Convergence and Absolute Convergence of Arithmetic Mean of a sequence Suppose $\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n |x_i|$ exists. Does $\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n x_i$ exist?

How about the converse?

My thoughts:


I guess for the sequence $\{x_1,-x_1,x_2,-x_2,\ldots\}$ the converse doesn't necessarily hold.
If I can show that the existence of $\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n x_i$ implies that $\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n x_i 1\{x_i\geq 0\}$ and $\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n |x_i| 1\{x_i\leq 0\}$ exist, then the first part would hold. But I'm not sure this is true, I need to find a counterexample.
A counterexample for the converse is $x_i = (-1)^i\sqrt i$.
A bit of an explanation of Cla's hint: Consider 
$$
S^1_n = \frac{x_1 + \ldots x_n}{n} &lt; \bigg| \frac{x_1 +\ldots x_n}{n} \bigg| &lt;\frac{|x_1| +\ldots |x_n|}{n} = S^2_{n}
$$
by triangle inequality. Since we know that $S^2_n \to_n a&lt; \infty$ and $S^1_n &lt; S^2_n$, $S^1_n$ converges by comparison test.