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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?

Jacob Willis
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2 Answers2

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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.

André Nicolas
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  • There's a much easier proof: the answer I've posted here. (Proving Bertrand's postulate takes some work.) ${}\qquad{}$ – Michael Hardy Jan 07 '16 at 17:44
  • I agree that the argument that uses the divergence of $\sum \frac{1}{p_k}$ is substantially more elementary. – André Nicolas Jan 07 '16 at 18:02
  • As $n$ gets larger, so does the gap between $n$ and $2n$. So how does that prove the sequence doesn't have limit $0$? – user19405892 Jan 07 '16 at 18:15
  • 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 $\gt \frac{1}{4}$. – André Nicolas Jan 07 '16 at 18:19
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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}

Michael Hardy
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  • 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? – Jacob Willis Jan 07 '16 at 17:48
  • @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 $\infty$. ${}\qquad{}$ – Michael Hardy Jan 07 '16 at 17:54
  • @JacobWillis : That question has been posted here: http://math.stackexchange.com/questions/15946/does-the-sum-of-reciprocals-of-primes-converge ${}\qquad{}$ – Michael Hardy Jan 07 '16 at 17:59
  • @JacobWillis : I've just added this answer to your question in your comment above. ${}\qquad{}$ – Michael Hardy Jan 07 '16 at 18:47