1
$\begingroup$

Consider $p_k$ the $k^{th}$ prime number.

It is well known that:

$\forall \epsilon>0$, there exists $n_0\in\mathbb{N}$ such that $\forall n\in\mathbb{N}$, $n\geq n_0$ we have $\frac{p_{n+1}}{p_n}<\epsilon+1$

For example, case $\epsilon\geq 1$ is Bertrand's postulate and in this case $n_0=1$.

I am looking for an elementary (or as elementary as it can get) proof for the following:

$\frac{p_{n+1}}{p_n}<\frac{3}{2}$ for $n\geq 5$ $\big($i.e. $\epsilon=\frac{1}{2}$ and we want to prove $n_0=5\big)$

Thank you.

$\endgroup$
5
  • 4
    $\begingroup$ So, you're asking for an elementary proof of something that's stronger than Bertrand. Maybe you should start with an elementary proof of Bertrand, and try to sharpen it. $\endgroup$ – Gerry Myerson Nov 2 '20 at 12:28
  • 1
    $\begingroup$ $11,13,19,23,29$, Nagura $\endgroup$ – Daniel Fischer Nov 2 '20 at 13:23
  • $\begingroup$ I know the proof for Bertrand, but I do not know how to improve it. $\endgroup$ – Primenumber Nov 2 '20 at 17:01
  • $\begingroup$ @Daniel, how elementary is that proof? $\endgroup$ – Gerry Myerson Nov 3 '20 at 21:45
  • $\begingroup$ @GerryMyerson If "elementary" refers to the advancedness of tools used, I'd say it's pretty elementary. One needs a few facts about the $\Gamma$-function, and the Chebyshev functions. The calculations are a bit tedious, though, so I wouldn't outright say it's easy. But if one knows Chebyshev's Mémoire sur les nombres premiers one should be able to follow it without big problems. Judge yourself. $\endgroup$ – Daniel Fischer Nov 3 '20 at 22:00
1
$\begingroup$

M. El Bachraoui, Primes in the interval $[2n, 3n]$, Int. J. Contemp. Math. Sci., Vol. 1, 2006, no. 13, 617 - 621, proves

Theorem 1.3. For any positive integer $n > 1$ there is a prime number between $2n$ and $3n$.

The proof is too long to give here, but it looks to be similar to, and about as elementary as, the Erdos proof of Bertrand's Postulate.

$\endgroup$
4
  • $\begingroup$ How does that result answer the question? $\endgroup$ – lhf Nov 3 '20 at 12:57
  • $\begingroup$ @lhf it says there's a prime between $p_n$ and $(3/2)p_n$, so $p_{n+1}\le(3/2)p_n$, as was requested. $\endgroup$ – Gerry Myerson Nov 3 '20 at 21:42
  • 1
    $\begingroup$ Not quite, if we take $n = \frac{p_k+1}{2}$ that theorem guarantees (since $3n$ is composite) $p_{k+1} \leqslant 3n - 1 = \frac{3}{2}p_k + \frac{1}{2}$. When $p_k \equiv 1 \pmod{4}$, that bound is even, and we're good. But for $p_k \equiv 3 \pmod{4}$ that bound can be prime and we need a slightly stronger result. I suppose it would be easy to modify Bachraoui's proof to yield a prime in $[2n,3n-2]$ for $n \geqslant 5$ (or two primes in $[2n,3n]$). $\endgroup$ – Daniel Fischer Nov 3 '20 at 22:20
  • $\begingroup$ @Daniel, I'm sure you're right on all counts. $\endgroup$ – Gerry Myerson Nov 4 '20 at 0:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.