Prove that if $$ y_n = \frac{x_1+\cdots + x_n}{n}$$ has a finite limit, then $x_n$ has the same limit as $y_n$.

Intuitively it is easy to understand. There are infinite number of $y_i$ which are extremely close to it’s limit (let us call it $a$), then it is really intuitive why limit of $x_n$ has to be $a$ too. If it wasn’t then either we would have different limit or we wouldn’t have limit at all.

I am pretty bad in such formal proofs, so all hints and help will be appreciated!

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    $\begingroup$ It is entirely possible for $x_n$ to have no limit. But if $x_n$ has a limit, it must be the same as the limit of $y_n$. $\endgroup$ – Arthur Oct 7 '20 at 7:18
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    $\begingroup$ This is false. We can even have an unbouded sequence $(x_n)$ such that $(y_n)$ has a finite limit. $\endgroup$ – Kavi Rama Murthy Oct 7 '20 at 7:19
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    $\begingroup$ @KaviRamaMurthy and what would the counter example be? Something like $(-1)^n$? $\endgroup$ – math-traveler Oct 7 '20 at 7:23
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    $\begingroup$ @math-traveler Yes, $(-1)^{n}$ is not convergent but the corresponding $y_n$'s tend yo $0$. $\endgroup$ – Kavi Rama Murthy Oct 7 '20 at 7:25
  • $\begingroup$ For an unbounded example, take a sequence that is mostly $0$, but occasionally (such as whenever $n$ is a power of $2$) is $\log n$. $\endgroup$ – Arthur Oct 7 '20 at 7:59

This is not true. If you take $x_n =(-1)^n $ then $$|y_n | =\left|\frac{x_1 + x_2 +... x_n }{n}\right|\leq \frac{1}{n}\to 0$$ and hence $$y_n \to 0$$ but $x_n $ does not converges to $0.$

But the reverse theorem is true. Namely if $x_n \to a$ then $y_n = \frac{x_1 + x_2 +... x_n }{n} \to a.$


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