Divergence of the Sum of the Reciprocals of Primes

Theorem

Let $P$ be the set of all prime numbers. Then

\sum_{p \in P} \frac{1}{p} = \infty .

There are many ways to prove this result, but the proof presented here is useful because it follows a very similar argument to one used in Dirichlet's theorem, but in the special case of the trivial character modulo $1$ , and thus it helps motivate that proof. It is however, not the simplest or most elementary.

Proof

Consider the Riemann zeta function defined by the infinite sum for real $s > 1$ . We will prove that

\ln ζ (s) = \sum_{p} \frac{1}{p^{s}} + g (s)

where $g (s) = O (1)$ as $s \to 1^{+}$ .

From this we can just take limits as $s \to 1^{+}$ where $\ln ζ (s)$ diverges and therefore so does $\sum_{p} \frac{1}{p^{s}}$ .

Then, using the Taylor series for $\ln (1 - x)$ we have

\begin{aligned} g (s) & = \ln ζ (s) - \sum_{p} \frac{1}{p^{s}} \\ = \ln (\prod_{p} {(1 - \frac{1}{p^{s}})}^{- 1}) - \sum_{p} \frac{1}{p^{s}} \\ = - \sum_{p} \ln (1 - \frac{1}{p^{s}}) - \sum_{p} \frac{1}{p^{s}} \\ = - \sum_{p} (\ln (1 - \frac{1}{p^{s}}) + \frac{1}{p^{s}}) \\ = - \sum_{p} (- \sum_{n = 1}^{\infty} \frac{1}{n p^{n s}} + \frac{1}{p^{s}}) & Taylor series \\ = \sum_{p} \sum_{n = 2}^{\infty} \frac{1}{n p^{n s}} \\ = \sum_{p} \sum_{n = 0}^{\infty} \frac{1}{(n + 2) p^{(n + 2) s}} \\ = \sum_{p} \frac{1}{p^{2 s}} \sum_{n = 0}^{\infty} \frac{1}{(n + 2) p^{n}} \\ \leq \sum_{p} \frac{1}{p^{2 s}} \sum_{n = 0}^{\infty} \frac{1}{2 p^{n}} \\ = \sum_{p} \frac{1}{2 p^{2 s}} \sum_{n = 0}^{\infty} \frac{1}{p^{n}} \\ \leq \sum_{p} \frac{1}{2 p^{2 s}} \sum_{n = 0}^{\infty} {(\frac{1}{2})}^{n} & p \geq 2 ⟹ \frac{1}{p} \leq \frac{1}{2} \\ = \sum_{p} \frac{1}{2 p^{2 s}} \frac{1}{1 - \frac{1}{2}} \\ = \sum_{p} \frac{1}{p^{2 s}} \\ \leq \sum_{n = 1}^{\infty} \frac{1}{n^{2 s}} \\ \leq \sum_{n = 1}^{\infty} \frac{1}{n^{2}} \end{aligned}

which converges by the $p$ -test. As such, $g (s)$ is bounded for $s > 1$ .

Now we must go from $lim_{s \to 1^{+}} \sum_{p} \frac{1}{p^{s}} = \infty$ to $\sum_{p} \frac{1}{p} = \infty$ .

From the fact that the limit diverges, we have that for any $M > 0$ , there exists an $s_{0}$ such that

1 < s < s_{0} ⟹ \sum_{p} \frac{1}{p^{s_{0}}} > 2 M .

Then, because the series $\frac{1}{p^{s_{0}}}$ converges (given the terms are positive and are a subset of the terms in a convergence series because $s_{0} > 1$ ) we know that the sequence of partial sums are Cauchy, and therefore for any $ϵ > 0$ (in particular we set $ϵ = M$ ) there exists an $N$ such that

\begin{aligned} m, n \geq N & ⟹ | \sum_{p \leq n} \frac{1}{p^{s_{0}}} - \sum_{p \leq m} \frac{1}{p^{s_{0}}} | < M \\ ⟹ \sum_{p \leq m} \frac{1}{p^{s_{0}}} - M \leq \sum_{p \leq n} \frac{1}{p^{s_{0}}} \leq \sum_{p \leq m} \frac{1}{p^{s_{0}}} + M \\ ⟹ M = 2 M - M < \sum_{p \leq m} \frac{1}{p^{s_{0}}} - M \leq \sum_{p \leq n} \frac{1}{p^{s_{0}}} . \end{aligned}

As such, setting $n = N$ , we have for any $M > 0$ , there exists an $N$ and $s_{0} > 1$ such that

\sum_{p \leq N} \frac{1}{p^{s_{0}}} \geq M .

Then we have because $s_{0} > 1$ ,

\sum_{p \leq N} \frac{1}{p} \geq \sum_{p \leq N} \frac{1}{p^{s_{0}}} \geq M

and hence $\sum_{p \leq N} \frac{1}{p}$ diverges.

The more general version of this argument for Dirichlet $L$ -functions can be found here.