- #1

evinda

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Let $(a_n)$ be a sequence of real numbers such that $a_n \to a$ for some $a \in \mathbb{R}$. I want to show that $\frac{a_1+a_2+\dots+a_n}{n} \to a$.

We have the following:

Let $\epsilon>0$.

Since $a_n \to a$, there is some positive integer $N$ such that if $n \geq N$, then $a-\epsilon<a_n<a+\epsilon$.

Let $b_n=\frac{a_1+a_2+\dots+a_n}{n}$, for $n \geq N$.

We have that $b_n=\frac{a_1+a_2+\dots+a_N}{n}+\frac{a_{N+1}+\dots+a_n}{n}$

and since

$\frac{(n-N)(a-\epsilon)}{n}<\frac{a_{N+1}+\dots+a_n}{n}<\frac{(n-N)(a+\epsilon)}{n}$

we have that

$\frac{C}{n}+\frac{(n-N)(a-\epsilon)}{n}<b_n<\frac{C}{n}+\frac{(n-N)(a+\epsilon)}{n}$

where $C=a_1+a_2+\dots+a_N$.

Can we now just let $n \to +\infty$ ?

Then we would get that $\lim_{n \to +\infty} \left( \frac{C}{n}+ \left( 1-\frac{N}{n}\right)(a-\epsilon)\right)< \lim_{n \to +\infty}b_n < \lim_{n \to +\infty} \left( \frac{C}{n}+\left( 1-\frac{N}{n}\right) (a+\epsilon)\right) \Rightarrow a-\epsilon<\lim_{n \to +\infty} b_n< a+\epsilon \Rightarrow \lim_{n \to +\infty} b_n=a$.

Is this right?

Because I found the proof online and there they pick $\lim_{n \to +\infty} \sup{b_n}$ in order to get the desired result. But is this necessary? (Thinking)