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We're given the series: $$\sum_{n=1}^\infty \frac{\cos(n\pi)}{n}$$ and told that we need to prove that the series is conditionally convergent. Here's what I'm confused about: What are the steps for determining that this is conditionally convergent? And what test do we use to determine if it really is convergent?... alternating series test, absolute ratio test, absolute convergence test? etc.

Mysterium
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Luna_II
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    Only need to do three things: [1] Verify that the terms are alternating in sign [2] Verify that the absolute values of the terms are strictly decreasing and converge at $0$ [3] Apply the Alternating Series test. – user2661923 Mar 24 '21 at 04:41
  • Falls down at the first step – this series ain't alternating. – Gerry Myerson Mar 24 '21 at 07:41
  • See https://math.stackexchange.com/questions/792046/examine-the-convergence-of-left-cos-n-over-n-right and https://math.stackexchange.com/questions/1190504/checking-convergence-of-sum-frac-sin-nxn and https://math.stackexchange.com/questions/3787384/convergence-of-sum-frac-sin-n-thetanr-and-sum-n-1-infty-u-n-cos-n and https://math.stackexchange.com/questions/2682621/prove-the-uniform-convergence-of-sum-n-1-infty-fraca-n-sinnxn-and and probably others. – Gerry Myerson Mar 24 '21 at 07:46
  • Also https://math.stackexchange.com/questions/2450212/if-sum-f-nx-uniformly-convergent-on-mathbb-r-then-there-is-m-n0? and https://math.stackexchange.com/questions/573720/convergence-on-sum-of-cos and https://math.stackexchange.com/questions/2079471/summing-a-trigonometric-series-sum-limits-n%ef%bc%9d1-infty-frac-sin-nxn? and https://math.stackexchange.com/questions/1143302/convergence-of-fourier-series-for-a-sum-which-is-not-uniform-convergent – Gerry Myerson Mar 24 '21 at 07:51
  • Have you had a look at those links, Luna? – Gerry Myerson Mar 25 '21 at 11:11
  • Sorry, I wasn't paying attention to your comments. I will check them out in a moment and get back to you. Thanks! – Luna_II Mar 25 '21 at 18:26
  • Okay, so I did eventually find out how to solve this problem after doing a little digging and talking with my professor. It does actually alternate, but you are correct to say that it falls down on the first step. We can define this series as $(-1)\sum \frac{(-1)^{n-1}}{n}$, where it alternates positively on the first step, instead of negative on a traditional harmonic alternating series. So this is conditionally convergent, but it diverges by $\sum \frac{1}{n}$. – Luna_II Mar 25 '21 at 19:34
  • Also, I should've emphasized that I am simply a calculus II student, and therefore I didn't understand some of the other examples you sent me, since they deal with more complex criteria like Dirichlet's Test, which is defined on complex numbers and something not covered until later on. – Luna_II Mar 25 '21 at 19:39
  • My mistake – I misread the $\pi$ in the sum as an $x$ and thought we were asking how to determine the values of $x$ for which it converged. All is well now. – Gerry Myerson Mar 27 '21 at 00:29

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