What is $\limsup\limits_{n\rightarrow\infty}\left\{\frac{p_{m+1}}{p_m}|m\in \mathbb{N},m\geq n\right\} $?

Question

What is $$\limsup\limits_{n\rightarrow\infty}\left\{\frac{p_{m+1}}{p_m}\middle|m\in \mathbb{N},m\geq n\right\} = ?$$ where $p_i$ is i'th prime number. We know that this limsup exists because of bertrand's postulate that says for every natural number $n>1$ there exists a prime like $p$ that $n<p<2n$.

Answer 1

Since there is always a prime between two consecutive, big enough, cubes (Ingham's theorem - it is quite an advanced result that goes in the direction of proving that there is always a prime between two consecutive squares, that is still a conjecture) such $\limsup$ is just one.

However, the fact that $p_{n+1}-p_n=o(p_n)$ dates back to $1896$ (year of first proof of the PNT by Hadamard and, indipendently, de la Vallée-Poussin), or to $1949$, when Erdos and Selberg found their brilliant elementary proof.

This because for any constant $c>1$, the PNT shows that $\pi(cn)-\pi(n)$ is eventually positive, hence that $\limsup$ is less or equal to $c$ for any $c>1$. Since it is obviously greater or equal to $1$, it is just one.