TITLE: Extensions of Bertrand's Postulate
QUESTION [7 upvotes]: Two questions came to mind when I was reading the proof for Bertrand's Postulate (there's always a prime between $n$ and $2n$):
(1) Can we change the proof somehow to show that: $\forall x > x_{0}$, there exists a prime $p$ $\in [x, ax]$, for some $a \in (1, 2)$? 
(2) Suppose the (1) is true, what is the smallest value of $x_{0}$?
I'm not sure how to prove either of them, any input would be greatly appreciated! And correct me if any of the above statement is wrong. Thank you!

REPLY [4 votes]: I think you would enjoy the page PRIME GAPS.  
My own version of the conjecture of Shanks, actually both a little stronger and a little weaker, is $$ p_{n+1} < p_n + 3 \,\left( \log p_n \right)^2,   $$
for all primes $p_n \geq 2.$ This is true as high as has been checked. 
Shanks conjectured that $$ \limsup \frac{p_{n+1} - p_n}{\left( \log p_n \right)^2} = 1,  $$ while Granville later corrected the number $1$ on the right hand side to $2 e^{- \gamma} \approx 1.1229,$ see CRAMER GRANVILLE. There is no hope of proving this, but I enjoy knowing what seems likely as well as what little can be proved. 
Here is a table from the third edition (2004) of Unsolved Problems in Number Theory by Richard K. Guy, in which $p = p_n$ is a prime but $n$ is not calculated, then $g = p_{n+1} - p_n,$ and $p = p(g),$ so $p_{n+1} = p + g.$ 
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