A question on prime density Let A = {c > 1 : there exists a natural number m, such that for every n > m, there is a prime between n and cn}. Bertrand's postulate says that A contains 2. My question is : Is inf A = 1 ? If not, what is the best lower bound known upto now?

To show this let $c = 1+\epsilon$  with $\epsilon>0.$
Using the prime number theorem once, 
$$\lim_{x \to \infty} \frac{\pi[(1+\epsilon)x] - \pi(x)}{x/ \ln x}$$ 
$$= \lim \frac{\pi[(1+ \epsilon)x]}{x/\ln x} - \lim \frac{\pi(x)}{x/ \ln x} $$
$$= (1+\epsilon) \lim \frac{\pi[(1+\epsilon)x]}{(1+\epsilon)x/\ln x}  - 1.$$  Then multiplying and dividing by $\ln[(1+\epsilon)x]$ and using the PNT again, 
$$= (1+\epsilon) \lim \frac{\pi[(1+\epsilon)x]}{(1+\epsilon)x/\ln[(1+ \epsilon)x]} \cdot \frac{\ln x}{\ln[(1+\epsilon)x]} -1$$
$$= (1+\epsilon)\cdot \lim \frac{\ln x}{\ln x + \ln (1+\epsilon )} - 1 =(1+\epsilon) - 1 = \epsilon$$
So 
$$\lim \frac{\pi[(1+\epsilon)x] - \pi(x)}{x/ \ln x} \sim \epsilon$$ or 
$$ \pi[(1+\epsilon)x] - \pi(x) \sim \frac{x}{\ln x }\epsilon. $$ 
Since $\epsilon\frac{x}{\log x}$ is monotone increasing there is some $x_0$ beyond which there are primes on $(x, (1+\epsilon)x).$ The above is basically due to Landau (any mistakes are due to me).

Schoenfeld proved in 1976 there is a prime on interval $(x,(1+1/16597)x)$ for all $x \geq 2010759.9.$ I guess the result has been superseded and that even the particular type of result is of mostly historical interest.
There is a result of Ramare and Yannick which I have not read from 2003 which claims a prime on $(x,(1+1/28313999)x$ for $x\geq 10726905041.$
A good summary of results about primes on short intervals is found in Yildirim's paper, A Survey of Results on Primes in Short Intervals, and he doesn't mention results of this sort at all. The theorem above gives that for any $\epsilon$ there is an $x_0$ out there somewhere. Schoenfeld's result may be a point of diminishing returns.
