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Suppose $\{ A_n \}_{n \geq 1} $ is a sequence of pairwise disjoint events and $P$ a probability. Im curious as to why

$$ \lim_{n \to \infty} \sum_{k=n}^{\infty} P(A_k) = 0 $$

I know that

$$ \sum_{k \geq n} P(A_k) = P\bigg( \bigcup_{k \geq n} A_k \bigg) \leq 1 $$

but does this implies that the sequence $\sum_{k \geq n} P(A_k) $ tends to $0$ ?

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  • $\begingroup$ This doesn't seem correct without additional restriction on the events. What if $P(A_k)=1$ for all $k$.... $\endgroup$ – grand_chat Mar 21 '15 at 3:45
  • $\begingroup$ Yeah, they need to be disjoint, I think. $\endgroup$ – Alan Mar 21 '15 at 3:45
  • $\begingroup$ I edited my question. $\endgroup$ – user222186 Mar 21 '15 at 3:46
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More generally, if $\sum_{k=1}^\infty a_k$ converges, then $\lim_{n\to\infty}\sum_{k=n}^\infty a_k=0$.

Proof: Let $S_n=\sum_{k=1}^{n-1} a_k$, and $S=\sum_{k=1}^{\infty} a_k$.

Then $S_n\to S$, or $S-S_n\to 0$. But $S-S_n=\sum_{k=n}^{\infty} a_k$.

In other words, this is a property of convergent series in general.

The reason that your series is necessarily convergent is that $S_n=\sum_{k=1}^n P(A_k)$ is bounded above by $1$ and non-decreasing, so it has to have a limit.

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Hint: If you know $y:=\sum_{k=1}^\infty P(A_k)$ is finite, then the sequence $y_n:=\sum_{k=1}^n P(A_k)$ converges to $y$ as $n\to\infty$. You're being asked to show that $\lim_{n\to\infty} (y-y_n) = 0$.

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