Context:
This was another question given to me by my math tutor:
Does the infinite series $$\sum_{n=1}^{\infty} \frac{\sin n}{n}$$ converge or diverge?
My working:
I intend to use Dirichlet's test to determine whether the series converges or diverges.We can rewrite the summand as $$\sum_{n=1}^{\infty} \overbrace{\sin n}^{a_n} \underbrace{\frac{1}{n}}_{b_n}$$ We can see that $b_n$ is a positive, decreasing sequence that approaches $0$. Therefore, we only need to verify that $$\Bigg| \sum_{n=1}^{N} \sin n \Bigg| \leq K$$ for some constant $K$. My argument using complex numbers is as follows:
Recall that $$\sin x = \frac{e^{ix} - e^{-ix}}{2i}$$ We can proceed to write out the summations and factor out the constant $\frac{1}{2i}$ yielding: $$\sum_{n=1}^{N} \sin n = \frac{1}{2i} \left(\sum_{n=1}^{N} e^{in} - \sum_{n=1}^{N} e^{-in}\right)$$ Notice that both sums are geometric. The first sum has a ratio of $e^i$ and the second sum has a ratio of $e^{-i}$. Using the formula for partial sums of a geometric series we obtain $$\sum_{n=1}^{N} \sin n = \frac{1}{2i} \left( \frac{e^{i} - e^{i \left(N + 1 \right)}}{1 - e^i} - \frac{e^{-i} - e^{-i \left(N + 1 \right)}}{1 - e^{-i}} \right)$$ Now the only terms that are affected by $N$ are $e^{i \left(N + 1 \right)}$ and $e^{-i \left(N + 1 \right)}$. Also recall that $|e^{ix}|=1$ for real $x$. So, $$|e^{i \left(N + 1 \right)}| = |e^{-i \left(N + 1 \right)}| = 1$$ Therefore$^*$, $$\Bigg| \sum_{n=1}^{N} \sin n \Bigg| = \Bigg| \frac{1}{2i} \left( \frac{e^{i} - e^{i \left(N + 1 \right)}}{1 - e^i} - \frac{e^{-i} - e^{-i \left(N + 1 \right)}}{1 - e^{-i}} \right) \Bigg|$$ is bounded above by some constant independent of $N$, as desired. Hence, the series converges.
$^*$If you cannot follow from how I went from the previous step to this one it is because I still have some work left to do. In particular, I should be using the triangle inequality but I haven't resolved this part yet. All I know is that my final answer is supposed to be(my math tutor told me) what I have shown above. I will edit it in as soon as I have figured it out.
My question: Are there other ways to show this? One of my friends pointed out that I could use the cosine sum identity and after using some properties and simplifications we get: $$\cos(n - \beta) - \cos(n + \beta) = 2 \sin n \sin \beta (\textrm{Full disclosure: I have no clue how she got this.})$$ The motivation here is to produce a telescoping sum on the left hand side by choosing some value for $\beta$. But I don't have the slightest idea on how to do that. Anyways, please do share your ideas.
Edit: I am aware of the other posts about this summation, however their answers are short and not very detailed. I do not understand answers like that very well as the writers tend to skip steps which confuses me even if it seems obvious and simple to others.
