The average of a bounded, decreasing-difference sequence (not sure if "decreasing-difference" is the right way to put it - please edit away if you know the proper technical term)
From my analysis homework:

Does there exist a bounded sequence
  $\{x_n\}_{n \in \mathbb{N}}$ such that
  $x_{n+1}-x_n \to 0$, but the sequence
  $(\sum_{i=1}^n x_i)/n$ does not have a
  limit? Give an example or show none
  exist.

In a previous part, I showed that there exists a bounded sequence $\{x_n\}_{n \in \mathbb{N}}$ such that $\left| x_{n+1}-x_n \right| \to 0$, yet $x_n$ does not converge, using a rearrangement of the alternating harmonic series. 
Intuitively, I believe none exist. I've tried showing that the sequence is Cauchy, but that does not seem to help:
$$\left| \frac{ \sum_{i=1}^m x_i}{m} - \frac{ \sum_{i=1}^n x_i}{n} \right| = \left| \frac{ (n-m) \sum_{i=1}^m x_i - m \sum_{i=m+1}^n x_i}{mn} \right| $$
I feel like I can't use my normal series tools because I am looking at an average, and the series does not necessarily converge.
 A: Edit 2: for your actual question, the answer is yes, you can find counterexamples. Consider the sequence $(1/n)$, and build your sequence as follows. 
Start at $x_1 = 1$, and let $\alpha_1$ be a running counter which we start at 2. We then alternate between a decreasing phase and an increasing phase. 
Decreasing phase: let $x_i = x_{i-1} - 1/\alpha_{i}$ if $x_{i-1} > 1/\alpha_i$. And increase $\alpha_{i+1} = \alpha_i + 1$. If $x_{i-1} < 1/\alpha_i$, set 
$$ x_i = x_{i+1} = \cdots = x_{100i} = x_{i-1}\qquad \qquad \alpha_{100i + 1} = \alpha_{i} $$
and then enter the increasing phase starting from the $100i+1$'th term. 
Increasing phase: let $x_i = x_{i-1} + 1/\alpha_i$ if $x_{i-1} + 1/\alpha_i < 1$. And increase $\alpha_{i+1} = \alpha_{i} + 1$. Else set 
$$ x_i = x_{i+1} = \cdots = x_{100i} = x_{i-1} \qquad \qquad \alpha_{100i + 1} = \alpha_{i} $$
and enter the decreasing phase starting from the $100i+1$'th term. 
Due to the long constant phases, you see that $\limsup_{n\to\infty} \frac{1}{n} \sum_1^n x_i = 1$, and the $\liminf = 0$. (The length of the constant phases grow exponentially, so that each constant phase completely dominates all previous terms in the average.)
The sequence looks like this:
1, 1/2, 1/6, {300 terms of 1/6}, 5/12, 37/60, 47/60, 389/420, {repeat 30700 times}, 673/840 ...

Edit: oh wait, the below doesn't actually address your question.

As a side note, the limit $\lim_{n\to\infty} \frac1n \sum_1^n x_i $ is known as the Cesàro mean of the sequence $x_i$. It is a fact that for any converging sequence $x_i\to \bar{x}$, the sequence $c_n = \frac{1}{n} \sum_1^n x_i$ also converges to $\bar{x}$. 
To do this, let $\delta > 0$. It suffices to find $N$ large enough that for all $m > N$, $|c_m - \bar{x}| < \delta$. Because $x_i\to \bar{x}$, we can find $I$ large such that for all $j > I$, $|x_i - \bar{x}| < \delta / 3$. Now choose $N > I$ large enough such that
$$ \sum_1^I x_i < \frac{\delta}{3}N  \qquad \textrm{and} \qquad  \frac{I}{N} < \frac{\delta}{3|\bar{x}|} $$
Then we have for all $m >  N$:
$$ m c_m = \sum_1^I x_i + \sum_{I+1}^m x_i $$
$$ |c_m - \bar{x}| \leq | \frac{1}{m} \sum_1^I x_i | + | \frac{1}{m} \sum_{I+1}^m x_i - \bar{x} | $$
The first term on the right hand side is bounded by $\delta/3$. The second term we estimate
$$ |\frac{1}m \sum_{I+1}^m x_i - \bar{x}| \leq |\frac{1}{m} \sum_{I+1}^m (x_i - \bar{x})| + |\frac{I}{m} \bar{x}| $$
and by construction both of the terms on the RHS are bounded by $\delta / 3$. So we have the desired inequality. 
