I found the following


Let $p_n$ denote the $n$-th prime number.

$S_1= \sum_{n \in \Bbb N} \frac 1 {p_{2n}} = \infty$ and $S_2=\sum_{n \in \Bbb N} \frac 1 {p_{2n+1}} =\infty$.


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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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 $$S_2=\sum_{n=1}^\infty\frac{1}{p_{2n+1}}\ge\sum_{n=1}^\infty\frac{1}{p_{2n+2}}=\sum_{n=1}^\infty \frac{1}{p_{2(n+1)}}=\sum_{n=2}^\infty\frac{1}{p_{2n}}=S_1-\frac{1}{p_1}$$

  • $\begingroup$ Yes, thanks! I will fix that. $\endgroup$ – Ruvi Lecamwasam Nov 24 '15 at 21:55

$p_{2n+1} > p_{2n}$ 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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.