TITLE: Let $(p_n)_{n \in \mathbb{N}}$ be the sequence of prime numbers, then $\lim_{n \to \infty}\frac{p_{n+1}}{p_n} = 1$?
QUESTION [9 upvotes]: The Bertrand postulate is that $p_{n+1} < 2p_{n}$ but it is far to be enough to conclude. 
I made a program in $R$, to see the convergence of this sequence for the first $500$ prime numbers and it seems to converge to 1. Is this true, and how to prove it ?

REPLY [3 votes]: Here's a different proof of that fact, which doesn't make a reference to the prime number theorem assuming the limit exists. We'll use the fact that the series
$$\sum_{n=1}^\infty \frac{1}{p_n}$$
diverges, which has a nice elementary proof due to Erdos. Now suppose that the limit is more than $1$. Then $p_{n+1}/p_n>1+a$ for some $a>1$ and all $n\geq N$ for some $N$. Thus 
$$\sum_{n=1}^\infty \frac{1}{p_n}\leq \sum_{n=1}^{N-1} \frac{1}{p_n} + \frac{1}{p_n}\sum_{i=0}^\infty \frac{1}{(1+a)^i}$$
which converges.
It would be interesting to give an elementary (not using the prime number theorem) proof that the limit exists.