Month: June 2021

• Dirichlet’s Prime Number Theorem – Part 3

So far in this series, we have proven that the sum of the inverses of all positive integers, i.e. the harmonic series, diverges and that the sum of the inverses of all squares is $\displaystyle \frac{\pi^2}{6}$. You can find parts 1 and 2 here and here.

In this part 3, we’ll prove that the sum of the inverses of all prime numbers diverges, i.e. $\displaystyle \lim_{N\rightarrow\infty} \sum_{p<=N} \frac{1}{p} = \infty$. So in that sense, it turns out, there are more prime numbers than there are square numbers, or the primes lie denser in the natural numbers than the squares.

Let’s dive right into it. The proof that I am going to present relies on the following Lemma 1:

$\displaystyle \lim_{n\rightarrow\infty} \left(1+\frac{1}{n}\right)^n = e\ \ \ \ \ \ \ \ \ \ \ (1)$

I will not present the proof to this Lemma now but I might cover it in a later blog post. We should emphasize the fact that $e$ is approached from below by this product.

Now for the proof of the main theorem:

From Lemma 1 which in particular states that $\displaystyle \left(1+\frac{1}{n}\right)^n < e$ we can derive the special case $\displaystyle \left(1+\frac{1}{p-1}\right)^{p-1} < e$. Taking the logarithm of both sides yields $\displaystyle \log \left(1+\frac{1}{p-1}\right) < \frac{1}{p-1}$. The latter can be slightly re-arranged like this: $\frac{1}{p-1} = \frac{1}{p} + \frac{1}{p(p-1)}$ by multiplying nominator and denominator by p and adding zero in the form of $0 = 1 - 1$. The same trick can also be used to show that $\displaystyle \frac{1}{1-\frac{1}{p}} = 1 + \frac{1}{p-1}$.

Now we have the following inequality for every N:

$\displaystyle \log\prod_{p\leq N}\frac{1}{1-\frac{1}{p}} = \sum_{p\leq N}\log\left(1+\frac{1}{p-1}\right) < \sum_{p\leq N} \frac{1}{p} + \sum_{p\leq N} \frac{1}{p(p-1)}$

To which we can apply the geometric series on the left hand side:

$\displaystyle \frac{1}{1 - \frac{1}{p}} = 1 + \frac{1}{p} + \frac{1}{p^2} + \cdots$

This results in the following inequality:

$\displaystyle \prod_{p\leq N} \frac{1}{1 - \frac{1}{p}} = \prod_{p\leq N}\left(1 + \frac{1}{p} + \frac{1}{p^2} + \cdots\right) = \sum_{p_{+}(n) \leq N} \frac{1}{n} > \sum_{n \leq N} \frac{1}{n}$

Now we can connect both inequalities via their left-most product terms to get:

$\displaystyle \log\sum_{n\leq N} \frac{1}{n} < \sum_{p\leq N}\frac{1}{p} + \sum_{p\leq N} \frac{1}{p(p-1)}\ \ \ \ \ \ \ \ \ \ \ (2)$

And we’re almost done! Remember that we wanted to show that the sum of the reciprocal primes diverges. All we have to observe now is that $\sum \frac{1}{p} > \sum\frac{1}{p(p-1)}$., i.e., if the sum of the prime reciprocals converged than so would the whole right-hand side of (2). But that’s a contradiction to the well known fact that the left-hand side (which is smaller) converges (see Part 1 of this series of posts). Therefore, $\displaystyle \lim_{n\rightarrow\infty} \sum_{p\leq N} \frac{1}{p} = \infty$.

• Euler-Maclaurin summation via the Basel problem

I recently started playing with the book 📖 “Riemann’s Zeta Function” by H.M. Edwards. In chapter 6 he introduces Euler-Maclaurin summation, a concept that has allowed Danish mathematician Jørgen Pedersen Gram to find the first list of 15 non-trivial zeros of the Riemann Zeta function in 1903. I will repeat parts of Edwards’ arguments here and use this blog post to introduce Euler-Maclaurin summation in my own words (in fact I will heavily quote from the book). I will not yet connect this with calculating zeros of the Zeta function, it might be the topic of a future post. I will, however, add some pictures and reference the Python code that I’ve used to generate those pictures (https://mybinder.org/v2/gh/ahartel/rzf.git/HEAD).

As the initial example, Edwards choses the sum $S = (\frac{1}{10})^2 + (\frac{1}{11})^2 + \ldots + (\frac{1}{100})^2$ which is part of the series $\zeta(2)$, whose elusive evaluation was known as the Basel problem before Euler solved it in 1735, discovering exactly Euler-Maclaurin summation on the way.

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