Why does $\sum_{i=1}^{\infty}\mathbb{P}\left(A_{i}\right)<\infty$ imply $\lim_{n\rightarrow\infty}\sum_{i=n}^{\infty}\mathbb{P}\left(A_{i}\right)=0$?

Question

I've encountered the fact in the title in a proof for the Borel–Cantelli lemma. Sure, it seems true, but can someone shed some light as to how to formally prove such a claim? (Rather then claim "obviously, we have..." as was done in the text).

Answer 1

We can rewrite the sum as,

$$\sum\limits_{i=n}^\infty\mathbb{P}(A_i) = \sum\limits_{i=1}^\infty\mathbb{P}(A_i)-\sum\limits_{i=1}^{n-1}\mathbb{P}(A_i)$$

and now take the limit as $n\rightarrow\infty$ on both sides,

$$\implies \lim\limits_{n\rightarrow\infty}\sum\limits_{i=n}^\infty\mathbb{P}(A_i) = \sum\limits_{i=1}^\infty\mathbb{P}(A_i)-\lim\limits_{n\rightarrow\infty}\sum\limits_{i=1}^{n-1}\mathbb{P}(A_i)$$

and notice that $\lim\limits_{n\rightarrow\infty} \sum\limits_{i=1}^{n-1}\mathbb{P}(A_i) = \sum\limits_{i=1}^{\infty}\mathbb{P}(A_i)$ to get,

$$\implies \lim\limits_{n\rightarrow\infty}\sum\limits_{i=n}^\infty\mathbb{P}(A_i) = 0$$

since $\lim\limits_{n\rightarrow\infty}\sum\limits_{i=1}^\infty\mathbb{P}(A_i)<\infty$

Answer 2

Let $\sum_{i=1}^{\infty}a_i$ be a convergent sequence and $S=\sum_{i=1}^{\infty}a_i$.

With $s_n:= \sum_{i=1}^{n}a_i$ and $r_n:=\sum_{i=n+1}^{\infty}a_i$ we have $s_n \to S$, $S=s_n+r_n$, hence $r_n=S-s_n$ and therefore $r_n \to S-S=0$.