4
$\begingroup$

Suppose that $\{ x_n \}_{ n = 1 }^\infty$ is a converging sequence of real numbers with $\lim_{ n \to \infty } x_n = x$. Define $$ y_n = \frac{ x_1 + \cdots + x_n }{ n } $$ Show that $\lim_{ n \to \infty } y_n = x$.

I'm not really sure where to start.

$\endgroup$
5
$\begingroup$

Fix $\varepsilon>0$. Then there exists $n_0$ such that $|x_k-x|<\varepsilon$ whenever $k>n_0$. Then $$ |y_n-x|=\left|\frac1n\,\sum_{k=0}^nx_k-x\right|\leq\frac1n\,\sum_{k=0}^n|x_k-x|=\frac1n\,\sum_{k=0}^{n_0}|x_k-x|+\frac1n\,\sum_{k=n_0+1}^n|x_k-x|\\ \leq \frac1n\,\sum_{k=0}^{n_0}|x_k-x|+\frac1n\,(n-n_0-1)\,\varepsilon \leq \frac1n\,\sum_{k=0}^{n_0}|x_k-x|+\varepsilon. $$ Note that the sum multiplying $1/n$ does not depend on $n$. So $$ \limsup_n|y_n-x|\leq\varepsilon. $$ As $\varepsilon$ was arbitrary, we conclude that $\limsup_n|y_n-x|=0$, which shows that $\lim_n|y_n-x|=0$, i.e. $\lim y_n=x$.

$\endgroup$
  • $\begingroup$ why the sum multiplying $1/n$ doesn't depends on n ? it should be decrease with $n$ isn't it? $\endgroup$ – Mathematics Dec 17 '12 at 5:09
  • $\begingroup$ With $\varepsilon$ fixed, $n_0$ is fixed, and so the sum is fixed. We are still free to increase $n$ so $1/n\to0$, and so the first term in the last sum goes to zero. $\endgroup$ – Martin Argerami Dec 17 '12 at 6:22
  • $\begingroup$ Like this approach. +1 $\endgroup$ – mrs Dec 17 '12 at 6:38
1
$\begingroup$

The following is basically Martin's argument but without messing with $\,\lim\sup\,$:

take any $\,\epsilon>0\Longrightarrow\,\exists\,N_\epsilon\in\Bbb N\,\,s.t.\,\,|x_n-x|<\epsilon\,\,,\,\forall\,n>N_\epsilon\,$ . Now, take $\,n>N_\epsilon\,$:

$$\left|\frac{x_1+\ldots+x_n}{n}-x\right|=\left|\frac{(x_1-x)+(x_2-x)+\ldots+(x_n-x)}{n}\right|\leq\left|\frac{(x_1-x)+\ldots+(x_{N_\epsilon}-x)}{n}\right|+\left|\frac{(x_{N_\epsilon+1}-x)+\ldots (x_n-x)}{n}\right|\leq$$

$$\leq \frac{k}{n}+\frac{n-N_\epsilon}{n}\epsilon$$

with $\,k\,$ a fixed positive constant. Well, now just make $\,n\to\infty\,$ ,and you'll get

$$0\leq\left|\frac{x_1+\ldots+x_n}{n}-x\right|\leq 0+\epsilon=\epsilon$$

and since $\,\epsilon\,$ was arbitrary we're done

Acclaration: As the above is a very common exercise at the start of limits of sequences, it may be it is given before $\lim\sup\,,\,\lim\inf\,$ and other beasts are studied, so perhaps the above approach is slightly more elementary.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.