Show that a sequence converges Suppose that $\{ x_n \}_{ n = 1 }^\infty$ is a converging sequence of real numbers with $\lim_{ n \to \infty } x_n = x$. Define
$$ y_n = \frac{ x_1 + \cdots + x_n }{ n } $$
Show that $\lim_{ n \to \infty  } y_n = x$.
I'm not really sure where to start.
 A: Fix $\varepsilon>0$. Then there exists $n_0$ such that $|x_k-x|<\varepsilon$ whenever $k>n_0$. Then
$$
|y_n-x|=\left|\frac1n\,\sum_{k=0}^nx_k-x\right|\leq\frac1n\,\sum_{k=0}^n|x_k-x|=\frac1n\,\sum_{k=0}^{n_0}|x_k-x|+\frac1n\,\sum_{k=n_0+1}^n|x_k-x|\\ \leq
\frac1n\,\sum_{k=0}^{n_0}|x_k-x|+\frac1n\,(n-n_0-1)\,\varepsilon
\leq \frac1n\,\sum_{k=0}^{n_0}|x_k-x|+\varepsilon.
$$
Note that the sum multiplying $1/n$ does not depend on $n$. So
$$
\limsup_n|y_n-x|\leq\varepsilon.
$$
As $\varepsilon$ was arbitrary, we conclude that $\limsup_n|y_n-x|=0$, which shows that $\lim_n|y_n-x|=0$, i.e. $\lim y_n=x$. 
A: The following is basically Martin's argument but without messing with $\,\lim\sup\,$:
take any $\,\epsilon>0\Longrightarrow\,\exists\,N_\epsilon\in\Bbb N\,\,s.t.\,\,|x_n-x|<\epsilon\,\,,\,\forall\,n>N_\epsilon\,$ . Now, take $\,n>N_\epsilon\,$:
$$\left|\frac{x_1+\ldots+x_n}{n}-x\right|=\left|\frac{(x_1-x)+(x_2-x)+\ldots+(x_n-x)}{n}\right|\leq\left|\frac{(x_1-x)+\ldots+(x_{N_\epsilon}-x)}{n}\right|+\left|\frac{(x_{N_\epsilon+1}-x)+\ldots (x_n-x)}{n}\right|\leq$$
$$\leq \frac{k}{n}+\frac{n-N_\epsilon}{n}\epsilon$$
with $\,k\,$ a fixed positive constant. Well, now just make $\,n\to\infty\,$ ,and you'll get
$$0\leq\left|\frac{x_1+\ldots+x_n}{n}-x\right|\leq 0+\epsilon=\epsilon$$
and since $\,\epsilon\,$ was arbitrary we're done
Acclaration: As the above is a very common exercise at the start of limits of sequences, it may be it is given before $\lim\sup\,,\,\lim\inf\,$ and other beasts are studied, so perhaps the above approach is slightly more elementary.
