If $z_n \to a$, then $\frac 1n \sum_{k=1}^n z_k \to a$

Question

If $\lim_{n \to \infty} z_n = a$, prove that $\lim_{n \to \infty} \frac 1 n \sum_{k = 1}^n z_k = a$.

Here is my attempt:

$$\left|\lim_{n \to \infty} z_n - a\right| \implies |z_n - a| < \epsilon \text{ for } n > N_1$$

$$\left|\frac 1 n \sum_{1}^n z_k - a\right| \le \frac 1n \sum_1^n |z_k - a|$$

I think I should make sum of $|(z_k-a)|$ less than epsilon but how can I do it?

You may find it helpful to look over the Mathjax formatting guide for writing math here/ — , Jan 05 '16 at 05:52

score 2 · Accepted Answer · answered Jan 05 '16 at 06:03

2

Note $$\Big | \sum_{k=1}^n \frac{z_k}{n} - a \Big | = \Big | \sum_{k=1}^n \frac{z_k-a}{n} \Big | \leq \sum_{k=1}^n\Big | \frac{z_k -a}{n} \Big | = \sum_{k=1}^{N_1} \Big | \frac{z_k -a}{n} \Big |+ \sum_{k=N_1+1}^n \Big | \frac{z_k -a}{n} \Big |$$ I believe you know how to make the second sum less than $\epsilon$, how about the first?

answered Jan 05 '16 at 06:03

GiantTortoise1729

3,048

the 2nd part should be less than epsilon/2, but how can i make the first part also less than epsilon/2,since k=1 to k=N_1-1 is annoying – Gillian Cheung Jan 05 '16 at 06:30
Let $M = \max{ |z_k -a| : k \in {1,\dots,N_1} }$. Then the first sum is $\leq MN_1/n$. How can you make this small? – GiantTortoise1729 Jan 05 '16 at 06:44

If $z_n \to a$, then $\frac 1n \sum_{k=1}^n z_k \to a$

1 Answers1