Alternating Harmonic Series Spin-off We know that the series
$\sum (-1)^n/n$
converges, and clearly every other alternating harmonic series with the sign changing every two or more terms such as
$$\left(1+\frac{1}{2}+\frac{1}{3}\right)-\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)+\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\right)-\cdots$$
must converge. My question here is that does the series below also converge?
$$\sum\frac{\textrm{sgn}(\sin(n))}{n}\quad\textrm{or}\quad\sum\frac{\sin(n)}{n|\sin(n)|}$$
Loosely speaking, the sign changes every $\pi$ terms. I'd be surprised if it doesn't converge. Wolfram Mathematica, after a couple of minutes of computing, concluded the series diverges but I can't really trust it. My first approach (assuming the series converges) was that if we bundle up terms with the same sign like the example above every bundle must have three or four terms, and since the first three terms of all bundles make an alternating series I was going to fiddle with the remaining fourth terms but they don't make an alternating series so I guess there's no point in this approach.
edit: I don't think we can use Dirichlet's test with
$$b_n=\textrm{sgn}(\sin(n)).$$
The alternating cycle here is $\pi$ and I don't believe it would bound the series. For example if the cycle was a number very slightly smaller than $3+1/4$, then $B_n$ (sum of $b_n$) would get larger and larger every four bundles for some time. I believe this should happen for $\pi$ as well since it is irrational. I'm not entirely sure why but $|B_n|\leq3$ for most small $n$ though I guess it's because $\pi-3$ is slightly smaller than $1/7$? Anyway $B_{312\ 692}=4$, $B_{625\ 381}=5$, $B_{938\ 070}=6$, $B_{166\ 645\ 135}=-7$, and $B_{824\ 054\ 044}=8$. $|B_n|$ does not hit $9$ up to $n=1\ 000\ 000\ 000$ with $B_{1\ 000\ 000\ 000}=-2$.
 A: Dirichlet's test is too weak. People should just forget it and just learn its proof.
Claim: If $\sum_{n=1}^N \text{sgn}(\sin(n)) = O(\frac{N}{\log^2 N})$, then $\sum_{n=1}^\infty \frac{\text{sgn}(\sin(n))}{n}$ converges.
Proof: $\sum_{n=1}^\infty \frac{\text{sgn}(\sin(n))}{n} = \lim_{N \to \infty} \left[\frac{\sum_{n=1}^N \text{sgn}(\sin(n))}{N}+\int_1^N \frac{\sum_{n \le t} \text{sgn}(\sin(n))}{t^2}dt\right]$ is obtained from summation by parts. The first term goes to $0$ by hypothesis of our claim, and $\int_1^\infty \frac{\sum_{n \le t} \text{sgn}(\sin(n))}{t^2} dt$ exists (and is finite) since the integrand is bounded above by $O(\frac{1}{t\log^2 t})$. $\square$
Note $\sum_{n=1}^N \text{sgn}(\sin(n)) = 2\#\{n \le N : \sin(n) > 0\}-N$, so, using that $\sin(n) > 0$ if and only if $\{\frac{n}{2\pi}\} \in (0,\frac{1}{2})$, we wish to show $\left|\#\{n \le N : \{\frac{n}{2\pi}\} \in (0,\frac{1}{2})\}-\frac{N}{2}\right| = O(\frac{N}{\log^2 N})$. Now, we use the fact that $\sup_{I \subseteq [0,1]} \left|\frac{\#\left\{n \le N : \{\frac{n}{2\pi}\} \in I\right\}}{N}-|I| \right| = O_\epsilon\left(N^{-\frac{1}{\mu-1}+\epsilon}\right)$ for any $\epsilon > 0$, where the supremum is over intervals $I$ and $\mu$ is the approximation exponent of $2\pi$. Since the approximation exponent of $2\pi$ is finite (since it is finite for $\pi$), we get $\left|\#\{n \le N : \{\frac{n}{2\pi}\} \in (0,\frac{1}{2})\}-\frac{N}{2}\right| = O(N^\alpha)$ for some $\alpha < 1$, which is obviously $O(\frac{N}{\log^2 N})$.
