# How to prove $\lim_{n \to \infty}\frac{a_1+a_2+\dots+a_n}{n}=l.$ [duplicate]

When Let ${\{a_n}\}$ be a convergent sequence, $\lim_{n \to \infty}a_n=l.$ Prove that $\lim_{n \to \infty}\dfrac{a_1+a_2+\dots+a_n}{n}=l.$ Hint : Prove that the sequence ${\{a_n}\}$ is bounded first.

Please provide me hint or full solution. Thanks.

## marked as duplicate by Mhenni Benghorbal, Inceptio, Namaste, Tom Oldfield, Lord_FarinMay 27 '13 at 16:30

Since $a_n \to l$, given $\epsilon > 0$, there exists $N(\epsilon)$ such that for all $n>N(\epsilon)$, we have $$\left\vert a_n - l \right\vert < \epsilon \implies l-\epsilon < a_n < l+\epsilon$$ We now have \begin{align} \dfrac{a_1+a_2+\cdots+a_{N-1}+a_N + a_{N+1} + \cdots + a_m}m &\in \left( \dfrac{a+(m-N)(l-\epsilon)}m, \dfrac{a+(m-N)(l+\epsilon)}m\right)\\ & = \left(l-\epsilon + \dfrac{a-N(l-\epsilon)}m, l+\epsilon + \dfrac{a-N(l+\epsilon)}m \right) \tag{$\star$} \end{align} where $a=a_1 + a_2 + \cdots + a_N$. Now given $\epsilon>0$, there exists $M(N,l,a,\epsilon)$ we have that for $m > M(N,l,a,\epsilon)$ such that $$\left \vert \dfrac{a-N(l+\epsilon)}m \right \vert < \epsilon$$ This gives us that $(\star)$, $\left(l-\epsilon + \dfrac{a-N(l-\epsilon)}m, l+\epsilon + \dfrac{a-N(l+\epsilon)}m \right) \subseteq \left(l-2\epsilon,l+2\epsilon\right)$.
Hence, given $\epsilon > 0$, there exists $M$ such that for $m>M$, we have $$\dfrac{\sum_{k=1}^m a_k}m \in \left(l-2\epsilon,l+2\epsilon\right)$$ Hence, $$\dfrac{\sum_{k=1}^m a_k}m \to l$$