# Convergence of series involving the prime numbers

Given the series $A=\sum\limits_{n=1}^{+\infty}\frac{p_n}{p_{n+1}}$ and $B=\sum\limits_{n=1}^{+\infty}\frac{p_{2n}}{p_{2n+1}}$, where $p_n$ is the sequence where the nth number are the nth prime number, is it possible to determine whether they are converging or diverging?

I think there are some ways using the fact that $\sum\limits_{p\in\mathbb{P}}\frac{1}{p}$ diverges, but I'm not sure of how I can use it.

• If one knows that $\sum\limits_p\frac1p$ diverges then the lower bound $$\frac{p_n}{p_{n+1}}\geqslant\frac1{p_{n+1}}$$ shows that $A$ diverges and the lower bound $$\frac{p_{2n}}{p_{2n+1}}\geqslant\frac{2}{p_{2n+1}}\geqslant\frac{1}{p_{2n+1}}+\frac{1}{p_{2n+2}}$$ shows that $B$ diverges. – Did Apr 14 '15 at 6:53
• My answer directly uses the divergence of the sum of the reciprocals of the primes, but now I see that "Did" posted the same answer in a comment. – Michael Hardy Jan 7 '16 at 17:47

For the first one, suppose $A<\infty.$ Then $p_n/p_{n+1} \to 0.$ By the ratio test, that implies $\sum_{n=1}^{\infty}1/p_n < \infty,$ contradiction.
Hint: Bertrand says $p_{n+1}/p_n < 2$.
You can consider the bounds, if $n$ sufficiently large, say for some $k>0$ $$n\log\left(n\right)<p_{n}<n\log\left(n\right)+n\log\left(\log\left(n\right)\right)$$ then $$\sum_{n\geq k}\frac{p_{n}}{p_{n+1}}>\sum_{n\geq k}\frac{n\log\left(n\right)}{\left(n+1\right)\log\left(n+1\right)+\left(n+1\right)\log\left(\log\left(n+1\right)\right)}>\frac{1}{2}\sum_{n\geq k}\frac{1}{\left(n+1\right)\log\left(n+1\right)}=\infty$$ and the same holds for $\sum_{n\geq k}\frac{p_{2n}}{p_{2n+1}}$.