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 $\epsilon$?


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.