# $\frac{p_{n+1}}{p_n}\leq\frac{3}{2}$ for $n\geq 5$

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.

• 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. – Gerry Myerson Nov 2 '20 at 12:28
• $11,13,19,23,29$, Nagura – Daniel Fischer Nov 2 '20 at 13:23
• I know the proof for Bertrand, but I do not know how to improve it. – Primenumber Nov 2 '20 at 17:01
• @Daniel, how elementary is that proof? – Gerry Myerson Nov 3 '20 at 21:45
• @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. – 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 $$3n$$.
• @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. – Gerry Myerson Nov 3 '20 at 21:42
• 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 $[2n,3n]$). – Daniel Fischer Nov 3 '20 at 22:20