show that if $x_n $ converges so does $\frac{x_1+x_2+x_3+...+x_n}{n}$

Question

If $x_n \to a$ then $$\lim_{n \to \infty}\frac{x_1+x_2+x_3+...+x_n}{n}=a$$

My attempt

$$=> \lim_{n \to \infty} x_1+x_2+x_3+...+x_n = na \\ => x_1-a +x_2-a +...+x_n-a = 0 \ where \ n \ is \ not \ finite$$

Since $x_n \to a$ Given $\epsilon >0$ there exists $N$ such that

$|x_n-a| < \epsilon$ whenever $n>N$

Applying to what we ended up with :

$n \epsilon = 0$

let $\epsilon \to 0$

$\lim ... = na$ this isn't really legal move. – rtybase Dec 28 '16 at 13:25 — rtybase, Dec 28 '16 at 13:25

score 0 · Answer 1 · answered Dec 28 '16 at 13:33

Let $c>0$, there exists $N$ such that $n>N$ implies that $|x_n-a|<c/2$, choose $M>N$ such that $(\sum_{i=1}^{i=N}|x_i-a|))/M<c/2$.

For $n>M$, you have $|(x_1+....+x_n)/n-a|\leq (\sum_{i=1}^{i=N}|x_i-a|)/n+\sum_{i=N+1}^{i=n}|x_i-a|$

$\leq (\sum_{i=1}^{i=N}|x_i-a|)/M+\sum_{i=N+1}^n|x_n-a|/n\leq c/2+(c/2)(n-N-1)/n<c$.

1 Answers1