How to deduce the inequality to show that $\frac{\pi(x)}{x} \to 0$?

Question

I'd like to answer (why?) question above.

What I've tried is:

First, I guess that $1- (\kappa/2)$ might have been related to $\frac{x_m - (\kappa/2) x_m}{x_m}$.

and we know that $(\kappa/2)x_m \lt \pi(x_m) - \pi(x_{m-1})$.

So $x_m - (\kappa/2) x_m \gt x_m - \pi(x_m) + \pi(x_{m-1})$.

On the right hand side, it can be interpreted as the number of composites $\le x_m$ and primes $\le x_{m-1}$.

I couldn't relate this with the product of primes in $(x_{m-1}, x_m]$.

Could you give me a hint?

Thank you.

Answer 1

Let $p_k$ denote the $k$-th biggest primes in $(x_{m-1}, x_m]$,

and each $p_k$ is distinct for $k = 1,2, \dots, \lfloor\kappa/2\rfloor+1$.

For each $k$, we can show that

$$\left(1-\frac{1}{p_k}\right) \lt \left(1- \frac{1}{x_m -(k-1)}\right)=\frac{x_m - k}{x_m -(k-1)}$$

Multiplying for all primes in $(x_{m-1}, x_m]$ gives us

$$\prod_{x_{m-1} \lt p \le x_m}\left(1-\frac{1}{p}\right)\le \prod_{k=1}^{\lfloor\kappa/2\rfloor+1}\left(1-\frac{1}{p}\right) \lt \prod_{k=1}^{\lfloor\kappa/2\rfloor+1}\left(1- \frac{1}{x_m - (k-1)}\right)\lt \frac{x_m - (\kappa/2) x_m}{x_m}$$

So it's done.