Methods for determining the convergence of $\sum\frac{\cos n}{n}$ or $\sum\frac{\sin n}{n}$ As far as I know, the textbook approach to determining the convergence of series like $$\sum_{n=1}^\infty\frac{\cos n}{n}$$ and $$\sum_{n=1}^\infty\frac{\sin n}{n}$$ uses Dirichlet's test, which involves bounding the partial sums of the cosine or sine terms. I have two questions:


*

*Are there any other approaches to seeing that these series are convergent? I'm mostly just interested to see what other kinds of arguments might be made.

*What's the best way to show that these two series are only conditionally convergent? I don't even know the textbook approach to that question.

 A: Hint for 2)
$$\sum_{n=1}^{\infty} \frac{|\cos n|}{n} \geq \sum_{n=1}^{\infty} \frac{\cos^2 n}{n}=\sum_{n=1}^{\infty} \frac{1+\cos {2n}}{2n}$$
Convergence of $\sum_{n=1}^{\infty}\frac{\cos{2n}}{2n}$, and divergence of $\sum_{n=1}^{\infty}\frac{1}{2n}$ gives the divergence. 
The same method applies to $\sum_{n=1}^{\infty}\frac{|\sin n|}{n}$. 
A: Thanks for the nice question, below is a sketch for part 1, which may be a few inches from a rigorous proof.
Let us consider the complex sum
$$
S \equiv \sum_{n = 1}^\infty \frac{ e^{in} }{n},
$$
The real and imaginary sums are the desired cosine and sine sums, as pointed out by i707107.
Then
$$
e^{-i} S = \sum_{n = 1}^\infty \frac{ e^{i(n-1)} }{n}
= 1 + \sum_{n = 1}^\infty \frac{ e^{i n} }{n + 1},
$$
and
$$
(1 - e^{-i}) \, S
= -1 + \sum_{n = 1}^\infty \left(\frac{1}{n} - \frac{1}{n + 1} \right) e^{i n}
= -1 + \sum_{n = 1}^\infty \frac{e^{i n}}{n \, (n + 1)}.
$$
So
$$
\begin{align}
|(1 - e^{-i}) \, S|
&\le 1 + \sum_{n = 1}^\infty \left| \frac{e^{i n}}{n \, (n + 1)} \right| \\
&= 1 + \sum_{n = 1}^\infty \frac{1}{n \, (n + 1)} \\
&= 1 + \sum_{n = 1}^\infty \left(\frac{1}{n} - \frac{1}{n + 1}\right) = 2.
\end{align}
$$
This means $|S|$ is finite, and so are the real and imaginary parts.
A: Since you specifically said "Are there any other approaches to seeing that these series are convergent" I will take the bait and give an extremely sketchy argument that isn't meant to be a proof at all.
The harmonic series $\sum_{n=1}^{\infty} \frac{1}{n}$ famously diverges, but very slowly. For large $n$ the numbers being added are so tiny that it takes exponential amounts of time for the millions of tiny terms to amount to much at all.
But $\cos(n)$ will be negative roughly half the time, so there's no possibility of any slow steady accumulation. Thus our sum $\sum_{n=1}^{\infty} \frac{\cos n}{n}$ will remain bounded.
And furthermore, the magnitude of the terms being added gets smaller and smaller, so  in a matter that's roughly analogous to the logic behind the alternating series test, the sum will oscillate [with a period of roughly 6], but with a decreasing amplitude around a value that is an artifact of the first handful of terms.
