As with many other occasions to prove series divergence, we will focus on the ‘tail’s beyond a point that keeps getting further out. In this case, we’ll choose the ‘tail’ beyond the first \(k\) primes,
\begin{equation*}
T = \sum_{n>k}\frac{1}{p_n}\text{.}
\end{equation*}
By examining certain terms in this, and assuming (falsely) that it can be made finite, we will obtain a contradiction.
First, let’s consider numbers of the form
\begin{equation*}
p_1 p_2 p_3\cdots p_k r+1=p_k\#\cdot r+1
\end{equation*}
\begin{equation*}
p_{n_1}p_{n_2}\cdots p_{n_{\ell}}\text{,}
\end{equation*}
where all \(n_i>k\) (some may be repeated).
Return to the ‘tail’. Since this \(p_k\#\cdot r+1\) factors with \(\ell\) factors, then somewhere in the \(\ell\)th power of the ‘tail’ we have the following term:
\begin{equation*}
T^\ell = \left(\sum_{n>k}\frac{1}{p_n}\right)^{\ell} = \frac{1}{p_1 p_2 p_3\cdots p_k r+1}+ \cdots \text{.}
\end{equation*}
Now assume that in fact the prime harmonic series converges; we will derive a contradiction.
First, for some \(k\text{,}\) the ‘tail’ \(T\) is less than \(\frac{1}{2}\text{,}\) i.e. \(T =\sum_{n>k}\frac{1}{p_n}<\frac{1}{2}\text{.}\) Since each term is positive, \(T>0\) and a geometric series involving the \(\ell\)th powers of \(T\) is very precisely bounded:
\begin{equation*}
0\leq \sum_{\ell=1}^{\infty}T^{\ell}=\sum_{\ell=1}^{\infty}\left(\sum_{n>k}\frac{1}{p_n}\right)^{\ell}\leq \sum_{\ell=1}^{\infty}\frac{1}{2^{\ell}}=2\text{.}
\end{equation*}
Now we bring in the first discussion in this proof; every single term of the form \(\frac{1}{p_1 p_2 p_3\cdots p_k r+1}\) will appear somewhere within this sum of the \(\ell\)th powers, though naturally \(\ell\) in each case will depend heavily upon \(r\text{.}\)
So the series of reciprocals of just these special terms is bounded.
\begin{equation*}
0<\sum_{r=1}^{\infty}\frac{1}{p_1 p_2 p_3\cdots p_k r+1}\leq \sum_{\ell=1}^{\infty}\left(\sum_{n>k}\frac{1}{p_n}\right)^{\ell}\leq 2\text{.}
\end{equation*}
A bounded series of all positive number should converge (e.g. by comparison).
Here comes the contradiction. The same series is bounded below as follows, for each integer \(k\text{.}\)
\begin{equation*}
\sum_{r=1}^{\infty}\frac{1}{p_1 p_2 p_3\cdots p_k r+1}>\sum_{r=1}^{\infty}\frac{1}{p_1 p_2 p_3\cdots p_k r+p_1 p_2 p_3\cdots p_k}
\end{equation*}
\begin{equation*}
=\frac{1}{p_1 p_2 p_3\cdots p_k}\sum_{r=1}^{\infty}\frac{1}{r+1}
\end{equation*}
This series certainly diverges, as a multiple of the tail of the harmonic series!
Since no matter how big \(k\) is (and hence how far out in the ‘tail’ we go) we report that a certain series both converges and diverges, we have a contradiction. Hence our original assumption that we could choose \(k\) to make \(T\) finite was false, and the prime harmonic series must diverge!