Show that $\lim_{n\to\infty} \frac{1}{n}\sum_{k=0}^{n} x_k=x $ if $\{x_n\} \to x$ If $\ \{x_n\}$ converge to $x$ show that: $$\lim_{n\to\infty} \frac{1}{n}\sum_{k=0}^{n} x_k =x.$$

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=0}^{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=0}^{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=0}^{N_1-1} |x_k-x| + \displaystyle\frac{1}{n}\sum_{k=N_1}^{n} |x_k-x|  \\
& = \displaystyle\frac{1}{n}\sum_{k=0}^{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=0}^{N_1-1} |x_k-x| + \displaystyle\frac{\epsilon}{2}
\end{array}
By taking the limit of the first sum, we will get : $$\lim_{n \to \infty}\displaystyle\frac{1}{n}\sum_{k=0}^{N_1-1} |x_k-x| = 0.$$
It follows that $\exists N_2 \in \mathbb{N}$ such that $n\geq N_2$ we have :
$$\displaystyle\frac{1}{n}\sum_{k=0}^{N_1-1} |x_k-x| < \displaystyle\frac{\epsilon}{2}.$$
Lets take $N=\max\{N_1,N_2\}$ then $\forall n\geq N$ 
$$\left|\displaystyle\frac{1}{n}\sum_{k=0}^{n} x_k-x \right| \leq \displaystyle\frac{\epsilon}{2} +\displaystyle\frac{\epsilon}{2} = \epsilon. $$
 A: 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.
A: 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 $|\frac 1n \sum_\limits{k=K}^{n} x_k-x|< \epsilon/2$  
