Would the following series be conditionally convergent, absolutely convergent or divergent?


Whereas for sine functions in series like this, you can usually say that it just alternates between 1 and -1 but would this one do the same for $k\geq1$? I don't think it would so that would mean I can't use the alternating series test. Perhaps, then, the ratio test or some comparison test?

  • $\begingroup$ You know this question wasn't solved though. $\endgroup$ – AustereTiger Mar 19 '18 at 16:48
  • $\begingroup$ @AusterTiger Did you verify with your instructor? $\endgroup$ – gimusi Mar 19 '18 at 16:57
  • $\begingroup$ I can't. This question is in a test. $\endgroup$ – AustereTiger Mar 19 '18 at 18:04


The strategy is to show that the necessary condition doesn't hold since $a_k \not \to 0$ and thus that $\sum a_k$ diverges.

Refer to Prove that $\sin(1 + k^3) \not \to 0$.

  • $\begingroup$ That just shows it is not absolutely convergent. It still could be conditionally convergent. $\endgroup$ – Mark Fischler Mar 15 '18 at 20:08
  • $\begingroup$ @MarkFischler: What? $\sum a_n$ convergent implies that $a_n \to 0$ $\endgroup$ – Martin R Mar 15 '18 at 20:09
  • $\begingroup$ @Martin R Consider the series $\sum (-1^k) (1-\frac1k)$, which is conditinally convergent. $\endgroup$ – Mark Fischler Mar 15 '18 at 20:11
  • $\begingroup$ @MarkFischler It would be interesting to see a full answer from you. $\endgroup$ – gimusi Mar 15 '18 at 20:14
  • 1
    $\begingroup$ @MarkFischler: Is it? A series is convergent if the sequence of partial sums $s_n = \sum_{k=1}^n a_k$ has a limit, and that implies that $a_n = s_n - s_{n-1} \to 0$. Compare en.wikipedia.org/wiki/Term_test. $\endgroup$ – Martin R Mar 15 '18 at 20:15

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