$ \pi (N)< B \ \frac{N}{\log N }$ where $ B\leq 4.$

Question

Let $p_{1}, p_{2},..., p_{r}$ are all the pimes not exceeding $ N$. We can say $ r< B \frac{N}{\log N }$ where $ B\leq 4$ .

I know Chebyshev proved the stronger version of it with $ B= 1.1055...$

But can we prove the above, which is not that much stronger as compared to Chebyshev's theorem, with some elementary arithmetic rather than following Chebyshev's line of proof ?

Any help would be appreciated. Thanks in advance.

Answer 1

The result holds for $N=1,2,3,4,5$. We now assume it is true for $N \ge 5$ and prove it for $2N-1$ (and hence $2N$ since $\pi(2N) = \pi(2N-1)$). Now

$$2^{2N-1} > \binom{2N-1}{N} = \frac{(2N-1)!}{N! (N-1)!}$$

and the RHS is divisible by all primes between $N+1$ and $2N-1$ because they divide the numerator but not the denominator. Suppose there are $m = \pi(2N-1) - \pi(N)$ primes in this range. Since each such prime is at least $N$, it follows that

$$2^{2N-1} > N^m,$$

and thus

$$m < \frac{(2N-1) \log 2}{\log N},$$

so, by induction,

$$\pi(2N-1) = \pi(N) + m \le \frac{4 N}{\log N} + \frac{(2N-1) \log 2}{\log N} < \frac{4 (2N-1)}{\log(2N-1)}.$$

where the inequality holds for $N \ge 5$. You can refine this argument to make i t work (for $N$ large enough) and any $B > \log 4$.