Assume that for every $n\geq 1 $ we are given a random variable $X_n$ such that $(X_n-n)/\sqrt n$ follows the standard normal distribution. Furthermore, assume that the $X_n$ are independent. Fix any $z\geq 0 $. Can we estimate the quantity $$ \frac{1}{n}\#\{1\leq i\leq n :X_{i+1}-X_i>z\}?$$ I am thinking it should be tending to $\mathrm e ^{-z}$ but would like to have a proof.

Assume that for every $n\geq 1 $ we are given a real random variable $X_n$ such that $(X_n-n)/\sqrt n$ follows the standard normal distribution. Furthermore, assume that the $X_n$ are independent. Fix any $z\geq 0 $. Can we estimate the quantity $$ \frac{1}{n}\#\{1\leq i\leq n :X^{(i+1)}-X^{(i)}>z\},$$ where $X^{(m)} $ is the $m$-th smallest number among all $\{X_n:n\geq 1 \}$. I am thinking it should be tending to $\mathrm e ^{-z}$ but would like to have a proof.

(Edited $X_i\to X^{(i)}$.)