We are interested in the following statement: 


For each $n>1$ and $x>2$ there is at least one prime $p$ satisfying $x<p<n x$.


For $n=2$ we get precisely the Bertrand's postulate which is true. As corollaries, the statements for arbitrary $n\geqslant 2$ are true. However, I am interested maybe there exist independent proofs (prossibly short and elementary) for some of the cases $n>2$.


Thanks for your help. 



Weak versions of Bertrand's postulate
Chebyshev used the divisibility of the middle binomial coefficient ${{2n}\choose{n}}$ to set upper and lower bounds on the number of primes in the form


$\frac{an}{\log(n)} > \pi(n) > \frac{bn}{\log(n)}$


for some constants $a$ and $b$. If $a$ is not too small and $b$ is not too large then the proofs can be elementary and quite short. See e.g. <http://www.fen.bilkent.edu.tr/~franz/nt/cheb.pdf> for proofs with $a = 6 \log(2)$ and $b = \log(2)/2$


There must then be a prime between $x$ and $nx$ if 


$ \frac{ax}{\log(x)} < \frac{bnx}{\log(nx)} $


which is equivalent to


$a \log(nx) < b n \log(x)$


so for the $a/b = 12$ as above you get


$12 \log(n) < (n-12) \log(x) $


$ x > \exp\left({\frac{12 \log(n)}{n-12}}\right)$.


So, for any value of $n > 12$ this gives a lower bound for $x$ above which there is always a prime between $x$ and $xn$; for smaller $x$ it can be checked by hand. 


