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I'm given a function $sin(nx)/(n^2)$ and I'm trying to find for which values of x the infinite series for this function would converge. It's easy to see that $sin(nx)$ is always between (-1,1), so then:

$-1/n^{-2} \le sin(nx)/n \le 1/n^{-2}$, so I could use the comparison test. But comparision test states that $a_n$ and $b_n$ must be positive, so would this series converge just for: $0 \le x \le$ $\pi/(2n)$ ?

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Hint: consider series $|\sin nx|/n^2$; do they converge? Why does it imply convergence of original series?

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  • $\begingroup$ Yes that series converges by the comparison test as it is always less than $1/n^{2}$ but to use the comparison test, I have to indicate that $a_n$ is positive, right? So wouldn't that only be for [0, $\pi/2n$] ? $\endgroup$ – user180708 Nov 28 '14 at 10:50
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    $\begingroup$ @user180708 the series I wrote are positive (notice the absolute value there) $\endgroup$ – Ilya Nov 28 '14 at 10:51
  • $\begingroup$ Okay, so then I could simply say that this series converges for all |x| ? This is what I also thought as an alternative answer. $\endgroup$ – user180708 Nov 28 '14 at 10:53
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    $\begingroup$ @user180708: this is indeed the case. If you want to check whether $\sum a_n$ converges, it is sufficient to check whether it converges absolutely, i.e. $\sum |a_n|<\infty$ $\endgroup$ – Ilya Nov 28 '14 at 10:54

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