Result of a Sum Up to $\pi(n)$

Question

I have been working with this sum recently:

$$\sum_{k=2}^{\pi(n)}\frac{\pi(n)-1}{p_k -1}$$In case the notation of $k=2$ is not clear, I mean this: $$\frac{\pi(n)-1}{2} + \frac{\pi(n)-1}{4} + \frac{\pi(n)-1}{6} + ...$$ I was wondering whether it converged, or whether there was a simpler way to express it. I know that the sum of the reciprocals of the primes diverges, but I'm not sure whether this is applicable here.

Answer 1

From $$p_k-1<p_k \Rightarrow \sum_{k=2}^{\pi(n)}\frac{\pi(n)-1}{p_k-1}>\left(\pi(n)-1\right)\sum_{k=2}^{\pi(n)}\frac{1}{p_k}\geq...$$ using this $$...\geq\left(\pi(n)-1\right)\left(\ln{\ln{(\pi(n)+1)}}-\ln{\frac{\pi^2}{6}}-\frac{1}{p_1}\right)$$ it should diverge.