# Bertrand's postulate

In number theory, Bertrand's postulate is the theorem that for any integer $n>3$, there exists at least one prime number $p$ with
    $n<p<2n-2.$
A less restrictive formulation is: for every $n>1$, there is always at least one prime $p$ such that
    $n<p<2n.$
Another formulation, where $p_{n}$ is the $n$-th prime, is: for $n\geq 1$
    $p_{n+1}<2p_{n}.$
This statement was first conjectured in 1845 by Joseph Bertrand (1822–1900). Bertrand himself verified his statement for all integers $2\leq n\leq 3\,000\,000$.
His conjecture was completely proved by Chebyshev (1821–1894) in 1852 and so the postulate is also called the Bertrand–Chebyshev theorem or Chebyshev's theorem. Chebyshev's theorem can also be stated as a relationship with $\pi (x)$, the prime-counting function (number of primes less than or equal to $x$):
    $\pi (x)-\pi {\bigl{\tfrac {x}{2}}{\bigr )}\geq 1,{\text{ for all }}x\geq 2.$

## Prime number theorem

The prime number theorem (PNT) implies that the number of primes up to x, π(x), is roughly x/log(x), so if we replace x with 2x then we see the number of primes up to 2x is asymptotically twice the number of primes up to x (the terms log(2x) and log(x) are asymptotically equivalent). Therefore, the number of primes between n and 2n is roughly n/log(n) when n is large, and so in particular there are many more primes in this interval than are guaranteed by Bertrand's postulate. So Bertrand's postulate is comparatively weaker than the PNT. But PNT is a deep theorem, while Bertrand's Postulate can be stated more memorably and proved more easily, and also makes precise claims about what happens for small values of n. (In addition, Chebyshev's theorem was proved before the PNT and so has historical interest.)
The similar and still unsolved Legendre's conjecture asks whether for every n ≥ 1, there is a prime p such that n$^2$ < p < (n + 1)$^2$. Again we expect that there will be not just one but many primes between n$^2$ and (n + 1)$^2$, but in this case the PNT does not help: the number of primes up to x$^2$ is asymptotic to x$^2$/log(x$^2$) while the number of primes up to (x + 1)$^2$ is asymptotic to (x + 1)$^2$/log((x + 1)$^2$), which is asymptotic to the estimate on primes up to x$^2$. So, unlike the previous case of x and 2x, we do not get a proof of Legendre's conjecture for large n. Error estimates on the PNT are not (indeed, cannot be) sufficient to prove the existence of even one prime in this interval. In greater detail, the PNT allows to estimate the boundaries for all ε > 0, there exists an S such that for x > S:
    $(1-\varepsilon ){\frac {x^{2}}{2\log x}}\;<\;\pi (x^{2})\;<\;(1+\varepsilon ){\frac {x^{2}}{2\log x}}\;,$
    $(1-\varepsilon ){\frac {(x+1)^{2}}{2\log(x+1)}}\;<\;\pi ((x+1)^{2})\;<\;(1+\varepsilon ){\frac {(x+1)^{2}}{2\log(x+1)}}\;.$
The ratio between the lower bound π((x+1)$^2$) and the upper bound of π(x$^2$) is
    ${\frac {(x+1)^{2}}{x^{2}}}\cdot {\frac {\log x}{\log(x+1)}}\cdot {\frac {1-\varepsilon }{1+\varepsilon }}\;.$
Note that since ${\frac {(x+1)^{2}}{x^{2}}}\rightarrow 1$ when $x\rightarrow \infty$, ${\frac {\log x}{\log(x+1)}}<1$ for all x > 0, and ${\frac {1-\varepsilon }{1+\varepsilon }}<1$ for a fixed ε, there exists an R such that the ratio above is less that 1 for all x > R. Thus, it does not ensure that there exists a prime between π(x$^2$) and π((x+1)$^2$). More generally, these simple bounds are not enough to prove that there exists a prime between π(x$^n$) and π((x+1)$^n$) for any positive integer n > 1.

