Limit involving prime and composite numbers

Question

Can someone help me to figure out what $$\lim_{n\to\infty} \frac {c_n}n-\frac{c_n}{p_n}-\frac {c_n}{n^2}$$ is equal to? I am pretty sure its $1$ and i tried many different things but i couldnt figure it out. $c_n$ is the nth composite number excluding $1$ and $p_n$ is the nth prime number. This limit is equal to $$\lim_{n\to\infty} c_n\frac {\gamma(n)}{n^2}$$ where $\gamma(x)$ is equal to how many numbers less or equal to $x$ are composite. Its kinda the inverse function of $c_n$

Answer 1

Let $p(x) = \frac{x}{\ln x}$, being the approximate prime counting function. That means there are approximately $\frac{x}{\ln x}$ primes less than or equal to x, and $x-\frac{x}{\ln x}$ composites less or equal to than x.

First, let's derive $p_n$. Since there are about $\frac{x}{\ln x}$ primes less than or equal to x, there are x primes less than or equal to $x\ln(x)$. So $p_n \sim n\ln n$. Now for $c_n$, there are $x(1-\frac{1}{\ln x})$ composites less than or equal to x, so there are x composites less than or equal to $\frac{x\ln x}{(\ln x) - 1}$. Therefore, $c_n \sim \frac{n\ln n}{(\ln n) - 1}$

So, your limit now becomes: $\frac{n\ln n}{n(\ln (n) - 1)} - \frac{1}{(\ln n) - 1} - \frac{n\ln n}{n^2(\ln (n) - 1)}$, and as n approaches infinity, the last two terms drop to zero, and a simple coefficient test shows the first term is equal to one. And, $$\lim_{n\to\infty} \frac {c_n}n-\frac{c_n}{p_n}-\frac {c_n}{n^2} = 1$$

Answer 2

$$\frac{(2m)!}{m!^2}={2m \choose m} \ge \prod_{m < p \le 2m} p, \qquad {2^{k+1} \choose 2^k}\le 8 {2^k \choose 2^{k-1}}, \qquad \prod_{p \le n} p \le \prod_{k \le \log_2(n)+1} {2^k \choose 2^{k-1}} \le 4^{4n}$$ If $\pi(n) \ge Cn$ then $\prod_{p \le n} p \ge q^{Cn-q}$ thus for $n$ large enough we must have $\pi(n) < Cn$ which means $$\lim_{n \to \infty} \frac{\pi(n)}{n} = 0$$ Whence $\lim_{n \to \infty} \frac{c_n}{n} = 1,\lim_{n \to \infty} \frac{c_n}{p_n} = \lim_{n \to \infty} \frac{n}{p_n} =0$ and $$\lim_{n\to\infty} \frac {c_n}n-\frac{c_n}{p_n}-\frac {c_n}{n^2} = 1-0-0$$