Limit inferior of the quotient of two consecutive primes

I have recently read an article about the prime number theorem, in which Mathematicians Erdos and Selberg had claimed that proving $\lim \frac{p_n}{p_{n+1}}=1$, where $p_k$ is the $k$th prime, is a very helpful step towards proving the prime number theorem, although I don't know how, primarily because I have not gone through the proof of the theorem even once.

Anyway, I was trying to prove the result $\lim \frac{p_n}{p_{n+1}}=1$ by really elementary methods, as is always my habit, and quite recently I learned of a theorem that the sum of the reciprocals of the primes diverges (I am only a beginner). The idea that suddenly struck me is that this theorem implies $\limsup\frac{p_n}{p_{n+1}}=1$, by a simple application of the ratio test and the fact that $\frac{p_n}{p_{n+1}} < 1$ for all $n$. So, is there any simple way to show that $\liminf\frac{p_n}{p_{n+1}}=1$ or at least $\ge1$, so that the result is proved?

Another beautiful result that came to me, follows from the theorem on divergence of $\sum\frac{1}{p}$, that given any $h > 1$ there are infinitely many $n$ such that $p_n < h^n$. Are there any other results with really simple proofs (easy to understand even for beginners like me) having deep consequences in the theory of the distribution of primes? I'd really like to hear them!

Was just about to ask/confirm the same question, this is a good place to put my thoughts then. So, from Rosser's theorem

$$\ln(n) + \ln(\ln(n)) -1< \frac{p_{n}}{n}<\ln(n) + \ln(\ln(n))$$ $$\frac{1}{\ln(n+1) + \ln(\ln(n+1))} < \frac{n+1}{p_{n+1}} < \frac{1}{\ln(n+1) + \ln(\ln(n+1)) -1}$$ Or $$\frac{\ln(n) + \ln(\ln(n))-1}{\ln(n+1)+\ln(\ln(n+1))}< \frac{n+1}{n}\cdot \frac{p_{n}}{p_{n+1}}<\frac{\ln(n)+\ln(\ln(n))}{\ln(n+1)+\ln(\ln(n+1))-1}$$

From which: $$\lim_{n \to \infty } \frac{p_{n}}{p_{n+1}}=\lim_{n \to \infty } \frac{\ln(n)}{\ln(n+1)}=1$$

Also, these bounds apply $1 > \frac{p_{n}}{p_{n+1}} > \frac{1}{2}$.