Here is the problem.

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I tried choosing $B_n = A_{mn}$ since it is an independent sequence for $m \geq 2$, but I am not quite sure how to guarantee that $\sum_{n=1}^{\infty} P(A_{mn}) = \infty$.

Is it true? If so, why? If not, what other subsequence can you suggest?

This is from Rosenthal, btw.


Hint It is not generally true that $\sum P(A_{nm})=+\infty$. But it must be true for one of the $m$ following sequences: \begin{equation*} (B^{i}_{n})_{n\in\mathbb{N}}=(A_{nm+i})_{n\in\mathbb{N}}, \quad i=0,1..,m-1 \end{equation*}

  • 2
    $\begingroup$ With $m=2$. $ $ $\endgroup$ – Did Aug 29 '14 at 9:59
  • $\begingroup$ Sergio, @Did, I am not quite sure I follow. $\sum_{n=1}^{\infty} P(A_{2n+1})$ is just as bad as $\sum_{n=1}^{\infty} P(A_{2n})$ right? $\endgroup$ – BCLC Aug 29 '14 at 10:10
  • $\begingroup$ Oh but one of those 2 necessarily diverges. THANKS!!! $\endgroup$ – BCLC Aug 29 '14 at 11:00
  • $\begingroup$ @DId Why does m have to be 2? This can extend to 3, 4, etc I think... $\endgroup$ – BCLC Aug 29 '14 at 11:14

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