Shifted sum of reciprocals of primes: $\sum \limits_{\forall p \in primes}\frac{1}{p+1}$

Question

I know that the sum:

$\sum \limits_{\forall p \in primes}\frac{1}{p}$ diverges.

I was wondering if:

$\sum \limits_{\forall p \in primes}\frac{1}{p+1}$ diverges as well.

And if there is a formula for the partial sum of primes up to an upperbound $n$.

My guess is that it does but I'm not quite sure what the quick way to show it is. Any help is greatly appreciated!

Answer 1

Let $p_n$ be the $n$-th prime. Then these two series diverge:

$$ \sum_{n=1}^{\infty} 1/p_n $$ $$ \sum_{n=1}^{\infty} 1/p_{n+1} $$ Since $p_{n+1} \geq p_n + 1$, we conclude that the series you described is divergent as well.

Answer 2

Hint: $$ \frac{1}{p+1} \ge \frac{1}{2p}. $$ For partial sum of reciprocals of primes asymptotic is well known: $$ \sum_{p< n} \frac{1}{p} = \log \log n + M_n, $$ where $\lim_{n\to \infty}M_n = M$ is Meissel–Mertens constant.