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Is the converse of the alternating series test true? In other words, given a sequence $a_n>0$, with neither $a_{2n}$ nor $a_{2n-1}$ constant, for which there exists no positive $N$ such that $a_n>a_{n+1}$ for all $n>N$, does $\displaystyle \sum_{k=1}^\infty (-1)^k a_k$ necessarily diverge? I've been unable to come up with any counterexamples, but I also can't figure out how to prove it.

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No, just have the even terms be $1/n^2$ and the odd terms all be zero - that's not eventually monotone and the alternating sum converges.

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  • $\begingroup$ Fair enough; what if $a_k>0$? Your example is non-monotone, but the even and odd terms, taken separately, are both monotone, and one is a constant. I'll edit the question to be more clear. $\endgroup$
    – Avi
    Commented Sep 24, 2014 at 10:50
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    $\begingroup$ $a_{2n} = 1/n^2$, $a_{2n+1} = 1/n^3$ has convergent alternating sum and fits the hypotheses of the edited question. I don't think there will be any meaningful converse to the alternating series test - if the alternating sum diverges, then in particular the sum is not absolutely convergent, so the hypotheses would have to also imply the series not being absolutely convergent. $\endgroup$
    – Matt Rigby
    Commented Sep 24, 2014 at 10:58
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    $\begingroup$ Also, the contrapositive of the current statement is that anything with positive-valued with divergent alternating sum is eventually increasing, which can't be true $\endgroup$
    – Matt Rigby
    Commented Sep 24, 2014 at 11:00
  • $\begingroup$ It could just be that it isn't eventually decreasing, rather than being eventually increasing, but point taken. $\endgroup$
    – Avi
    Commented Sep 24, 2014 at 11:19

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