What is $\limsup_{n\to\infty} \frac{p_{n+1}}{p_n}$? Let $(p_n)_{n\in\mathbb N}$ be the strictly increasing sequence of all primes.
I'm wondering what
$$S:=\limsup_{n\to\infty} \frac{p_{n+1}}{p_n}$$
is. Is the result already known? By Bertrand's postulate we get $S\le 2$. Can that bound be improved? Do we maybe have $S=1$?
Remark: I'm very thankful to all your answers.
 A: Yuan-You Fu-Rui Cheng, Explicit estimate on primes between consecutive cubes, showed that for $x\ge\exp(\exp(15))$ there is at least one prime between $x^3$ and $(x+1)^3$. Thus, for sufficiently large $n$ we must have
$$\frac{p_{n+1}}{p_n}<\frac{\left(p_n^{1/3}+1\right)^3}{p_n}<1+\frac7{p_n^{2/3}}\;,$$
and $S=1$.
A: There's no need to quote papers on prime gaps - this follows directly from the prime number theorem which states that $$\pi(x)=\sum_{p\leq x} 1=\frac{x}{\log x}+O\left(\frac{x}{\log ^2 x}\right).$$ Indeed, if there existed a $c>1$ such that $p_{n+1}>cp_n$ infinitely often, then we would have $\pi(cx)=\pi(x)$ for infinitely many values of $x$, however this is impossible since $$\frac{cx}{\log cx}-\frac{x}{\log x}\sim \frac{(c-1)x}{\log x}\neq O\left(\frac{cx}{\log^2 x}\right).$$
A: the limsup is one, see the corollary to Theorem 3 in Rosser and Schoenfeld (1962). The information also allows you to find the maximum of the ratio, which is likely 3/2, but you can find out for sure. In any case, for $n \geq 6,$
$$ \frac{p_{n+1}}{p_n} < 1 + \frac{\log \log n}{\log n}, $$
so the limit is $1.$

A: The value is $1$. If $c>1$ is any constant, then
$$\pi(nc) - \pi(n) \approx \frac{nc}{\log(nc)} - \frac{n}{\log(n)}$$
converges to $\infty$ when $n\to \infty$ (one can make the "$\approx$" precise). Thus for large enough $n$, there is always a prime between $n$ and $nc$, and so for all sufficiently large $n$, $p_{n+1} < cp_n$. This is enough to show that the lim sup is $1$. 
A: It follows from the Prime Number Theorem that 
$$
\lim_n \frac{p_{n+1}}{p_n}=\lim_n\left( \frac{p_{n+1}}{(n+1)\ln(n+1)}\cdot \frac{n\ln(n)}{p_n}\cdot \frac{(n+1)\ln(n+1)}{n\ln(n)}\right)=1 \cdot 1 \cdot 1
$$
Now, since $ \lim_n \frac{p_{n+1}}{p_n}=1$ then the limsup exists and is the same.
A: Suppose $r>0$ and that  $p_{n+1}/p_n>(1+r)$ for infinitely many $n.$ We have $p_{n+1}/\log p_{n+1}\sim (n+1)$ and $p_n/\log p_n\sim n.$
Then for infinitely many $n$ we have $$1+r/2<(n+1)(\log p_{n+1})/(n\log p_n)<$$ $$<(1+1/n)(\log (1+r)+\log p_n)/\log p_n=(1+1/n)(1+(\log (1+r))/\log p_n).$$ But this last expression tends to $1$ as $n\to \infty.$
