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I have a question which is giving me a hard time.

I want to show that $$ \lim_{n\to \infty}\frac{1}{n} \sum_{k=1}^n x_k =x$$ given that $\lim_{n\to \infty} x_n= x$.

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3 Answers 3

up vote 1 down vote accepted

Or you could slog through a tedious proof:

Choose $\epsilon>0$. Let $N$ be such that $n\geq N$ means $|x_n-x| < \frac{\epsilon}{2}$. Now choose $N'\geq N$ so that $n\geq N'$ means $\frac{1}{n} \sum_{k=1}^N |x_n-x| < \frac{\epsilon}{2}$.

Then, if $n\geq N'$, we have the estimate:

$$|\frac{1}{n} \sum_{k=1}^n (x_n-x)| \leq \frac{1}{n} \sum_{k=1}^n |x_n-x| \leq \frac{1}{n} \sum_{k=1}^N |x_n-x| + \frac{1}{n} \sum_{k=N+1}^n |x_n-x| < \frac{\epsilon}{2}+n\frac{1}{n}\frac{\epsilon}{2}= \epsilon$$

Hence $\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^n (x_n-x) = 0$ from which the result follows.

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Hint: Lookup Cesàro summation.

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thanks for the hint: is this right? as $\lim x_n =x$ $$ \frac{x_1 +\ldots +x_n}{n} =\frac{nx}{n}=x$$ –  Mike Nov 9 '12 at 23:09
    
@Mike: no. You are using notation in a sloppy way to disguise (even to you) the that you are interchanging two limits (you are saying that $$\lim_{n\to\infty}\frac1n\sum_{k=0}^nx_k=\lim_{k\to\infty}\lim_{n\to\infty} \frac1n\sum_{k=0}^nx_k=\lim_{n\to\infty}\frac1n\sum_{k=0}^n\lim_{k\to\infty} x_k$$ –  Martin Argerami Nov 9 '12 at 23:31

Since $x_n \to x$, we have that, given any $\epsilon > 0$, there exists $N(\epsilon)$ such that for all $n > N(\epsilon)$, we have that $\vert x_n - x \vert < \epsilon/2$. Hence, for all $n > N(\epsilon)$, we have that $$\left \vert \dfrac{x_1 + x_2 + \cdots + x_n}{n} - x \right \vert\\ = \left \vert \dfrac{(x_1-x) + (x_2-x) + \cdots + (x_N - x) + (x_{N+1}-x) + \cdots + (x_n-x)}{n} \right \vert\\ = \left \vert \dfrac{(x_1-x) + (x_2-x) + \cdots + (x_N - x)}{n} + \dfrac{(x_{N+1}-x) + \cdots + (x_n-x)}{n} \right \vert\\ \leq \left \vert \dfrac{(x_1-x) + (x_2-x) + \cdots + (x_N - x)}{n} \right \vert + \left \vert \dfrac{(x_{N+1}-x) + \cdots + (x_n-x)}{n} \right \vert\\ \leq \left \vert \dfrac{(x_1-x) + (x_2-x) + \cdots + (x_N - x)}{n} \right \vert + \dfrac{\left \vert (x_{N+1}-x) \right \vert + \cdots + \left \vert (x_n-x) \right \vert }{n}\\ \leq \left \vert \dfrac{(x_1-x) + (x_2-x) + \cdots + (x_N - x)}{n} \right \vert + \dfrac{\epsilon + \cdots + \epsilon }{2n}\\ = \underbrace{\left \vert \dfrac{(x_1-x) + (x_2-x) + \cdots + (x_N - x)}{n} \right \vert}_{\text{Choose $n$ large enough such that this quantity is less than } \epsilon/2} + \dfrac{(n-N) \epsilon }{2n}\\ \leq \dfrac{\epsilon}2 + \dfrac{\epsilon}2 = \epsilon $$ Since this is true for any $\epsilon > 0$, we have that $$\lim_{n \to \infty} \dfrac{\displaystyle \sum_{k=1}^n x_k}n = x$$

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