Bounded sum of reciprocals of primes.

Question

How can one adapt Apostol's proof that the bounded sum of the reciprocals of the first primes is $$\log\log x + C + O(1/ \log x) $$ to conclude the same about $$ \sum\limits 1/(p+1) $$ ? I just need a little push.

Answer 1

The approach above does not provide the explicit value of $C_0$ in the estimate $$\sum_{p\leq x}\frac{1}{p+1}=\log \log x+C_0+O(\frac{1}{\log x}).$$ Here is how to remedy this: We have $$ \sum_{p\leq x}\frac{1}{p+1}= \sum_{p\leq x}\frac{1}{p}- \sum_{p\leq x}\frac{1}{p(p+1)} . $$ The estimate $$ \sum_{p\leq x}\frac{1}{p(p+1)} = \sum_{p}\frac{1}{p(p+1)} +O\left(\frac{1}{x\log x}\right) $$ shows that $$ C_0=C-\sum_{p}\frac{1}{p(p+1)} $$ where $C$ is defined by $$\sum_{p\leq x}\frac{1}{p}=\log \log x+C+O(\frac{1}{\log x}).$$