5
$\begingroup$

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

$$\sum_{p<n} 1/p \sim \log \log n$$

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:

We start with the logarithm of the partial harmonic series.

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

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

$$\sum_{p<n} 1/p \sim \ln(\ln(n))$$

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.

$\endgroup$
1
  • 3
    $\begingroup$ 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 $\ln\ln x$. $\endgroup$
    – anon
    Mar 14, 2021 at 6:03

1 Answer 1

4
$\begingroup$

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" $C$.

The rest of the reasoning in Step 4 is even less justified (or more flawed, as you suggest may be the case). Analysis is full of limits taken in different ways, and proving the two methods can be related may be a very hard problem or even an unsolved problem. One source of heuristics in number theory is to take an iterated limit and just switch the order without justification to see what happens, and if the result makes sense, claim that should be the result of carrying out the limits in the initial order even if there is no proof of this yet.

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.

$\endgroup$
1
  • $\begingroup$ Paul Pollack's paper is very interesting because it doesn't require a lot of paths training, but does require a lot of bookwork. Here is my writeup which attempts to explain each step in Pollack's reasoning - which perhaps some readers might find useful fromprimestoriemann.blogspot.com/2021/03/… $\endgroup$
    – Penelope
    Mar 28, 2021 at 21:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .