dourtuntellorvl

2022-06-29

Recall that Bertrand's postulate states that for $n\ge 2$ there always exists a prime between $n$ and $2n$. Bertrand's postulate was proved by Chebyshev. Recall also that the harmonic series

$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots $

and the sum of the reciprocals of the primes

$\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots $

are divergent, while the sum

$\sum _{n=0}^{\mathrm{\infty}}\frac{1}{{n}^{p}}$

is convergent for all $p>1$. This would lead one to conjecture something like:

For all $\u03f5>0$, there exists an $N$ such that if $n>N$, then there exists a prime between $n$ and $(1+\u03f5)n$.

Question: Is this conjecture true? If it is true, is there an expression for $N$ as a function of $\u03f5$?

$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots $

and the sum of the reciprocals of the primes

$\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots $

are divergent, while the sum

$\sum _{n=0}^{\mathrm{\infty}}\frac{1}{{n}^{p}}$

is convergent for all $p>1$. This would lead one to conjecture something like:

For all $\u03f5>0$, there exists an $N$ such that if $n>N$, then there exists a prime between $n$ and $(1+\u03f5)n$.

Question: Is this conjecture true? If it is true, is there an expression for $N$ as a function of $\u03f5$?

odmeravan5c

Beginner2022-06-30Added 20 answers

That result follows from the Prime Number Theorem. You can make it effective with a result of Dusart: for $n\ge 396738,$ there is always a prime between $n$ and $n+n/(25{\mathrm{log}}^{2}n)$. So in particular this holds for

$N\ge max(\mathrm{exp}\left(\sqrt{\frac{1}{25\epsilon}}\right),\text{}396738).$

On the Riemann hypothesis (using the result of Schoenfeld) there is a prime between $x-\frac{{\mathrm{log}}^{2}x\sqrt{x}}{4\pi}$ and $x$ for $x\ge 599$ and this should give a better bound.

$N\ge max(\mathrm{exp}\left(\sqrt{\frac{1}{25\epsilon}}\right),\text{}396738).$

On the Riemann hypothesis (using the result of Schoenfeld) there is a prime between $x-\frac{{\mathrm{log}}^{2}x\sqrt{x}}{4\pi}$ and $x$ for $x\ge 599$ and this should give a better bound.

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