Divergence of the Sum of the Reciprocals of Primes
Theorem
Let \(\mathbb{P}\) be the set of all prime numbers. Then
\[ \sum_{p \in \mathbb{P}} \frac{1}{p} = \infty.\]
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
\[ \ln \zeta(s) = \sum_p \frac{1}{p^s} + g(s)\]
where \(g(s) = O(1)\) as \(s \to 1^+\).
From this we can just take limits as \(s \to 1^+\) where \(\ln\zeta(s)\) diverges and therefore so does \(\sum_p \frac{1}{p^s}\).
Then, using the Taylor series for \(\ln(1 - x)\) we have
\[\begin{align*}
g(s) &= \ln \zeta(s) - \sum_p \frac{1}{p^s} \\
&= \ln \left(\prod_p \left(1 - \frac{1}{p^s}\right)^{-1}\right) - \sum_p \frac{1}{p^s} \\
&= -\sum_p \ln\left(1 - \frac{1}{p^s}\right) - \sum_p \frac{1}{p^s} \\
&= -\sum_p \left(\ln\left(1 - \frac{1}{p^s}\right) + \frac{1}{p^s}\right) \\
&= -\sum_p \left(- \sum_{n = 1}^\infty \frac{1}{np^{ns}} + \frac{1}{p^s}\right) & \text{Taylor series} \\
&= \sum_p \sum_{n = 2}^\infty \frac{1}{np^{ns}} \\
&= \sum_p \sum_{n = 0}^\infty \frac{1}{(n + 2)p^{(n + 2)s}} \\
&= \sum_p \frac{1}{p^{2s}} \sum_{n = 0}^\infty \frac{1}{(n + 2)p^{n}} \\
&\leq \sum_p \frac{1}{p^{2s}} \sum_{n = 0}^\infty \frac{1}{2p^{n}} \\
&= \sum_p \frac{1}{2p^{2s}} \sum_{n = 0}^\infty \frac{1}{p^{n}} \\
&\leq \sum_p \frac{1}{2p^{2s}} \sum_{n = 0}^\infty \left(\frac{1}{2}\right)^n & p \geq 2 \implies \frac{1}{p} \leq \frac{1}{2}\\
&= \sum_p \frac{1}{2p^{2s}} \frac{1}{1- \frac{1}{2}} \\
&= \sum_p \frac{1}{p^{2s}} \\
&\leq \sum_{n = 1}^\infty \frac{1}{n^{2s}} \\
&\leq \sum_{n = 1}^\infty \frac{1}{n^2} \\
\end{align*}\]
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
\[ 1 < s < s_0 \implies \sum_p \frac{1}{p^{s_0}} > 2M.\]
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 \(\epsilon > 0\) (in particular we set \(\epsilon = M\)) there exists an \(N\) such that
\[\begin{align*}
m, n \geq N &\implies \left|\sum_{p \leq n} \frac{1}{p^{s_0}} - \sum_{p \leq m}\frac{1}{p^{s_0}}\right| < M \\
&\implies \sum_{p \leq m}\frac{1}{p^{s_0}} - M \leq \sum_{p \leq n} \frac{1}{p^{s_0}} \leq \sum_{p \leq m} \frac{1}{p^{s_0}} + M \\
&\implies M = 2M - M < \sum_{p \leq m}\frac{1}{p^{s_0}} - M \leq \sum_{p \leq n} \frac{1}{p^{s_0}}. \\
\end{align*}\]
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.
