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I encountered a problem that I can't figure out. I need to see whether the following series converges or diverges: $\frac{\sin^2(n)}{n}$, with n from 1 to infinity. The problem is that sin is defined on complex numbers, so this time sin can take values outside the interval $[-1,1]$. How do you solve this problem? Thank you in advance!

$$\sum_{n=1}^\infty\frac{\sin^2(n)}n$$

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    Did you mean $$\sum_{n=1}^\infty\frac{\sin^2(n)}n$$ – 2017-02-25
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    yes, that's the series I need – 2017-02-25
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    How is $\sin$ being defined on complex numbers relevant here since you're summing over $\mathbb{N}$? – 2017-02-25
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    When you define the sine function over the complex numbers, $\sin z$ has the usual value when $z$ happens to be a real number. Here, $n = 1, 2, 3, \dots$ are real values. – 2017-02-25
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    That's what I can't figure out, I've been told I need to use the complex form of sin, but I don't see for what. – 2017-02-25

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Notice that

$$\sin^2(n)=\frac{1-\cos(2n)}2$$

And it's easy to use $\cos(2n)=\Re(e^{2in})$ to get

$$\sum_{n=1}^\infty\frac{\sin^2(n)}n=\frac12\Re\left[\sum_{n=1}^\infty\left(\frac1n-\frac{e^{2in}}n\right)\right]$$

The right most part of the sum converges by the Dirichlet test while the first part diverges by the p-series, hence, your series diverges.

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    How do you apply Dirichlet on e^(2in)/n ? I'm sorry if I'm asking something simple. – 2017-02-25
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    Note that $\sum e^{2in}$ is bounded and $1/n\to0$. – 2017-02-25
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    Why is e^2in bounded? Isn't its limit infinite? I tried writing it as a geometric progression and got to the same conclusion. – 2017-02-25
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    It's merely a geometric series. Solve the geometric series and show that it is bounded. – 2017-02-25
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    I arrived at e^(2i) * (e^(2in) - 1) / (e^(2i) - 1) and I don't know how to continue. I tried writing e^2in as cos n + i cos n, but I didn't arrive to anything. – 2017-02-25
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    * i sin n, sorry for the typo – 2017-02-25
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    @Jay_Peters Note that $$|e^{2in}-1|\le|e^{2in}|+|-1|=2$$Thus, it is bounded... – 2017-02-25
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    Ah, I understand! Thank you very much for helping me out and sorry for asking so many questions! – 2017-02-25
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    No problem! :-) – 2017-02-25