# Recall that Bertrand's postulate states that for n &#x2265;<!-- ≥ --> 2 there always exi

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 $ϵ>0$, there exists an $N$ such that if $n>N$, then there exists a prime between $n$ and $\left(1+ϵ\right)n$.

Question: Is this conjecture true? If it is true, is there an expression for $N$ as a function of $ϵ$?
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odmeravan5c
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 $n+n/\left(25{\mathrm{log}}^{2}n\right)$. So in particular this holds for

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.