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I understand that given the absolute convergence test, if I am able to prove that the absolute of the series converges, then the series itself will converge itself as well.

What if I want to prove for conditional convergence? Is it sufficient to prove straight away that the absolute does not converge, or do I have to first prove that the series converges, and then prove that absolute does not converge, hence it must converge conditionally?

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  • $\begingroup$ You have to prove that the series converges and also prove that it does not converge absolutely. $\endgroup$ – Kavi Rama Murthy Oct 9 '18 at 9:06
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Yes, of course you need to prove both. If you prove the series of absolute values does not converge it doesn't mean that the original series converge. For example the series $\sum_{n=0}^\infty (-1)^n$ simply diverges.

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A series converges conditionally if both of the following conditions are true:

  1. The series converges
  2. The series does not converge absolutely.

To prove a series converges absolutely, you need to prove both conditions. It is not enough to prove that the series does not converge absolutely, since, for example, $$\sum_{n=1}^\infty (-1)^n n$$ does not converge absolutely, but that doesn't mean it converges conditionally.

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  • $\begingroup$ I have an example on hand. 1 + sin$(\frac{1}{2})$ + sin$(\pi + \frac{1}{2}) + ...$. Since sin$(\frac{1}{n})$ is simply the negative of sin$(\pi + \frac{1}{n})$, I see that the series converges to 1. In this case, am I right to say that this series converges conditionally but not absolutely? $\endgroup$ – statsguy21 Oct 9 '18 at 9:11
  • $\begingroup$ @statsguy21 Did you prove that the series doesn't converge absolutely? So far, you only proved (well, sort of proved, there's still some work to do) that the series converges. $\endgroup$ – 5xum Oct 9 '18 at 9:15

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