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Let $f_n(x) = \frac{\sin(nx)}{n^2}$. I want to show that the infinite series $\sum f_n$ converges for all $x$.

After trying the ratio test and getting nowhere, I attempted to use the comparison test for absolute convergence using the series $\sum \frac{1}{n^2}$, which converges. Further,

$$0 \leq \frac{|\sin(nx)|}{n^2} \leq \frac{1}{n^2}$$

So $\sum f_n$ converges absolutely and therefore converges.

In determining the interval of convergence, I reasoned that because the above inequality is true for all real $x$, it must be that the interval of convergence is $\mathbb{R}$, but I'm not sure.

Would someone please show whether or not the above comparison test is sufficient for proving absolute convergence, and if my reasoning for the interval of convergence is also valid? Thank you.

Edit: my other concern is that the comparison test only works with series consisting of positive terms, yet my series has nonnegative terms. I don't think it's a problem, but I'm not sure about that either.

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    $\begingroup$ You showed that the series converges absolutely for all $x\in\Bbb R$, so the interval on which this series converges is indeed all of $\Bbb R$. Although, this is not a power series, so "interval of convergence" is not typically used here. $\endgroup$ – Dave Aug 7 '17 at 19:00
  • $\begingroup$ That makes sense. As you noted, the series is not a power series, but because I've only been exposed to that type of series, I didn't know what else to call the series in question haha. Does an infinite series with a variable $x$ that is not a power series like the one in my post have a name? $\endgroup$ – Benedict Voltaire Aug 7 '17 at 19:48
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    $\begingroup$ In this case, the series is a Fourier series. And from the Weierstrass M-test, the series converges uniformly for all $x\in \mathbb{R}$ $\endgroup$ – Mark Viola Aug 7 '17 at 20:35
  • $\begingroup$ @Dave also the theorem for comparison tests require the terms be positive, but my series has nonnegative terms. Can it still be used? $\endgroup$ – Benedict Voltaire Aug 7 '17 at 20:41
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    $\begingroup$ It requires non-negative terms, not necessarily positive. $\endgroup$ – Rab Aug 7 '17 at 20:47

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