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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.

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Real case was solved, for example, in this question. And here is a similar question for normed spaces. –  Martin Sleziak Mar 4 '13 at 21:26
    
BTW Why is your question called convergence of weighted average? –  Martin Sleziak Mar 4 '13 at 21:28
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marked as duplicate by Martin Sleziak, Asaf Karagila, rschwieb, Dominic Michaelis, Thomas Mar 4 '13 at 22:46

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2 Answers

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 $n$.

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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.

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