Prove that if $\lim_n x_n = 0$ then $\lim_n \frac{x_1+x_2+...x_n}{n}=0$

Question

I need help proving that if $\lim_n x_n = 0$ then $\lim_n \frac{x_1+x_2+...x_n}{n}=0$

I know $x_1+x_2+...+x_n = \sum x_n$ is a series that is a converges uniformly if and only if for every $\varepsilon>0$ there exists M.

Answer 1

Fix $\epsilon$ to be as small as you please. Then (by definition) there exists an $N$ sufficiently large such that $|x_n| < \epsilon$ for all $ n >N$.

Then we have that $$ \bigg| \frac{x_1 + x_2 + ... x_N + x_{N+1} + ...+ x_n}{n} \bigg| \leq \bigg|\frac{x_1 + x_2 + ... x_N}{n}\bigg| + \frac{|x_{N+1}| + ... + |x_n|}{n} \\< \bigg|\frac{x_1 + x_2 + ... x_N}{n} \bigg| + \frac{n - (N+1)}{n} \epsilon \\ = \bigg|\frac{x_1 + x_2 + ... x_N}{n} \bigg| - \frac{(N+1)\epsilon}{n} + \epsilon $$

Letting $n \to \infty$ gives you that this the desired quantity is smaller than $\epsilon$ and since $\epsilon$ is chosen to be arbitrary, that quantity must be equal to $0$.

For practice, you should try reproducing the same argument where, instead of $x_n \to 0$ we have $x_n \to c$ for some real number $c$ and the problem is to show $$\frac{x_1 + x_2 + ... x_n}{n} \to c $$