convergence of weighted average. proof It is well known that for any sequence $\{x_n\}$ in a normed space which converges to a limit $x$, the sequence of averages of the first $n$ terms is also convergent to $x$.  That is, the sequence $\{a_n\}$ defined by 
$$a_n = \frac{x_1+x_2+\ldots + x_n}{n}$$
converges to $x$. Does anyone know how to prove this?
Thanks.
 A: Given $\epsilon>0$, there are at most finitely many $k$ with $|x_k-x|>\frac\epsilon2$, say only for $k\le N$. The contribution $\frac{(x_1-x)+\ldots+(x_N-x)}{n}$ goes $\to0$ as $n\to\infty$, especially it is absolutely $<\frac\epsilon2$ for big $n$. The remaining $\frac{(x_{N+1}-x)+\ldots +(x_n-x)}{n}$ is also absolutely $<\frac\epsilon2$, hence $\left|\frac{x_1+\ldots+x_n}{n}-x\right|<\epsilon$ for big $n$.
A: First, the intuition for this result is that if $x_n$ is close to $x$, then for large $n$, the expression $\frac{x_1+\cdots x_n}{n}$, consists in the numerator mostly of elements very close to $x$, so that the numerator is approximated by $const + (n-k)x$, which cancels with the denominator to give $x$. This intuition can be turned into a rigorous proof, as follows. 
Proof sketch: Let $\epsilon >0$ be given. Then there is some $n_0$ such that for all $n>n_0$ holds that $|a_n-x|<\epsilon/2$. For every $m>n_0$, split the sum $a_n$ as $b_n+c_n$, where $b_n=\frac{x_1+\cdots x_{n_0}}{n}$ and $c_n=\frac{x_{n_0+1}+\cdots x_{m}}{n}$. Now


*

*Argue why you don't care about what happens for $m\le n_0$.

*Use the fact that the numerator in $b_n$ is constant to show that there is some $n_1$ such that for all $n>n_1$ holds that $|b_n|<\epsilon/2$.

*Use the choice of $n_0$ to evaluate $|c_n-x|$.

*Wrap things up to conclude the result. 
