# Validity of simple intuition for $\sum 1/p \sim \log \log n$ work?

I want to show the following using the simplest arguments for students without university training in mathematics.

$$\sum_{p

I'd like to check the validity of my rationale.

Step 1:

We start with the Euler product formula which we derive elsewhere [ref].

$$\sum_{n}\frac{1}{n^{s}}=\prod_{p}(1-\frac{1}{p^{s}})^{-1}$$

We take the logarithms of both sides.

Step 2:

On the RHS we have:

\begin{align} \ln\left(\prod_{p}\frac{1}{(1-\frac{1}{p^{s}})}\right) &= \sum_{p}\ln\frac{1}{(1-\frac{1}{p^{s}})}\\ &=-\sum_{p}\ln(1-\frac{1}{p^{s}})\end{align}

We can use $$\ln(1-x)=-x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\ldots$$ to expand $$\ln(1-\frac{1}{p^s})$$.

\begin{align} -\sum_{p}\ln(1-\frac{1}{p^{s}})&=\sum_{p}(\frac{1}{p^{s}}+\frac{1}{2p^{2s}}+\frac{1}{3p^{3s}}+\frac{1}{4p^{4s}}\ldots)\\&=\sum_{p}\frac{1}{p^{s}}+C \end{align}

We know $$C$$ converges because we know $$\sum1/n^s$$ converges for $$s>1$$, and the sum over primes $$\sum 1/p^s$$ is a subset of the sum over integers $$\sum 1/n^s$$.

Step 3:

$$\ln(\ln(n+1)) < \ln\left(\sum_{1}^{n}\frac{1}{x}\right) < \ln(\ln(n)+1)$$

The inequality is from the integral comparison tests for finite harmonic partial sums (ref). The lower and upper bounds are like $$\ln(\ln(n))$$

The sum $$\ln\left(\sum_{1}^{n}\frac{1}{x}\right)$$ is the LHS of the Euler product in the limit as $$n\rightarrow \infty$$, and $$s\rightarrow1^{+}$$ for the Euler product.

Step 4: This is the step I am not sure about.

In this step we try to combine the results from step 2 and 3.

We consider the limit $$s\rightarrow1^{+}$$ for the expression from step 2:

$$\sum_{p}\frac{1}{p}+C$$

Similarly we consider the limit $$n\rightarrow \infty$$ for the expression in step 3 and conclude that the logarithm of the harmonic series diverges but it does so as $$\ln(\ln(n))$$.

Therefore we can say the prime harmonic series diverges as

$$\sum_{p

The key idea I think I'm using is to enable comparison of two expressions by taking the limits $$n\rightarrow \infty$$ and $$s\rightarrow1^{+}$$ to make them equivalent.

Is there a flaw in this argument? I would appreciate explanations that didn't assume advanced knowledge.

• If you want intuition for students who know a bit of calculus, you can hand them the prime number theorem $\pi(x)\sim{\rm li}(x)$ to show the density of primes around $x$ is something like $(\ln x)^{-1}$, then heuristically we can guess $\sum_{p\le x}f(p)$ for "nice" $f$ ought to be asymptotic to $\int^x_c f(t) \,d{\rm li}$, which in the case of $1/x$ is $\ln\ln x$.
– anon
Mar 14, 2021 at 6:03

In Step 2, you should write $$C$$ as $$C_s$$: it depends on $$s$$. You gloss over this point in Step 4 where you write $$C$$ and let $$s \to 1$$. You in fact should check (i) $$C_s$$ converges for $$s > 1/2$$ and (ii) $$C_s$$ is continuous in $$s$$, so as $$s \to 1^+$$, $$C_s$$ tends to $$C_1$$. They're not all the "same" $$C$$.
In the setting you are writing about, there are two kinds of series with a limit that you are interested in: $$\lim_{s \to 1^+} \sum_p \frac{1}{p^s} \text{ and } \lim_{n \to \infty} \sum_{p < n} \frac{1}{p}.$$ The first series runs over all prime numbers for each $$s > 1$$, while the second runs over finitely many primes for each $$n$$, so these are fundamentally very different kinds of series over the primes. Using the pole of the zeta-function at $$s = 1$$, it is "not hard" to show $$\sum_{p} 1/p^s \sim \log(1/(s-1))$$. Converting knowledge of the $$s$$-limit of a series over all prime numbers to knowledge of the $$n$$-limit of a series of finitely many prime numbers (for each $$n$$) is not something I thought you would be able to do at the level you are trying to pitch your presentation. But I was mistaken. See Paul Pollack's paper here, which uses estimates on $$\zeta(s)$$ and $$\log \zeta(s)$$ for $$s$$ near $$1$$ from the right to show $$\sum_{p < n} 1/p$$ differs from $$\log(\log n)$$ by at most 6 for large $$n$$, and having $$\sum_{p < n} 1/p - \log(\log n)$$ be bounded is stronger than $$\sum_{p < n} 1/p \sim \log(\log n)$$. You'll have to judge how much of Pollack's argument can be made accessible to your students.