Fix a (large) number N and suppose that Bob chooses a random number n in the interval $1/2N ≤ n ≤ 3/2N$. If he repeats this process many times, prove that approximately $1/ ln(N)$ of his numbers will be prime. More precisely, define the probability that an integer n in the interval 1/2N ≤ n ≤ 3/2N is a prime number
$P(N)$ = (number of primes between $1/2N$ and $3/2N$) / (number of integers between $1/2N$ and $3/2N$)
and prove that
$ lim(n to ∞) P(N)/(1/ln(N)) = 1$
This shows that if $N$ is large, then $P(N)$ is approximately $1/ln(N)$.
What I have tried:
$P(N)$ = (π (3/2N) - π(1/2n))/N
= (3/2)/ln(3/2N)-(1/2)/ln(1/2N) as n goes to ∞
How to calculate the next steps in order to get 1/ln(N)?
Use that $\ln(\frac32N)=\ln(\frac32)+\ln N$ and similarly for the other denominator. Then use that, for any constant $a$ (like $\ln\frac32$ or $\ln\frac12$), the ratio $$ \frac{a+\ln N}{\ln N } $$ approaches $1$ as $N\to\infty$.