# If $\lim a_n = L$, then $\lim s_n = L$

If $\lim a_n = L$, then $\lim s_n = L$, where $s_n = \frac{ a_1 + \dots + a_n}{n}$

I know that:
$|s_n – L| = |\frac{a_1 + \dots + a_n}{n} – L| = \frac{1}{n}|a_1 + \dots + a_n – nL| \leq \frac{1}{n}\sum_{i=1}^{n}|a_i -L|$

#### Solutions Collecting From Web of "If $\lim a_n = L$, then $\lim s_n = L$"

Let $\varepsilon > 0$. You know that for some $N$, $|a_n – L| < \varepsilon/2$ for $n > N$.

We also know that there exists some $K$ such that $|a_n – L| < K$ for all $n$. For example, you can take $K = 1 + \text{max}(|a_1 – L|, \cdots, |a_N – L|,\varepsilon/2)$.

Now take any $m > N$. We have:

$$|s_m – L| \leq \frac{1}{m}\sum_{i = 1}^m|a_i – L| = \frac{1}{m}\left(\sum_{i = 1}^N|a_i – L| + \sum_{i = N + 1}^{m}|a_i – L|\right) < \frac{1}{m}(NK + (m – N)\frac{\varepsilon}{2}) < \frac{NK}{m} + \frac{\varepsilon}{2}$$

Now just take $M \geq \text{max}(N,\frac{2NK}{\varepsilon})$. So, for $m > M$, $\frac{NK}{m} < \frac{\varepsilon}{2}$, which implies $|s_m – L| < \varepsilon$.

let $\epsilon>0$ has given.$$\exists N\in\mathbb{N}~~ \text{s.t.}~~~ n > M, |a_n − L| <\frac{\epsilon}{2}$$ so
$$|s_n-l| = \left|\frac{ a_1 + \dots + a_n}{n}-l\right|= \left|\frac{ a_1 + \dots + a_n-nl}{n}\right|= \left|\frac{ a_1-l + \dots+a_N-l}{n}\right| + \left|\frac{a_N+l+\dots + a_n-l}{n}\right| < \left|\frac{ a_1-l + \dots+a_N-l}{n}\right| +\frac{(n-N)\epsilon}{2n}$$ suppose $c=a_1-l + \dots+a_N-l$ thus first fraction=$\frac{c}{n}\to 0$ this will give you $N_2>N ~~\text{s.t.}~~\left|\frac{c}{n}\right|<\frac{\epsilon}{2}~~~ \forall n>N_2$