What series approaches to $\log(\log(\log(n)))$?

Question

Background: Harmonic series approaches to $\lim_{n->\infty}\log(n)$. This gives the Euler–Mascheroni constant. In addition, the harmonic series summed only over the primes approaches to $\lim_{n->\infty}\log(\log(n))$. This gives the Meissel–Mertens constant.

Question: What similar series approaches to $\lim_{n->\infty}\log(\log(\log(n)))$.

My failed attempts: I thought about further filtering out primes to get an even slower-growing series. For example, I only choose the paired primes (twin primes or $k$-twin primes). But then I learned that these series converge using Brun's theorem.

There ought to be someone who has thought about this problem possibly with answers. Can anyone please point me in the right direction? Thanks.

Answer 1

The heuristic argument for why the sum of prime reciprocals diverges as $\log\log n$ is that we may model an integer $n$ as being "prime with probability $\frac1{\log n}$", by the prime number theorem: so its contribution to the sum of prime reciprocals is on average $\frac1{n \log n}$. Further approximating this sum by an integral, we notice that $$\int \frac{1}{x \log x}\,\mathrm dx = \log\log x + C.$$ Similarly, it is true that $$\int \frac{1}{x \log x \log\log x}\,\mathrm dx = \log\log\log x + C,$$ so we want a sum over reciprocals where an integer $n$ has a probability of $O(\frac1{\log n\log\log n})$ of contributing to the sum.

So we could, for example, say: "Take the sum of reciprocals of all primes which have a prime number of digits." Heuristically, a number $n$ has a $\frac1{\log n}$ chance of being prime, but it has $O(\log n)$ digits, and he number of digits has an $O(\frac1{\log \log n})$ chance of being prime.

Now let's try to figure this out carefully. What is the sum of reciprocals of all prime-digit primes less than $n$?

Well, for every prime $p$ such that $10^p < n$, we have the sum of reciprocals of all primes from $10^{p-1}$ to $10^p-1$. By the Meissel–Mertens result, for sufficiently large $p$, this should contribute $(1+o(1))(\log \log 10^p - \log \log 10^{p-1})$, which simplifies to $(1+o(1))\log \frac{p}{p-1})$. As $x \to 0$, $\log(1+x) \sim x$, and we can simplify this to $(1+o(1))\frac1p$.

Therefore for each prime $p$ such that $10^p < n$, the contribution to the sum is $(1+o(1))\frac1p$, which means that the overall sum is asymptotically equal to the sum of $\frac1p$ over all $p$ up to $\log_{10} n$. By the Meissel–Mertens asymptotics again, this should be $\log\log\log n$.

Answer 2

You have

• $\frac{d}{dx} \log(x)=\frac1x$ so $\sum\limits_{m=1}^n\frac1m$ will be $O(\log(n))$
• $\frac{d}{dx} \log(\log(x))=\frac1{x\log(x)}$ so $\sum\limits_{m=2}^n\frac1{m\log(m)}$ will be $O(\log(\log(n)))$
• $\frac{d}{dx} \log(\log(\log(x)))=\frac1{x\log(x)\log(\log(x))}$ so $\sum\limits_{m=16}^n\frac1{m\log(m)\log(\log(m))}$ will be $O(\log(\log(\log(n))))$ at least when $n> e^{e}$

Answer 3

There is a generic way to look at this in terms of integers in residue classes.

Definitions:

1. Let $p,q$ represent primes and let $n$ be a positive integer.
2. $\pi(n)$ has its usual meaning.
3. Let $p_1, p_2,\cdots$ represent the primes starting with $2,3,\cdots$.

Let $\{a_1\}$-primes represent integers $\leq n$ that are $\not\equiv a_1 \bmod p$ ($p_1 \leq p \leq \sqrt{n}$). Slight deviation from definition but we will assume $\{0\}$-primes as the usual primes.

Let $\{a_1,a_2\}$-primes represent $\{a_1\}$-primes that are $\not\equiv a_2 \bmod p$ ($p_2 \leq p \leq \sqrt{n}$).

Let $\{a_1,a_2,a_3\}$-primes represent $\{a_1,a_2\}$-primes that are $\not\equiv a_3 \bmod p$ ($p_3 \leq p \leq \sqrt{n}$).

and so on.

The sum of the reciprocal of $\{a_1\}$-primes is $\approx \log\log(n)$

The sum of the reciprocal of $\{a_1,a_2\}$-primes is $\approx \log\log\log(n)$

The sum of the reciprocal of $\{a_1,a_2,a_3\}$-primes is $\approx \log\log\log\log(n)$ and so on. Obviously $n$ has to be sufficiently large to accommodate more logs.

ps: There are beautiful relationships between the counts of such primes as well. Happy to share if there is interest.