and is also answered :) – Rei Henigman Mar 19 '23 at 20:37
  • I voted to close the question. While the linked question is not exactly the same as mine, after realizing the inconsistencies of my question I would probably want to edit it to be very similar to the linked question so... – Rei Henigman Mar 19 '23 at 20:42

1 Answers1

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It is well known (and easily to prove) that for positive $a_n$ and decreasing the series $\sum a_n$ converges if and only if $\sum 2^n a_{2^n}$ converges. This result has a name, something like "condensation theorem" or something. Once you know this, just use it as many times as needed.

Salcio
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  • Hmm the problem with this argument, which looks fine in theory, is that you cannot really apply that test (Cauchy condensation test is the name) an "infinite number of times"? Or I guess you'd have to take the limit in the process? – Bruno B Mar 19 '23 at 20:47