Proving that if $\sum_{k = m}^{\infty}P(A_k) < \infty$ then $\lim_{m \rightarrow \infty}\sum_{k = m}^{\infty}P(A_k) = 0$.

Question

I want to prove that if $\sum_{k = m}^{\infty}P(A_k) < \infty$ then $\lim_{m \rightarrow \infty}\sum_{k = m}^\infty P(A_k) = 0$.

Bu I am not quite there, I will write where I got to trying to do this.

So $\sum_{k = m}^\infty P(A_k)$ converges because the sequence of partial sums is increasing and bounded.

Denote the sequence of partial sums as $b_a = P(A_1)+ P(A_2)+ \cdots + P(A_a) $

This means that for every $\epsilon >0$ there exists a $p \in N$ s.t. whenever $a \ge p$ then $|b_a - L| < \epsilon \implies -\epsilon < b_a - L < \epsilon \implies -\epsilon - b_a < - L \implies L < \epsilon + b_a $.

Now we know that $P(A_m)+ P(A_{m+1})+ \cdots \le L < \epsilon + b_a $ so, focusing now on the $\lim_{m \rightarrow \infty}$, if we choose $m = p$ we get that for every $\epsilon > 0$:

$$\left|\sum_{k = m}^\infty P(A_k) \right| \cdots$$

It seems pretty straightforward but I can't seem to pick the right $m$ to do the job.

Answer 1

Let $S_n = \Sigma_{k=1}^n P(A_k) $ and as $\Sigma_{k=m}^{\infty} P(A_k) <\infty$, we can deduce that $S_n$ converges and lets say it converges to $S < \infty$. Now, observe that $\Sigma_{k=m}^{\infty} P(A_k) = S-S_m$ and let $m \to \infty $ on both sides. So, we have

$$\lim_{m \to \infty}\Sigma_{k=m}^{\infty} P(A_k) = \lim_{m \to \infty}(S-S_m) = 0$$

Answer 2

I think this is true for any convergent series:

$$\sum_{n=1}^\infty a_n\;\;\;\text{converges}\;\;\iff\;\text{its partial sums sequence is Cauchy}\;\iff$$

$$\forall\;\epsilon>0\;\exists\; N\in\Bbb N\;\;s.t.\;\;m>N\;,\;p>0\implies |s_{m+p}-s_m|<\epsilon\iff$$

$$\left|\sum_{n=1}^{m+p} a_n-\sum_{n=1}^m a_n\right|=\left|\sum_{n=m+1}^{m+p}a_n\right|<\epsilon$$

Added: As written in my comment below, one can directly show the claim in the question without using Cauchy's Condition for series.

First, the series is positive as $\;P(A_k)\ge 0\;\;\forall\,k\;$ , so that the series converges iff it is bounded. Suppose

$$\begin{cases}\sum\limits_{n=1}^\infty a_n=A\\{}\\s_m:=\sum\limits_{n=1}^m a_n\end{cases}\;\;\;\implies \sum_{n=m+1}^\infty a_n= A-s_m\xrightarrow[m\to\infty]{}A-A=0$$

since $\;\{s_m\}\;$is the sequence of partial sums of the series.