Problem:
Let $(x_n)_n \subseteq \mathbb{R}$. Suppose there exists $x\in \mathbb{R}$ such that $x_n\rightarrow x$.
Show that $\frac{x_1+x_2....+x_n}{n}\rightarrow x$.
Attempt:
Let $\epsilon>0$. Since $x_n\rightarrow x$, there exists natural number $N$, for which $n> N$ implies $|x_n-x|<\epsilon$. Hence, whenever $n> N$, we have:
$|\frac{x_1+....+x_n}{n}-x|=|\frac{(x_1-x)+....+(x_n-x)}{n}|\leq \frac{|x_1-x|}{n}+...+|\frac{x_n-x}{n}|<n\frac{\epsilon}{n}=\epsilon$.
Is this correct?
Note, the above is non-sense, without looking at previous answers, I have come up with my own solution:
Let $\epsilon>0$. Since convergent sequences are bounded, there exists $U>0$, for which, $|x_n-x|<U$ for all $n$. Since $x_n\rightarrow x$, there exists $N^1$, for which, $n>N^1$ implies $|x_n-x|<\epsilon$. Put $N$ to be such that $N>N^1$ and $N>\frac{UN'}{\epsilon}$. Then, whenever $n>N$, we have:
$|\frac{x_1+....+x_n}{n}-x|$ $\leq$ $|\frac{(x_1-x)}{n}|+.....+|\frac{x_{N'+1}}{n}|+.....|\frac{x_n-x}{n}|<\frac{UN'}{n}+\epsilon<2\epsilon$
