$ \pi(n) > \frac{n}{22 \ln(n)} $ simple proof?

Question

What is the easiest way to show that

$$ n > 2 $$

$$ \pi(n) > \frac{n}{2 \ln(n)} $$

Where $\pi(n)$ is the prime counting function.

I read a proof of the PNT with the zeta function but this statement is much weaker !!

What is the shortest proof ? The simplest ? The most elementary ?

Do we use results from Mertens ? ( $\Pi ( 1 - 1/p) $ or $ \sum 1/p $ ) Do we need to use results of Mertens ? Do we need to estimate $\sum \ln(p) $ ?

How about the even weaker

$$ \pi(n) > \frac{n}{22 \ln(n)} $$

Is that even easier ? Or not ?

Answer 1

I think, the proof given by Dusard in his thesis is the best one: we have $$ \pi(x) \ge \frac{x}{\log(x)}\left( 1+\frac{1}{\log(x)}+\frac{1.8}{\log^2(x)}\right) $$ for all real $x\ge 32299$, and this bound is sharp. For all $x\ge x_0$ we obtain the bounds in general, and then we test for $\frac{1}{2}$ or $\frac{1}{22}$ for all $x\le x_0$ by a direct computation.

There is also an elementary proof by Chebychev, see here, with the variant Daniel is suggesting above. But it is also a bit tedious to do this and a general results perhaps seems to be useful then.

Answer 2

Too long for a comment:

In Section 4.5 of Apostol's Introduction to Analytic Number Theory (page 82-84), we have

Theorem 4.6 For every integer $n\ge2$, we have

$${1\over6}{n\over\log n}\lt\pi(n)\lt6{n\over\log n}$$

The proof is elementary. This takes care of the OP's question with $22$ in the denominator, but leaves open the question whether there's a simple proof if you replace Apostol's $6$ with a $2$. Apostol does, however, introduce the theorem saying, "Although better inequalities can be obtained with greater effort (see [60]) the following theorem is of interest because of the elementary nature of its proof." Reference [60] is:

Rosser, J. Barkley, and Schoenfeld, Lowell (1962) Approximate formulas for some functions of prime number theory. Illinois J. Math., 6:69-94; MR 25, #1139.

Answer 3

I think this $f(x)$ and $g(x)$ might help answer the question :

$f(x) + f(x/2) + f(x/3) + f(x/4) + ... = x$ and $\lim_{n \to \infty} \frac{f(n)}{\pi(n)} = 1$?

In particular combining the answer I gave and the related asymptotic to the harmonic sum

$$ H(x) = 1 + 1/2 + 1/3 + 1/4 + ... 1/x = \ln(x) - 1$$

Although the details are not yet clear to me.

The idea is to relate $f(x)$ and $g(x)$.

An easy proof of conjecture $A$ (not using the PNT) would be very useful. But maybe it is not required.

Just my 50 cent.

Answer 4

The following works for integer $n$ larger than $82$.

We desire to prove the PNT up to a multiplicative constant. This is much easier than proving the PNT and was in fact done by Chebyshev(first) and others.

using https://en.wikipedia.org/wiki/Proof_of_Bertrand%27s_postulate

We need the following two lemmas from that page.

Lemma 1: ${2n \choose n} > \frac{4^n}{2n+1}$.

Lemma 2: The greatest power $R(p, n)$ of a prime $p$ dividing ${2n \choose n}$ satisfies $p^{R(p, n)} \le 2n$.

From these two lemmas it follows that

$$\frac{4^n}{2n+1} < {2n \choose n} \le (2n)^{\pi(2n)}$$

which is a contradiction for large $n$ if $\pi(2n)$ is bounded. This gets us within a constant of the PNT:

$$\pi(2n) \ge \frac{n \log 4 - \log (2n+1)}{\log 2n}.$$