4

If $\ \{x_n\}$ converge to $x$ show that: $$\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^{n} x_k =x.$$

Let choose an $\epsilon>0$, then $\exists N_1\in \mathbb{N}$ such that $\forall n\geq N_1$ we have $|x_n-x|<\epsilon/2.$

\begin{array}{} \displaystyle\frac{1}{n}\sum_{k=1}^{n} x_k-x &= \displaystyle \frac{x_1+x_2+...+x_n}{n}-x\\ &=\displaystyle \frac{x_1+x_2+...+x_n-nx}{n} \end{array} Lets take the absolute value: \begin{array}{} \left|\displaystyle\frac{1}{n}\sum_{k=1}^{n} x_k-x \right| & =\left|\displaystyle \frac{x_1+x_2+...+x_n-nx}{n} \right| \\ & =\left| \displaystyle\frac{(x_1-x)+(x_2-x)+...(x_n-x)}{n} \right| \\ & \leq \left| \displaystyle\frac{x_1-x}{n}\right|+...+\left|\displaystyle\frac{x_n-x}{n}\right| & by \ the \ triangle \ inequality.\\ & = \displaystyle\frac{1}{n}\sum_{k=1}^{N_1-1} |x_k-x| + \displaystyle\frac{1}{n}\sum_{k=N_1}^{n} |x_k-x| \\ & = \displaystyle\frac{1}{n}\sum_{k=1}^{N_1-1} |x_k-x| + \displaystyle\frac{\epsilon}{2n}(n-N_1) & (\forall n\geq N_1: |x_n-x|<\epsilon/2.)\\ &\leq \displaystyle\frac{1}{n}\sum_{k=1}^{N_1-1} |x_k-x| + \displaystyle\frac{\epsilon}{2} \end{array}

By taking the limit of the first sum, we will get : $$\lim_{n \to \infty}\displaystyle\frac{1}{n}\sum_{k=1}^{N_1-1} |x_k-x| = 0.$$

It follows that $\exists N_2 \in \mathbb{N}$ such that $n\geq N_2$ we have : $$\displaystyle\frac{1}{n}\sum_{k=1}^{N_1-1} |x_k-x| < \displaystyle\frac{\epsilon}{2}.$$

Lets take $N=\max\{N_1,N_2\}$ then $\forall n\geq N$ $$\left|\displaystyle\frac{1}{n}\sum_{k=1}^{n} x_k-x \right| \leq \displaystyle\frac{\epsilon}{2} +\displaystyle\frac{\epsilon}{2} = \epsilon. $$

Luigi Traino
  • 876
  • 5
  • 14
Elina
  • 227
  • 2
  • 9
  • 1
    I think your last inequality needs to be $ max(|x_k-x|, $for$ 1 \leq k \leq n)<\epsilon$ – JasonM Jun 10 '16 at 23:59
  • Furthermore, I think your condition at the top needs to be $|x_k-x|<\epsilon$ for all $k$, $1 \leq k \leq n$ – JasonM Jun 11 '16 at 00:02
  • is it good now? – Elina Jun 11 '16 at 00:51
  • I don't think so. You have the basic idea down, but there's some clean up that needs to happen with the sum. Like carmichael561 said, you have to break it up into two sums, and choose $n$ large enough to make the first sum small. The second sum should be small by our choice of $N$. – JasonM Jun 11 '16 at 00:56
  • I think it's good now. Thank you guys – Elina Jun 13 '16 at 14:43

2 Answers2

4

Everything looks good up until the second-to-last line. How did you get this from the previous line?

I suggest trying the following: given $\varepsilon>0$, choose $N$ such that $|x_n-x|<\varepsilon$ for $n>N$. Then divide up the terms $\frac{1}{n}|x_k-x|$ into two groups, corresponding to $1\leq k\leq N$ and $N<k\leq n$.

In the first group, we have no control over the size of $|x_k-x|$, but there are only $N$ terms in the group and $\frac{1}{n}$ can be made as small as we like. In the second group, there are $n-N$ terms, each of which is of size at most $\frac{\varepsilon}{n}$. So both groups can be made small by choosing $n$ large enough.

carmichael561
  • 53,688
1

Your second to last step does not proceed from the one before. And you have $n$'s pulling double duty, which is a little bit confusing.

But what I think you really need to say is that for only finitely many $x_k, |x_k-x|>\epsilon/2$

$\forall \epsilon>0,\exists K>0$ such that $k>K \implies |x_k - x| < \epsilon/2$

$|\sum_\limits{k=0}^K x_k-x| < M$
Choose $N > \max(K, 2M/\epsilon)$

Then:

$n>N\implies|\frac 1n \sum_\limits{k=0}^K x_k-x| <\epsilon/2$ and $|\frac 1n \sum_\limits{k=K}^{n} x_k-x|< \epsilon/2$

Doug M
  • 57,877