# $\lim_{n\to \infty}\frac{1}{n} \sum_{k=1}^n x_k =x\;$ given $\;\lim_{n\to \infty} x_n= x\;?$

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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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. –  amWhy Nov 9 '12 at 22:22

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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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 $$\lim_{n \to \infty} \dfrac{\displaystyle \sum_{k=1}^n x_k}n = x$$