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I've been trying to prove the following proposition: $$\lim _{n\to \infty }\left(x_n\right)=0\:\:\rightarrow \:\lim _{n\to \infty }\left(\frac{x_1+x_2+...+x_n}{n}\right)=0$$

This is a proof sketch, about which I feel a bit uncertain:

$$\left|\frac{x_1+x_2+...+x_n}{n}-0\right|=\left|\frac{x_1+x_2+...+x_n}{n}\right|=\frac{\left|x_1+x_2+...+x_n\right|}{n}$$ Applying the triangle inequality:

$$\frac{\left|x_1+x_2+...+x_n\right|}{n}\:\le _{ }\:\frac{\left|x_1\right|+\left|x_2\right|+...+\left|x_n\right|}{n}$$ $x_n$ is bounded, thus there exists some $M>0$ such that $\left|x_n\right|<M$ for all $n\in \mathbb{N}$, and therefore: $$\frac{\left|x_1\right|+\left|x_2\right|+...+\left|x_n\right|}{n}\le _{\:\:}\frac{\left(n-1\right)M\:+\left|x_n\right|}{n}\:\le M\:+\frac{\left|x_n\right|}{n}\le M+\left|x_n\right|$$

$x_n\rightarrow 0,\:$thus$\:\forall \:\left(ϵ-M\right)\:>0,\:\:\exists \:N\in \mathbb{R}\:$such that$\:\left|x_n\right|<ϵ-M$, and therefore:

$$\left|\frac{x_1+x_2+...+x_n}{n}\right|<...<M\:+\left|x_n\right|<ϵ$$

I'd be really glad for reviews and corrections if I've had any mistakes somewhere, it would be very helpful. Thanks!

Blabla
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    That would only apply for when $\epsilon>M,$ and you need to prove it for all $\epsilon>0.$ So this doesn't work. – Thomas Andrews Apr 12 '23 at 14:45
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    Basically, you need a lot of small terms in the numerator to balance out the terms that are closer to $M.$ – Thomas Andrews Apr 12 '23 at 14:47
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    Try to break down the sum in the numerator into two pieces: one whose contribution is small because of the denominator $n\to\infty$ and another one whose contribution is small because the addends are small. – GReyes Apr 12 '23 at 15:16
  • See https://math.stackexchange.com/questions/155839/on-ces%c3%a0ro-convergence-if-x-n-to-x-then-z-n-fracx-1-dots-x-nn-to – student91 Apr 12 '23 at 15:44

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