TITLE: Convergence of $\sum_{n=1}^{\infty} \frac{\sin(n)}{n}$
QUESTION [5 upvotes]: 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?

REPLY [5 votes]: 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.