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?

$p_{2n+1} > p_{2n}$ for all $n$, thus $\frac{1}{p_{2n+1}} < \frac{1}{p_{2n}}$ for all $n$. Thus if $S_1$ converges so does $S_2$. Note that the sum of the reciprocals of sequence $(p_{2n+2})$ if and only if $S_1$ converges, because $$\sum_{n \in \mathbb N} \frac{1}{p_{2n+2}} = S_1 - \frac{1}{p_2}$$
$p_{2n+2} > p_{2n+1}$ for all $n$, so using similar logic as before $S_2$ converges if $S_1$ converges. 
