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Let $(s_n)$ be a sequence with $s_n = \frac{x_1 + x_2 + \ldots + x_n}{n}, n\ge 1$. Show that if $(x_n)$ converges to $x$, then also $(s_n)$ does.

Since $(x_n) \to x$, given $\varepsilon > 0$, there exist $M \in \Bbb N$ such that for all $n \ge M$, we have $|x_n - x| < \varepsilon$.

To show: $|\frac{x_1 + x_2 + \ldots + x_n}{n} - x| < \varepsilon$.

How to get the approach? Thanks in advance.

Dan Asimov
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lap lapan
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  • You are not trying to show |(x_1+...+x_n)/n - x| < . Instead, you are trying to show that for any > 0, there exists an integer N > 0 (depending on ) such that n ≥ N implies that |(x_1+...+x_n)/n - x| < is true. – Dan Asimov Sep 21 '21 at 00:21
  • @DanAsimov Cause I do not know how to start since it is the partial sum of $x_k$. – lap lapan Sep 21 '21 at 00:23
  • This must be a duplicate of an existing question on this site.... – Greg Martin Sep 21 '21 at 00:45
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    As a brief conceptual (hand-waving) summary of the answer referenced by @angryavian : suppose that $M$ is chosen so that for $n \geq M, |x_n - x| < \epsilon/2.$ Further suppose that $|x_1 + \cdots x_M - Mx| = (k+1)\epsilon.$ In effect, you have an overage of $k\epsilon.$ Each term $x_n ~: n > M$ results in an undercut of $\epsilon/2$. If you take on an additional $2k$ terms, you will have undercut enough, so that the overall average is now at or below $\epsilon.$ So, you let $M_1 = M + 2k$. Then, for any $n \geq M_1$ the desired inequality will be produced. – user2661923 Sep 21 '21 at 00:54
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    Solved and I get it right now. Thanks Sir. – lap lapan Sep 21 '21 at 00:55

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