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$\sum_{n=1}^{\infty}\frac{(-1)^n}{n^{\frac{1}{10}}}$

According to my understanding, if $\sum\left|a_n\right|$ diverges but $\sum a_n$ converges, then the series is conditionally convergent.

For $\sum\left|a_n\right|$ my series can be test via the p-series test and since $\frac{1}{10} \lt 1$ it diverges.

So next I test $\sum a_n$ using the alternating series test and find that it is a decreasing series and the limit converges to 0.

Thus, I came to the conclusion that this is conditionally convergent. Is this correct?

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    $\begingroup$ @user1952009 This is not what "absolutely convergent" means. Grouping the terms as one pleases is not allowed, it becomes another series. cf. Riemann's Rearrangement Theorem. $\endgroup$
    – Clement C.
    Commented Jul 16, 2016 at 16:16
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    $\begingroup$ It does change things. You are looking at another series, $\sum_n b_n$ with $b_n = a_{2n-1}+a_{2n}$. That other series is absolutely convergent, but $\sum_n a_n$ is not. $\endgroup$
    – Clement C.
    Commented Jul 16, 2016 at 16:18
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    $\begingroup$ But that statement is utterly confusing, then. What does your "it" refer to? Given that the OP is asking about a specific series, while your statement is only true if "it" refers to a different (related) series you implicitly define after using the pronoun, it can only bring confusion. (Especially since it's about a point the OP was not asking about, and most likely not considering to begin with) $\endgroup$
    – Clement C.
    Commented Jul 16, 2016 at 16:21
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    $\begingroup$ True, but then so is the (divergent, so not even well-defined) series $\sum_n (-1)^n$... :) $\endgroup$
    – Clement C.
    Commented Jul 16, 2016 at 16:24
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    $\begingroup$ (as a last comment: basically, any conditionally convergent series you can use the alternating series test on will allow the same grouping you used: pairing even and odd terms will result in an absolutely convergent series, as all terms have the same sign) $\endgroup$
    – Clement C.
    Commented Jul 16, 2016 at 16:29

2 Answers 2

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Yes, it is. (Note however that to be fully correct, the statement you make should read that "$(\lvert a_n\rvert)_n$ is a decreasing sequence," not that "$\sum_n a_n$ is a decreasing series.")

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  • $\begingroup$ All too easy... +1 $\endgroup$
    – Mark Viola
    Commented Jul 16, 2016 at 19:23
  • $\begingroup$ @Dr.MV "I like my proofs like I like my eggs."? $\endgroup$
    – Clement C.
    Commented Jul 16, 2016 at 19:26
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    $\begingroup$ Over easy I presume. $\endgroup$
    – Mark Viola
    Commented Jul 16, 2016 at 19:27
  • $\begingroup$ Or scrambled so nobody can refute? $\endgroup$
    – Mark Viola
    Commented Jul 16, 2016 at 19:27
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    $\begingroup$ And you would be correct. It was actually African or European. But it was only an homage, and thus need not be identical. $\endgroup$
    – Mark Viola
    Commented Jul 16, 2016 at 19:50
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A monotonically descending alternating series converges, is that series a A monotonically descending alternating series ?

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