# Quick question about lim and sup

I've got a question regarding a step in a proof, the situation is following:

Let $X_{1},\dots,X_{n}$ be independent, symmetric stochastic variables so that $\sum\limits_{n=1}^{\infty}X_{n}$ exists in probability. If I use the fact that convergent in probability implies the same convergence of the corresponding Cauchy sequence i can write: $$\lim_{n,m\to\infty}P(|S_{m}-S_{n}|>t)=0$$ where $S_{n}=\sum\limits_{i=1}^{n}X_{i}$. My textbook then says that from the triangle inequality we can write that as: $$\lim_{n\to\infty}\sup_{m\geq n+1}P(|S_{m}-S_{n}|>t)=0$$ Could someone please help me fill in the blanks?

I hope I didn't leave anything relevant out, let me know if you feel i did.

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This uses the property of Cauchy sequences. One has that for every $\epsilon>0$, there exists an $N$ such that for all $n,m\geq N$, $P(|S_n-S_m|>t)<\epsilon$. So if we temporarly anchor $n>N$, the inequality holds even after taking the supremum over $m\geq n+1$, so long as $n\geq N$.
As far as triangle inequality goes, likely the book is referring to the fact that $|S_n-S_m|\leq |S_n-S_N|+|S_m-S_N|$ which implies that if $|S_n-S_m|>t$ then either $|S_n-S_N|>t/2$ or $|S_m-S_N|>t/2$, so that one can write:
$P(|S_n-S_m|>t) \leq P(|S_n-S_N|>t/2) + P(|S_m-S_N|>t/2)$
and so if we again think of $N$ as any large anchored quantity, then the two terms on the right can be made uniformly small for $n,m\geq N$. In other words if you take the supremum in $m$ over $m\geq n+1$, then the right hand side can be made arbitrarely small by picking a large enough $n$ and $N$. Essentially this "trick" decouples $n$ and $m$ from eachother.