Convergence and Absolute Convergence of Arithmetic Mean of a sequence

Question

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.

Answer 1

A bit of an explanation of Cla's hint: Consider $$ S^1_n = \frac{x_1 + \ldots x_n}{n} < \bigg| \frac{x_1 +\ldots x_n}{n} \bigg| <\frac{|x_1| +\ldots |x_n|}{n} = S^2_{n} $$ by triangle inequality. Since we know that $S^2_n \to_n a< \infty$ and $S^1_n < S^2_n$, $S^1_n$ converges by comparison test.

Answer 2

A counterexample for the converse is $x_i = (-1)^i\sqrt i$.

Answer 3

Another trivial counter example for the converse could be this:

$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n x_i 1\{x_i\geq 0\}$ does not exist and $\frac{1}{n}\sum_{i=1}^n |x_i| 1\{x_i\leq 0\}=\frac{1}{n}\sum_{i=1}^n x_i 1\{x_i\geq 0\}$.