3
$\begingroup$

This question already has an answer here:

Denote the prime numbers $2,3,5,7,\ldots$ as $p_1,p_2,\ldots$. Determine whether the infinite series $\dfrac{p_1}{p_2}+\dfrac{p_3}{p_4}+\cdots = \dfrac{2}{3}+\dfrac{5}{7}+\cdots$ converges.

I was wondering about this question because we can't really use the ratio test here because the ratio is constantly changing. So how would we determine if the series converges or not?

$\endgroup$

marked as duplicate by Martin R, Leucippus, JonMark Perry, jameselmore, Alex M. Jan 7 '16 at 18:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

7
$\begingroup$

By Bertrand's Postulate (a theorem since the mid nineteeth century) there is always a prime between $n$ and $2n$. As a consequence, $p_{2k-1}/p_{2k}$ does not have limit $0$, so the series cannot converge.

$\endgroup$
  • $\begingroup$ There's a much easier proof: the answer I've posted here. (Proving Bertrand's postulate takes some work.) ${}\qquad{}$ $\endgroup$ – Michael Hardy Jan 7 '16 at 17:44
  • $\begingroup$ I agree that the argument that uses the divergence of $\sum \frac{1}{p_k}$ is substantially more elementary. $\endgroup$ – André Nicolas Jan 7 '16 at 18:02
  • $\begingroup$ As $n$ gets larger, so does the gap between $n$ and $2n$. So how does that prove the sequence doesn't have limit $0$? $\endgroup$ – user19405892 Jan 7 '16 at 18:15
  • $\begingroup$ Let $n=p_{2k-1}$. Even if we interpret the Bertrand Postulate weakly, the prime $p_{2k}$ is $\lt 4p_{2k-1}$, so the ratio is $\gt \frac{1}{4}$. $\endgroup$ – André Nicolas Jan 7 '16 at 18:19
4
$\begingroup$

It diverges. \begin{align} \frac{p_1}{p_2} + \frac{p_3}{p_4} + \cdots \ge \frac 1 {p_2} + \frac 1 {p_4} + \cdots & = \frac 1 2 \left( \frac 2 {p_2} + \frac 2 {p_4} + \cdots \right) \\[10pt] & \ge \frac 1 2 \left( \left( \frac 1 {p_2} + \frac 1 {p_3} \right) + \left( \frac 1 {p_4} + \frac 1 {p_5} \right) + \cdots \right) \\[10pt] & = \infty. \end{align}

$\endgroup$
  • $\begingroup$ Why does $\left( \left( \frac{1}{p_2} + \frac{1}{p_3} \right) + \left( \frac{1}{p_4} + \frac{1}{p_5} \right) + \cdots \right)$ diverge to infinity? $\endgroup$ – Jacob Willis Jan 7 '16 at 17:48
  • $\begingroup$ @JacobWillis : Maybe you should post that as a separate question. Suppose $A\subseteq\{1,2,3,\ldots\}$ and $\sum\limits_{n\in A} \dfrac 1 n<\infty$ then $\sum\limits_{n\in B} \dfrac 1 n <\infty$ where $B$ is the closure of $A$ under multiplication. The closure of the set of primes under multiplication is all of $\{1,2,3,\ldots\}$, and the sum of the reciprocals of those diverges to $\infty$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jan 7 '16 at 17:54
  • $\begingroup$ @JacobWillis : That question has been posted here: math.stackexchange.com/questions/15946/… ${}\qquad{}$ $\endgroup$ – Michael Hardy Jan 7 '16 at 17:59
  • $\begingroup$ @JacobWillis : I've just added this answer to your question in your comment above. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jan 7 '16 at 18:47

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