Let $\{x_n\}_{n=1}^\infty$ be a convergent sequence in R with limit x. Show that $\lim_{n\to\infty} 1/n \sum_{k=1}^{n} x_k = x$ Please explain this proof, I'm so lost, I thought I knew what I was doing but this looks completely different to my answer. I don't know how to get better at this.
Let $\{x_n\}_{n=1}^\infty$ be a convergent sequence in R with limit x. 
Show that $\lim_{n\to\infty} 1/n \sum_{k=1}^{n} x_k = x$
Solution: Let ε>0. 
As lim$_{n\to\infty}$ $x_n$ $=x$,
there is $n_1 ∈ N$ such that
$|x_n − x| < ε/ 2$
for $n≥n_1$. 
Choose $n_2 ∈ N$ such that
$\lim_{n\to\infty} 1/n_2 |\sum_{k=1}^{n_1-1} (x_k-x)| < \epsilon/2$
Set $n_ε$ := max ${(n_1, n_2)}$. For $n ≥ n_ε$, we then have:
$1/n \sum_{k=1}^{n} x_k-x|$ = $1/n \sum_{k=1}^{n} (x_k-x)|$ 
$\leq 1/n |\sum_{k=1}^{n_1-1} x_k-x|$ + $1/n \sum_{k=n_1}^{n} (x_k-x)|$ 
$\leq 1/n_2 |\sum_{k=1}^{n_1-1} x_k-x|$ + $1/n \sum_{k=n_1}^{n} (x_k-x)|$
<$\epsilon$/2 + $((n+1-n_1)/n$ ) $max_{k=n_1..n} |(x_k-x)|$
<$\epsilon$/2 + $\epsilon$/2
=$\epsilon$
 A: The proof in the OP is flawed as written.  We present, therefore, a corrected proof of the theorem in that which follows.
Let $S_n$ be given by
$$\begin{align}
S_n&=\left(\frac1n\sum_{k=1}^n x_k\right)-x\tag1\\\\
&=\frac1n\sum_{k=1}^n (x_k-x)\tag2
\end{align}$$
where in going from $(1)$ to $(2)$ we used the fact that $x=\frac1n\sum_{k=1}^n(x)$.
Now, let $\epsilon>0$ be given.  Inasmuch as $\lim_{n\to\infty}x_n=x$, there exists an integer $n_1$ such that for all $n>n_1$, $|x_n-x|<\epsilon/2$.
Next, with $n_1$ fixed we split the sum in $(2)$ into the two sums
$$S_n=\frac1n\left(\sum_{k=1}^{n_1}(x_k-x)+\sum_{k=n_1+1}^n(x_k-x)\right)\tag3$$
Taking the absolute value on both sides of $(3)$ and applying the triangle inequality reveals
$$\begin{align}
|S_n|&\le \left|\frac1n\sum_{k=1}^{n_1}(x_k-x)\right|+\left|\sum_{k=n_1+1}^n(x_k-x)\right|\\\\
&\le \frac1n\sum_{k=1}^{n_1}|x_k-x|+\frac1n\sum_{k=n_1+1}^n|x_k-x|\tag4
\end{align}$$
Let's look at the second term on the right-hand side of $(4)$.  Note that $k>n_1$ for all of the terms.  And hence $|x_k-x|<\epsilon/2$ for all of the terms in the summation.  Therefore, we have
$$\frac1n\sum_{k=n_1+1}^n|x_k-x|<\frac\epsilon2 \left(\frac{n-n_1}{n}\right)<\frac\epsilon2\tag5$$
Now let's take a look at the first sum on the right-hand side of $(4)$.  Since the series $x_n$ converges, its terms are bounded.  And with $n_1$ fixed, we can find a number $n_2$ so large that whenever $n>n_2$ we have 
$$\frac1n \sum_{k=1}^{n_1}|x_k-x|<\frac\epsilon2\tag6$$
Putting together $(5)$ and $(6)$ completes the proof.
