# Convergence of a sequence… [duplicate]

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.

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

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