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.
 A: Hint: Bertrand says $p_{n+1}/p_n < 2$.
A: 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.
A: \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}
A: 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}}
 $.
