Let $(p_n)_{n \in \mathbb{N}}$ be the sequence of prime numbers, then $\lim_{n \to \infty}\frac{p_{n+1}}{p_n} = 1$?

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 ?

Hint: Use the fact that $p_{n} /n\log n \to 1$ via Prime Number Theorem and your desired limit gets converted to $$\lim_{n\to\infty}\frac{(n+1)\log(n+1)}{n\log n}$$ which you can easily prove to be equal to $1$. Bounding the prime gap $p_{n+1}-p_{n}$ is a difficult problem and is not really needed here.

Yes, this is true. The current "state-of-the-art" estimate for prime gaps is $$p_{n+1} - p_n = O(p_n^{0.525}) $$ (due to Baker, Harman, Pintz), which guarantees that your limit is 1.

yes, for sufficiently large prime $p,$ the next prime is no larger than $$ p + p^{0.7} $$ Let me get link and correct exponent.

https://en.wikipedia.org/wiki/Prime_gap#Upper_bounds

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.

Let $p_{n+1}=(1+x_n)p_n .$ From the Prime Number Theorem we have, as $n\to \infty,$ $$1+o(1)=\frac {\pi(p_{n+1})}{p_{n+1}/\log p_{n+1}}=\frac {n(\log p_n+\log (1+x_n))}{(1+x_n)p_n}$$ $$\text {and }\quad 1+o(1)=\frac {\pi (p_n)}{p_n/\log p_n}=\frac {(n-1)\log p_n}{p_n}.$$

Taking the ratio of these two formulas, since $x_n>0$ we have $$1+o(1)=\frac {n}{n-1}\cdot \frac {1+(\log (1+x_n))/\log p_n}{1+x_n}<$$ $$<\frac {n}{n-1}\cdot \frac {1+x_n/\log p_n}{1+x_n}=$$ $$=\frac {n}{n-1}\left(1+\frac {x_n}{1+x_n}(-1+1/\log p_n)\right)=V(n).$$

Now $x_n>0,$ while $1/\log p_n\to 0$ as $n\to \infty$. So if $k>0$ and $x_n>k$ for infinitely many $k,$ we would have $$\lim_{m\to \infty}\inf_{n\geq m}V(n) \leq 1-\frac{k}{1+k}<1$$ contrary to $V(n)=1+o(1).$

Therefore $\lim_{n\to \infty}x_n=0.$