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.

  • 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
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    $\begingroup$ $11,13,19,23,29$, 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

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.

  • $\begingroup$ 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
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    $\begingroup$ 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

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