calculus - Convergence of $sum_{n=1}^{infty} frac{sin(n)}{n}$

I am trying to argue that
$$
\sum_{n=1}^{\infty} \frac{\sin(n)}{n}
$$

is divergent. It get that it must be divergent because $\sin(n)$ is bounded and there is an $n$ on the bottom. But I have to use one of the tests in Stewart's Calculus book and I can't figure it out. I can't use the Comparison Tests or the Integral Test because they require positive terms. I can't take absolute values, that would only show that it is not absolutely convergent (and so it might still be convergent). The Divergence Test also doesn't work.

I see from this question:

Evaluate $ \sum_{n=1}^{\infty} \frac{\sin \ n}{ n } $ using the fourier series

that the series is actually convergent, but using some math that I don't know anything about. My questions are

(1) Is this series really convergent?

(2) Can this series be handled using the tests in Stewart's Calculus book?

Answer

One may apply the Dirichlet test, noticing that

• $\displaystyle \frac1{n+1} \le \frac1{n}$




• $\displaystyle \lim_{n \rightarrow \infty}\frac1{n} = 0$



• $\displaystyle \left|\sum^{N}_{n=1}\sin
  n\right|=\left|\text{Im}\sum^{N}_{n=1}e^{in}\right| \leq
  \left|e^i\frac{1-e^{iN}}{1-e^i}\right| \leq \frac{2}{|1-e^i|}<\infty,\qquad N\ge1,$

giving the convergence of the given series.