Weak versions of Bertrand's postulate 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.  
 A: 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.  
A: Pal Erdös' version can be found in "proofs from the book".  This was rather elementary.
A: This statement is false. For example, if $n=1.5$ and $x=3$, then there is no prime $p$ satisfying $x<p<nx$. 
On the other hand, it is easy to derive from the prime number theorem that, for a given $n>1$, there is a prime $p$ satisfying $x<p<nx$ for any sufficiently large $x$.
In fact much more is true. By a result of Baker-Harman-Pintz (Proc. London Math. Soc. 83 (2001), 532-562), there is a prime $x<p<x+x^{0.526}$ for any sufficiently large $x$.
