Let $(x_n)$ be a sequence that converges to a number x. Define a new sequence $(s_n)$ by $s_n= \frac{1}{n}\sum\limits_{k=1}^n x_k$
Show that $s_n$ converges to $x$ as well.
Asked
Active
Viewed 217 times
0
-
This has been asked before. – Pedro Feb 16 '15 at 06:42
-
I just realized the linked duplicate covers a converse. I'm looking for a duplicate. – Pedro Feb 16 '15 at 06:55
1 Answers
2
Given $\varepsilon >0,$ there is $n_0\in \mathbb{N}$, such that for all $n>n_0$ implies that $\left|x_n-x\right|<\varepsilon.$ So, for $n>n_0$, $$\left|\frac{1}{n}\sum_{i=1}^nx_n-x\right|=\frac{1}{n}\left|\sum_{i=1}^nx_n-nx\right|\leq\frac{1}{n}\sum_{i=1}^n \left|x_n-x\right|\leq \frac{1}{n}n\varepsilon=\varepsilon.$$
Then, $s_n=\frac{1}{n}\sum_{i=1}^nx_n$ coverges to $x$.
-
In the last inequality you don´t have $|x_k - x| < \epsilon$ for all $k$, only for those where $k>n_0$. – Aug 08 '18 at 14:02