## Generalizations

In 1919, Ramanujan (1887–1920) used properties of the Gamma function to give a simpler proof than Chebyshev's. His short paper included a generalization of the postulate, from which would later arise the concept of Ramanujan primes. Further generalizations of Ramanujan primes have also been discovered; for instance, there is a proof that
    $2p_{i-n}>p_{i}{\text{ for }}i>k{\text{ where }}k=\pi (p_{k})=\pi (R_{n})\,,$
with p$_k$ the kth prime and R$_n$ the nth Ramanujan prime.
Other generalizations of Bertrand's postulate have been obtained using elementary methods. (In the following, n runs through the set of positive integers.) In 1973, Denis Hanson proved that there exists a prime between 3n and 4n. In 2006, apparently unaware of Hanson's result, M. El Bachraoui proposed a proof that there exists a prime between 2n and 3n. El Bachraoui's proof is an extension of Erdős's arguments for the primes between n and 2n. Shevelev, Greathouse, and Moses (2013) discuss related results for similar intervals.
Bertrand’s postulate over the Gaussian integers is an extension of the idea of the distribution of primes, but in this case on the complex plane. Thus, as Gaussian primes extend over the plane and not only along a line, and doubling a complex number is not simply multiplying by 2 but doubling its norm (multiplying by 1+i), different definitions lead to different results, some are still conjectures, some proven.

## Sylvester's theorem

Bertrand's postulate was proposed for applications to permutation groups. Sylvester (1814–1897) generalized the weaker statement with the statement: the product of k consecutive integers greater than k is divisible by a prime greater than k. Bertrand's (weaker) postulate follows from this by taking k = n, and considering the k numbers n + 1, n + 2, up to and including n + k = 2n, where n > 1. According to Sylvester's generalization, one of these numbers has a prime factor greater than k. Since all these numbers are less than 2(k + 1), the number with a prime factor greater than k has only one prime factor, and thus is a prime. Note that 2n is not prime, and thus indeed we now know there exists a prime p with n < p < 2n.

## Erdős's theorems

In 1932, Erdős (1913–1996) also published a simpler proof using binomial coefficients and the Chebyshev function $\vartheta$, defined as:
    $\vartheta (x)=\sum _{p=2}^{x}\log(p),$
where p ≤ x runs over primes. See proof of Bertrand's postulate for the details.
Erdős proved in 1934 that for any positive integer k, there is a natural number N such that for all n > N, there are at least k primes between n and 2n. An equivalent statement had been proved in 1919 by Ramanujan (see Ramanujan prime).

## Better results

It follows from the prime number theorem that for any real $\varepsilon >0$ there is a $n_{0}>0$ such that for all $n>n_{0}$ there is a prime $p$ such that $n<p<(1+\varepsilon )n$. It can be shown, for instance, that
    $\lim _{n\to \infty }{\frac {\pi ((1+\varepsilon )n)-\pi (n)}{n/\log n}}=\varepsilon ,$
which implies that $\pi ((1+\varepsilon )n)-\pi (n)$ goes to infinity (and, in particular, is greater than 1 for sufficiently large $n$).
Non-asymptotic bounds have also been proved. In 1952, Jitsuro Nagura proved that for $n\geq 25$ there is always a prime between $n$ and ${\bigl1+{\tfrac {1}{5}}{\bigr )}n$.
In 1976, Lowell Schoenfeld showed that for $n\geq 2\,010\,760$, there is always a prime $p$ in the open interval $n<p<{\bigl1+{\tfrac {1}{16\,597}}{\bigr )}n$.
In his 1998 doctoral thesis, Pierre Dusart improved the above result, showing that for $k\geq 463$, $p_{k+1}\leq \left(1+{\frac {1}{2\log ^{2}{p_{k}}}}\right)p_{k}$, and in particular for $x\geq 3\,275$, there exists a prime $p$ in the interval $x<p\leq \left(1+{\frac {1}{2\log ^{2}{x}}}\right)x$.
In 2010 Pierre Dusart proved that for $x\geq 396\,738$ there is at least one prime $p$ in the interval $x<p\leq \left(1+{\frac {1}{25\log ^{2}{x}}}\right)x$.
In 2016, Pierre Dusart improved his result from 2010, showing (Proposition 5.4) that if $x\geq 89\,693$, there is at least one prime $p$ in the interval $x<p\leq \left(1+{\frac {1}{\log ^{3}{x}}}\right)x$. He also shows (Corollary 5.5) that for $x\geq 468\,991\,632$, there is at least one prime $p$ in the interval $x<p\leq \left(1+{\frac {1}{5\,000\log ^{2}{x}}}\right)x$.
Baker, Harman and Pintz proved that there is a prime in the interval $[x-x^{0.525},\,x]$ for all sufficiently large $x$.
Dudek proved that for all $n\geq e^{e^{33.3}}$, there is at least one prime between $n^{3}$ and $(n+1)^{3}$.
Dudek also proved that the Riemann hypothesis implies that for all $x\geq 2$ there is a prime $p$ satisfying
    $x-{\frac {4}{\pi }}{\sqrt {x}}\log x<p\leq x.$

## Consequences

* The sequence of primes, along with 1, is a complete sequence; any positive integer can be written as a sum of primes (and 1) using each at most once.
* The only harmonic number that is an integer is the number 1.