# If $y_n = \frac{x_1 + x_2 + \cdots + x_n}{n}$ converges to $L$, than so is $x_n$

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!

• 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 $y_n$. – Arthur Oct 7 '20 at 7:18
• This is false. We can even have an unbouded sequence $(x_n)$ such that $(y_n)$ has a finite limit. – Kavi Rama Murthy Oct 7 '20 at 7:19
• @KaviRamaMurthy and what would the counter example be? Something like $(-1)^n$? – math-traveler Oct 7 '20 at 7:23
• @math-traveler Yes, $(-1)^{n}$ is not convergent but the corresponding $y_n$'s tend yo $0$. – Kavi Rama Murthy Oct 7 '20 at 7:25
• For an unbounded example, take a sequence that is mostly $0$, but occasionally (such as whenever $n$ is a power of $2$) is $\log n$. – 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 $$y_n = \frac{x_1 + x_2 +... x_n }{n} \to a.$$