# Why does this interval contain at least $x/(5\log(x))$ primes?

Let $$A_x = (x/2,x] \cup (x/4,x/3] \cup (x/6,x/5] \cup \cdots.$$ I can prove that $$\sum_{n \in A_x} \Lambda(n) = \log(2)x+O(\log(x)),$$ and that further, the error term cannot be improved.

Given this, an exercise in the book "A course in Analytic Number Theory" by Marius Overholt asks to deduce from this that the interval $$(x/2,x]$$ contains at least $$x/(5\log(x))$$ primes for all $$x$$ sufficiently large. How do I do this?

• The Wikipedia entry for "proof of Bertrand's postulate" may provide some insight... – abiessu Aug 2 at 20:15
• @abiessu Where exactly? – inequalitynoob2 Aug 2 at 20:36
• en.m.wikipedia.org/wiki/Proof_of_Bertrand%27s_postulate – Roddy MacPhee Aug 2 at 23:55
• I'm thinking of the section detailing why $\binom {2n}n$ doesn't have any prime factors arising from the interval $[\frac 23n,n]$. – abiessu Aug 3 at 0:07
• @abiessu I do not see how this helps here. If you have an answer, I would appreciate it if you wrote it down. – inequalitynoob2 Aug 3 at 8:23