# How can I show the following sequence converge to $0$?

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 $0$ ?

• This doesn't seem correct without additional restriction on the events. What if $P(A_k)=1$ for all $k$.... – grand_chat Mar 21 '15 at 3:45
• Yeah, they need to be disjoint, I think. – Alan Mar 21 '15 at 3:45
• I edited my question. – user222186 Mar 21 '15 at 3:46

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 $S-S_n=\sum_{k=n}^{\infty} a_k$.
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.
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 $\lim_{n\to\infty} (y-y_n) = 0$.