Sum of reciprocals of every nth prime

Question

I'm looking for a proof that $\displaystyle\sum_{n\mathop=1}^{\infty}\frac{1}{p_{kn}}$ diverges, where $p_n$ denotes the $n$th prime number and $k$ is a natural number.

I know the proof that $\displaystyle\sum_{n\mathop=1}^{\infty}\frac{1}{p}$ diverges so this can be assumed.

Answer 1

If $a_k$ is a decreasing sequence of positive numbers, such that $\sum_k a_k$ diverges, then so does $\sum_k a_{nk}$. This is because you can estimate $a_{nk}$ from below by $1/n$ times a (suitably chosen, 'near' $a_{nk}$) partial sum of the divergent series (using monotonicity).

Edit: as an answer to a comment: $$n a_{nk} \ge \sum\limits_{l=nk}^{n(k+1)-1} a_l $$ by monotonicity. So $$ \sum_k a_{nk} \ge\frac{1}{n}\sum_k \sum\limits_{l=nk}^{n(k+1)-1} a_l= \frac{1}{n}\sum\limits_{l}a_l$$