Explanation for Terry T. post I read here that :  " If one inserts these inequalities into the Legendre sieve and optimises the parameter, one can improve the upper bound for the number of primes in $[N,2N]$ to $$O \left(\frac{N \log \log N}{ \log N} \right)$$
My question is:How?
Am i missing something obvious?
I understand the estimate $\frac{N}{\log\log N}$ that he mentions earlier but the one above i don't.
Thank you in advance.
 A: This bound is derived by Murty and Saradha in their paper "On the Sieve of Eratosthenes", which can be downloaded from Murty's website here. Their result is to derive a slightly tighter bound for the error term of the sieve.
With the sieve of Eratosthenes-Legendre, to count all the prime numbers on the interval $[\sqrt{x}, x]$ one has to sieve with all the prime numbers less than $\sqrt{x}$. However this leads to a main term completely dominated by a large error term. If we instead pick a value $z < \sqrt{x}$ and sieve with all the primes less than $z$, then we can reduce the error term at a cost of not completely counting all the prime numbers in $[\sqrt{x}, x]$. The sieve then has the form:
$x \prod_{p < z} (1 - \frac{1}{p}) + \mathcal{O}(2^z)$
Using Rankin's trick, Murty and Saradha were able to improve the error term to
$x \prod_{p < z} (1 - \frac{1}{p}) + \mathcal{O}(x(\log z)^2 exp(-\frac{\log x}{\log z})) $
They were able to then deduce the bound
$\pi(x) \ll \frac{x}{\log x}(\log \log x) $
by setting $\log z = \epsilon \frac{\log x}{\log \log x}$ for sufficiently small $\epsilon$.
This is also covered in greater detail in Murty and Cojocaru's book "An Introduction to Sieve Methods and their Applications."
