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Let $\{a_n\}$ be a sequence of real numbers. Define $\sigma_n = 1/n(a_1 + \dots + a_n)$. Suppose that $\lim a_n = a \in \mathbb{R}$. Show that $\lim \sigma_n = a$.

Here is my work so far...

Fix $\epsilon > 0$. We are given that for $N \in \mathbb{N}$ we have that $|a_n - a| < \epsilon/\alpha$ whenever $n \geq N$, and $\alpha$ will be choosen later. We must find $N \in \mathbb{\widehat{N}}$ so large such that $|\sigma_n - a| < \epsilon$ whenever $n \geq N$. Note that

$$|\sigma_n - a| = |1/n(a_1 + \dots + a_n) - a|...$$

Then I got stuck trying to work out the details. Should I try showing that the $\sigma_n$ is a Cauchy sequence, and somehow use the convergence of the $a_n's$? Thanks for your help.

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marked as duplicate by kccu, Math1000, Shubham Johri, Martin R, Community May 20 at 20:41

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Let's bring in another number $M=\max_{i=1..\infty}{|a_i-a|}$. So $$ |\sigma_n - a| = \frac{1}{n}| \sum_i(a_i -a)| \leq \frac{1}{n}\sum_i |a_i-a| \leq \frac{N}{n}M + \frac{n-N}{n} \epsilon $$

Now define $N_2 = \frac{NM}{\epsilon}$

For all $n > N_2$ $$ |\sigma_n - a| \leq \frac{N}{n}M + \frac{n-N}{N} \epsilon \leq \frac{NM}{N_2} + \epsilon = 2\epsilon $$

Thus, $\lim_{n\to\infty} \sigma_n = a$

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