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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

$$ \sum_{n=0}^{\infty} \frac{1}{n^{p}} $$

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

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

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

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That result follows from the Prime Number Theorem. You can make it effective with a result of Dusart: for $n\ge396738,$ there is always a prime between $n$ and $n+n/(25\log^2 n)$. So in particular this holds for $$ N\ge\max\left(\exp\left(\sqrt{\frac{1}{25\varepsilon}}\right),\ 396738\right). $$

On the Riemann hypothesis (using the result of Schoenfeld) there is a prime between $x-\frac{\log^2x\sqrt x}{4\pi}$ and $x$ for $x\ge599$ and this should give a better bound.

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