A result on sequences: $x_n\to x$ implies $\frac{x_1+\dots+x_n}n\to x$ without using Stolz-Cesaro If $x_n \to x$, how might we prove  
$$\lim_{n \to \infty} \frac{\sum_{i=1}^{n} x_i}{n} = x$$
Of course, one has $\limsup x_n = \liminf x_n = x$, and thus, using the Stolz-Cesaro theorem:
$$\liminf x_n \le \liminf \frac{\sum_{i=1}^{n} x_i}{n} \le \limsup \frac{\sum_{i=1}^{n} x_i}{n} \le \limsup x_n$$ which shreds this easily. However, I'm wondering whether one could do this without the Stolz-Cesaro theorem. 
Also, apparently the converse of this statement is not true. If we take $x_n = 0,\, n$ even, and $x_n =1,\, n$ odd, this suffices, correct? As the second expression tends to $1/2$, while the first has no limit? 
 A: Let $\epsilon > 0$. Then there is an $N \in \mathbb{N}$ such that $|x_n - x| < \epsilon$ for all $n > N$. Therefore, if $n > N$,
$$\begin{align}
\left|\left(\frac{1}{n}\sum_{i=1}^{n}x_i\right) - x\right|
&= \left|\frac{1}{n}\sum_{i=1}^{n}(x_i - x)\right| \\
&= \left|\frac{1}{n}\sum_{i=1}^{N}(x_i-x) + \frac{1}{n}\sum_{i=N+1}^{n}(x_i-x)\right|\\
&\leq \frac{1}{n}\sum_{i=1}^{N}|x_i-x| + \frac{1}{n}\sum_{i=N+1}^{n}|x_i-x|\\
&\leq \frac{1}{n}\sum_{i=1}^{N}|x_i-x| + \left(1 - \frac{N}{n}\right)\epsilon \\
&\leq \frac{1}{n}\sum_{i=1}^{N}|x_i-x| + \epsilon \\
\end{align}$$
Holding $N$ fixed and letting $n \rightarrow \infty$, the first term on the right-hand side goes to zero, so
$$\lim_{n \rightarrow \infty}\left|\left(\frac{1}{n}\sum_{i=1}^{n}x_i\right) - x\right|\leq \epsilon$$
Or equivalently,
$$\left|\left(\lim_{n \rightarrow \infty}\frac{1}{n}\sum_{i=1}^{n}x_i\right) - x\right|\leq \epsilon$$
Since this is true for any $\epsilon$, the result follows.
A: Let $M = \limsup x_n$ (which is also $\lim x_n$, of course), and let $\varepsilon > 0$ be given. There is some $N$ such that $x_n \leq M + \epsilon$ for all $n > N$. Letting $C = \sum_{i=1}^{N} x_i$, we have 
$$\frac{1}{n} \sum_{i=1}^n x_i \leq \frac{1}{n} [C + (n-N)(M+ \varepsilon)],$$
from which it follows that $\limsup \frac{1}{n} \sum_{i=1}^n x_i \leq M + \varepsilon$. Since $\varepsilon$ was arbitrary, we have in fact $\limsup \frac{1}{n} \sum_{i=1}^n x_i \leq M$.
An analogous result can be proved for the $\liminf$, and then one can finish the argument as you did. Only minor modifications are needed when $\lim x_n = \pm \infty$. (Namely, when $M = -\infty$, replace $M + \epsilon$ in the proof with an arbitrary real number. Nothing needs to be proved about the lim sup when $M = +\infty$.)
Your counterexample is correct.
