Proof of $\sum \frac 1 {p_{2n}} = \infty$ I found the following
Theorem

Let $p_n$ denote the $n$-th prime number.
$S_1= \sum_{n \in \Bbb N} \frac 1 {p_{2n}} = \infty$ and $S_2=\sum_{n \in \Bbb N} \frac 1 {p_{2n+1}} =\infty$.

Proof

If one of $S_1,S_2$ converges, then so does the other, but then $S = \sum_{n=1}^\infty p_n^{-1} < \infty$, which Euler showed that diverges, q.e.d.

I don't understand why the convergence of $S_1$ would imply the convergence of $S_2$. Could someone explain that bit?

As $p_{2n+1}>p_{2n}$, we have $S_1\ge S_2$, so if $S_1$ converges then so does $S_2$.
For the other direction, note
$$S_2=\sum_{n=1}^\infty\frac{1}{p_{2n+1}}\ge\sum_{n=1}^\infty\frac{1}{p_{2n+2}}=\sum_{n=1}^\infty \frac{1}{p_{2(n+1)}}=\sum_{n=2}^\infty\frac{1}{p_{2n}}=S_1-\frac{1}{p_1}$$
