Are there any divergent series that approximate to O(log log log n)?

Question

$$1 + {1 \over 2} + {1 \over 3} + {1 \over 4} + \cdots \sim O(\log n)$$

$${1 \over 2} + {1 \over 3} + {1 \over 5} + {1 \over 7} + \cdots \sim O(\log\log n)$$

$$\text{???} \sim O(\log\log\log n)$$

Are there any divergent series (each of terms is greater than zero) with upper condition?

+Added] Each of terms must be a rational number.

(Sorry, I'm not familiar with English and MathJax, so there may be awkward expressions.)

Does $(\log\log\log 101 - \log\log\log 100) + (\log\log\log 102 - \log\log\log 101) + (\log\log\log 103 - \log\log\log 102) + (\log\log\log 104 - \log\log\log 103) + (\log\log\log 105 - \log\log\log 104) + \ldots$ count? — Milo Brandt, Jan 04 '20 at 05:52
You might want to see this post on ways to show there is no slowest diverging series. — Toby Mak, Jan 04 '20 at 05:59
Your MathJax code was a mess. See my edits to your question. — Michael Hardy, Jan 04 '20 at 06:02
@MiloBrandt I've never think about that, but I think the format is a little different from what I want. Thank you, I'll revise my article. — Antithesis, Jan 04 '20 at 06:09
@MichaelHardy Thanks! How could I learn more MathJax coding? — Antithesis, Jan 04 '20 at 06:12
Partly by learning about LaTeX. MathJax is quite a different thing from LaTeX, but their ways of writing mathematical notation are almost the same. — Michael Hardy, Jan 04 '20 at 06:14
@Antithesis: You can also right click on any MathJax and choose Show Math As->TeX commands to see how it was done. I learned a lot that way. — Ross Millikan, Jan 04 '20 at 06:27
https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference — Hans Lundmark, Jan 04 '20 at 09:57

score 3 · Answer 1 · edited Jan 04 '20 at 18:03

3

Given any sequence $b_n$ of positive numbers increasing to $\infty$ there is always a series $\sum a_n$ of positive terms such that $a_1+a_2+\cdots+a_n=b_n$ for all $n$. [Take $a_1=b_1$ and $a_n=b_n-b_{n-1}$ for $n \geq 2$]

edited Jan 04 '20 at 18:03

Michael Hardy

1

answered Jan 04 '20 at 05:52

Kavi Rama Murthy

311,013

score 3 · Accepted Answer · answered Jan 04 '20 at 06:19

3

Yes. Since $e^{e^e} \approx 3.8\cdot 10^6$ we would like to have our sum $1$ after, say, $3.9\cdot 10^6$ terms, so we start with $3.9 \cdot 10^6$ terms of $\frac 1{3.9\cdot 10^6}$. Then for each successive term $a_n$, take $\frac 1k$ with $k$ the smallest number possible so that the sum does not exceed $\log \log \log n$. Roughly speaking you want $a_n$ to be about $\frac 1{n \log n \log \log n}$. You can continue the pattern with more logs as far as you want.

answered Jan 04 '20 at 06:19

Ross Millikan

374,822

Thanks very much! Could you tell me any specific series for example? – Antithesis Jan 04 '20 at 06:26
Without the restriction to rational numbers, once you get started $\frac 1{n \log n \log \log n}$ works. I just used rationals close by because you asked for them. One could write a program to come up with them, but I haven't done so. The general process will give a series that diverges as slowly as you want. – Ross Millikan Jan 04 '20 at 06:31

score 1 · Answer 3 · answered Jan 04 '20 at 06:51

1

If we have a $C^1$ function $f:\mathbb{R} \to \mathbb{R}$ which is reasonably well-behaved as $x \to \infty$, then $\sum_{k=1}^{n} f'(k) = O(f(n))$. In your case, $f(x)=\log \log \log x$ and $f'(x)=\frac{1}{x \log x \log \log x}$.

answered Jan 04 '20 at 06:51

MathematicsStudent1122

12,462

Are there any divergent series that approximate to O(log log log n)?

3 Answers3