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.)

Answer 1

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.

Answer 2

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 $n \geq 2$]

Answer 3

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 $f'(x)=\frac{1}{x \log x \log \log x}$.