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Prove that if $a_n$ is convergent, then $M_n:=\frac{1}{n}\sum_{1}^{n} a_n$ satisfies $\lim M_n=\lim a_n$.

(sorry for English)

Lucas Rosas
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  • What have you tried? By the way, it looks like it is Cauchy's First Theorem on limits whose proof is available online. –  Apr 23 '15 at 11:49
  • If $\lim a_n = a$, then for every $\varepsilon > 0$ only finitely many terms of the sequence $(a_n)$ differ from $a$ by more than $\varepsilon$. What implication does this have on $\sum_{i = 1}^m a_m$ for some positive integer $m$? – A.P. Apr 23 '15 at 11:56

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The goal is to show that for all $\varepsilon > 0$ there is an index $n$ such that $|M_n-a|<\varepsilon$, where $a=\lim_na_n$. Fix $N$ such that $|a_n-a|<\varepsilon/2, n>N$. Then for $n>N$: $$\begin{aligned} \left|\frac{1}{n}\sum_{i=1}^na_i-a\right|&\leqslant\left|\frac{1}{n}\sum_{i=1}^n(a_i-a)\right|\\ & \leqslant\left|\frac{1}{n}\sum_{i=1}^N(a_i-a)\right|+\left|\frac{1}{n}\sum_{i=N+1}^n(a_i-a)\right|=O\left(\frac{1}{n}\right)+\varepsilon/2<\varepsilon \end{aligned}$$ provided $n$ is large enough.

frog
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  • Hi, frog, can you explain me why $$\left|\frac{1}{n}\sum_{i=1}^na_i-a \right| \leq \left|\frac{1}{n}\sum_{i=1}^n(a_i-a)\right| $$. I think that is equal. :/ And, what is O(1/n)?? – Lucas Rosas Apr 23 '15 at 15:01
  • Sure you are right, it is equal $\frac{1}{n}\sum_{i=1}^n(a_i-a)=\frac{1}{n}\sum_{i=1}^n a_i-\frac{1}{n}\sum_{i=1}^n a=\frac{1}{n}\sum_{i=1}^n a_i-\frac{1}{n}n a = \frac{1}{n}\sum_{i=1}^n a_i-a$. But the second inequality is an inequality by the triangle inequality. And $O(1/n)$ is a Landau symbol. It means that the first term in my last line of the inequality is bounded by a constant divided by $n$ and can be made arbitrary small provided $n$ is large enough. This holds since the finite sum is (of course) finite. – frog Apr 25 '15 at 11:26