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I just don't know what test to use for $\sum_{n=1}^\infty \frac{\cos^2(n\pi)}{n\pi}$ to see if it converges or diverges? I wanted to do the comparison test with $\frac{1}{n}$ if I factor out the $\frac{1}{\pi}$. Can I do this? And then it is divergent?

Thanks so much!

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    $\begingroup$ Hint: $\cos(n\pi) = (-1)^n$. $\endgroup$ – user474330 Oct 3 '17 at 21:19
  • $\begingroup$ I don't see a reason to downvote this question, the OP put effort into it, also used MathJax. To the OP welcome to Math.SE, don't get discouraged by the down votes. You can pull $\frac1{\pi}$ out of the summation, one also needs to know that $\cos^2(n\pi)=1$ to conclude that the series diverge. $\endgroup$ – kingW3 Oct 3 '17 at 22:07
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Notice that $\cos(n\pi) = (-1)^n$ and therefore $\cos^2(n\pi) = 1$, and try using the Limit Comparison Test.

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  • $\begingroup$ Why would you use the limit comparison test? And if so, what would you compare it to? $\endgroup$ – Simply Beautiful Art Oct 3 '17 at 23:01
  • $\begingroup$ I would compare it with $\frac 1 n$, which proves the series diverges. $\endgroup$ – AsafHaas Oct 5 '17 at 16:38
  • $\begingroup$ Now I think it is not needed-the series is a divergent series multiplied by a constant, therefore divergent. $\endgroup$ – AsafHaas Oct 5 '17 at 16:39
  • $\begingroup$ =P That's what I meant to point out. $\endgroup$ – Simply Beautiful Art Oct 5 '17 at 18:35
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$cos(n\pi)$ is either $+1$ or $-1$ but there is a square on the cosine term, so numerator is $1$. Denominator is of linear nature. (The $\pi$ upfront has got nothing to do with it). So your series behaves like the Harmonic series. Based upon this result, we arrive at KingW's result. More important, do you see why?

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