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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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  • $\begingroup$ Did you mean $$\sum_{n=1}^\infty\frac{\sin^2(n)}n$$ $\endgroup$ – Simply Beautiful Art Feb 25 '17 at 15:22
  • $\begingroup$ yes, that's the series I need $\endgroup$ – Jay_Peters Feb 25 '17 at 15:24
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    $\begingroup$ How is $\sin$ being defined on complex numbers relevant here since you're summing over $\mathbb{N}$? $\endgroup$ – Zain Patel Feb 25 '17 at 15:26
  • $\begingroup$ 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. $\endgroup$ – user49640 Feb 25 '17 at 15:28
  • $\begingroup$ 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. $\endgroup$ – Jay_Peters Feb 25 '17 at 15:29
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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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  • $\begingroup$ How do you apply Dirichlet on e^(2in)/n ? I'm sorry if I'm asking something simple. $\endgroup$ – Jay_Peters Feb 25 '17 at 16:55
  • $\begingroup$ Note that $\sum e^{2in}$ is bounded and $1/n\to0$. $\endgroup$ – Simply Beautiful Art Feb 25 '17 at 16:55
  • $\begingroup$ Why is e^2in bounded? Isn't its limit infinite? I tried writing it as a geometric progression and got to the same conclusion. $\endgroup$ – Jay_Peters Feb 25 '17 at 17:02
  • $\begingroup$ It's merely a geometric series. Solve the geometric series and show that it is bounded. $\endgroup$ – Simply Beautiful Art Feb 25 '17 at 17:03
  • $\begingroup$ 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. $\endgroup$ – Jay_Peters Feb 25 '17 at 17:22

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