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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.
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    $\begingroup$ The natural observation is that $1/n\sum_ix_i\leq 1/n\sum_i|x_i|$. For the other question, observe that $\sum_i(-1)^i/i$ is a converging sequence by the Leibniz (alternating) criterion, but the absolute series doesn't converge. $\endgroup$
    – User3773
    Mar 4, 2015 at 12:29
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    $\begingroup$ @Cla: your second example is not relevant, because $\frac{1}{n}\sum_{i=1}^n1/i$ does converge. $\endgroup$
    – TonyK
    Mar 4, 2015 at 12:33
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    $\begingroup$ @Cla We have that $\sum_{i=1}^ni^{-1}\sim\log n$ as $n\to\infty$. Hence, $n^{-1}\sum_{i=1}^ni^{-1}\to0$ as $n\to\infty$. $\endgroup$
    – Cm7F7Bb
    Mar 4, 2015 at 12:34
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    $\begingroup$ You were right. I forgot the $1/n$ multiplication term..sorry for that. $\endgroup$
    – User3773
    Mar 4, 2015 at 15:01

3 Answers 3

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

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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 $\frac{1}{n}\sum_{i=1}^n |x_i| 1\{x_i\leq 0\}=\frac{1}{n}\sum_{i=1}^n x_i 1\{x_i\geq 0\}$.

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  • $\begingroup$ Does such a sequence exist? $\endgroup$
    – TonyK
    Mar 4, 2015 at 13:07
  • $\begingroup$ I was thinking about $x_1, -x_1,x_2,-x_2,\ldots$ where $x_i\geq 0$ but this doesn't seem to work, good point.. $\endgroup$
    – radIQ
    Mar 4, 2015 at 13:46
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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.

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  • $\begingroup$ "$S_n^1$ converges by the comparison test": not necessarily. You are claiming that if $(u_n)$ and $(v_n)$ are sequences (not series), and $v_n \to a$, and $|u_n| \le v_n$ for all $n$, then $u_n$ converges. Which is obviously false. $\endgroup$
    – TonyK
    Mar 4, 2015 at 12:39
  • $\begingroup$ perhaps I'm missing something. What I meant was $S^1 <S^2 <a$ as $n \to \infty$ by comparison test because $x_k <|x_k|$. $\endgroup$
    – Alex
    Mar 4, 2015 at 12:51
  • $\begingroup$ But you claimed that $S_{1,n}$ converges! Where does that come from? (By $S_{1,n}$ I mean your $S_n^1$. Similarly, I suggest using $S_{2,n}$ instead of $S_n^2$.) $\endgroup$
    – TonyK
    Mar 4, 2015 at 13:01
  • $\begingroup$ It is upper-bounded by $S^2_n$ which converges, so it converges by direct comparison test. Where's the mistake? I honestly don't understand. $\endgroup$
    – Alex
    Mar 4, 2015 at 13:07
  • $\begingroup$ For instance, the sequence $(0,1,0,1,.\ldots)$ is upper-bounded by $(1,1,1,1,\ldots)$. But the second sequence converges, while the first sequence doesn't. The comparison test applies to series, not sequences. $\endgroup$
    – TonyK
    Mar 4, 2015 at 13:11

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