Question about proof that $\lim_{n \rightarrow \infty}a_n=A\implies\lim_{n \rightarrow \infty}\frac{a_1+...+a_n}{n}=A$

Question

I am trying to follow the proof that $$\lim_{n \rightarrow \infty}a_n=A\implies\lim_{n \rightarrow \infty}\frac{a_1+...+a_n}{n}=A$$ given in this answer to another MSE question.

We understand that since $(a_n)$ is convergent, it must be bounded. Let $|a_n|\leq K\ \forall\ n\in \mathbb{N}$. Now for $\epsilon >0$, let $N\in \mathbb{N}$ be such that $|a_n-A|<\epsilon\ ,\forall\ n>N$.

Consider $\left|\frac{a_1+a_2+...+a_n}n-A\right|=\left|\frac{(a_1-A)+(a_2-A)+...+(a_n-A)}n\right|\leq\frac{|a_1-A|}n+\frac{|a_2-A|}n+...+\frac{|a_n-A|}n$.

We are not sure what to do now. We know we need to show that the right side is less than $\epsilon$ I think. We understand that for $n$ large, the terms go to zero but how do we know that the first finite number of terms are less than $\epsilon$ too?

Answer 1

Here's your hint. If $a_n=A$ for all $n$, it would be easy, right? Now what if $A-\epsilon < a_n < A+\epsilon$ for all $n$? You could get that $|\frac{1}{n}\sum_{k=1}^n a_k - A | < \epsilon$. If the limit exists, it is similarly bounded. To avoid saying "if the limit exists", take the limsup, which always exists, and say $|\limsup \frac{1}{n}\sum_{k=1}^n a_k - A | < \epsilon$.

So here's the problem. You only have $|a_n-A|< \epsilon$ for $n \geq N$. What to do? Split it into two cases and work the rest from there: $$\frac{1}{n}\sum_{k=1}^n a_k = \frac{1}{n}\sum_{k=1}^N a_k + \frac{1}{n}\sum_{k=N+1}^n a_k.$$

Now see what you can do.