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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?

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  • $\begingroup$ What are $p_n$? Are they arbitrary, or a particular set? $\endgroup$ – Ruvi Lecamwasam Nov 24 '15 at 21:47
  • $\begingroup$ Oops, forgot to write that, it's the sequence of prime numbers. $\endgroup$ – YoTengoUnLCD Nov 24 '15 at 21:48
  • $\begingroup$ I imagine that $(p_n)$ is the sequence of prime numbers? That would be valuable to be precised in your question. $\endgroup$ – mathcounterexamples.net Nov 24 '15 at 21:48
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You can always say that $p_{2n}\le p_{2n+1}$, hence $\frac{1}{p_{2n}}\ge \frac{1}{p_{2n+1}}$ and thus $$S_1=\sum_{n=1}^\infty \frac{1}{p_{2n}}\ge \sum_{n=1}^\infty \frac{1}{p_{2n+1}}=S_2.$$

On the other hand, you have $p_{2n}\le p_{2n-1}$, which would imply that (recall that $p_1=2$) the

$$S_1=\sum_{n=1}^\infty \frac{1}{p_{2n}}\le \sum_{n=1}^\infty \frac{1}{p_{2n-1}}=2+\sum_{n=1}^\infty \frac{1}{p_{2n+1}}=\frac 12+S_2.$$

These two relations, together with that all terms in the sums are positive, imply that the series $S_1$ and $S_2$ diverge or converge simultaneously.

Suppose that $S_1$ and $S_2$ converge, then they converge absolutely, which allows to say that $$S_1+S_2 = -\frac 12 +\sum_{n\ge 1}\frac{1}{p_n}=+\infty,$$ which leads to a contradiction.

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  • $\begingroup$ So in general if $(X_n)$ is a decreasing positive sequence and $\sum_nX_n $ is infinite then so are $\sum_nX_{2n}$ and $\sum_nX_{2n+1}$. $\endgroup$ – DanielWainfleet Nov 24 '15 at 22:13
  • $\begingroup$ @user254665, yes, you can say that. $\endgroup$ – TZakrevskiy Nov 24 '15 at 22:16
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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 $$S_2=\sum_{n=1}^\infty\frac{1}{p_{2n+1}}\ge\sum_{n=1}^\infty\frac{1}{p_{2n+2}}=\sum_{n=1}^\infty \frac{1}{p_{2(n+1)}}=\sum_{n=2}^\infty\frac{1}{p_{2n}}=S_1-\frac{1}{p_1}$$

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  • $\begingroup$ Yes, thanks! I will fix that. $\endgroup$ – Ruvi Lecamwasam Nov 24 '15 at 21:55
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$p_{2n+1} > p_{2n}$ 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.

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  • $\begingroup$ And similarly $\sum_np_{3n}=\infty$, etc. $\endgroup$ – DanielWainfleet Nov 24 '15 at 22:14

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