$\sin n>0$, $a_n=1/n \, ,\sin n<0$, $a_n=-1/n \, , \sum_{n=1}^{\infty} a_n$ converge? Define sequence $a_n$ as: If $\sin n>0$, $a_n=1/n$, and if $\sin n<0$, $a_n=-1/n$.
Does the series $\displaystyle \sum_{n=1}^{\infty} a_n$ converge?
 A: Use dirichlets test, which says
$$\sum_{i=1}^\infty a_i \epsilon_i$$ converges when $a_i$ is a monotone decreasing sequence converging to zero, and the
$$\left|\sum_{i=1}^n \epsilon_i \right|< M$$ for all $n$, (The sum is bounded),
where $\epsilon_i$ is $-1$ if $\sin(i)<0$ and $1$ if $\sin(i)>0$.
As $\sin(n)$ is uniform distributed in $(-1,1)$ you should be able to chose $M=10$ (my own guess is that $4$ would be enough too).
A: Define the sign function $\mathrm{sgn} : \Bbb{R} \to \{-1, 0, 1\}$ as
$$ \mathrm{sgn}(x) = \begin{cases} 1 & x > 0 \\ 0 & x = 0 \\ -1 & x < 0 \end{cases}. $$
Then we can rephrase $(a_n)$ as
$$ a_n = \frac{\mathrm{sgn}(\sin n)}{n}. $$
Now let $s_n = \sum_{k=1}^{n} \mathrm{sgn}(\sin k)$ be the sum of signs. Then by summation by parts,
\begin{align*}
\sum_{k=1}^{n} a_k
&= \sum_{k=1}^{n} \frac{s_{k}-s_{k-1}}{k}
 = \sum_{k=1}^{n} \frac{s_{k}}{k} - \sum_{k=1}^{n-1} \frac{s_{k}}{k+1} \\
&= \frac{s_{n}}{n} + \sum_{k=1}^{n-1} \frac{s_{k}}{k(k+1)}.
\end{align*}
From the Weyl's criterion, it follows that
$$ \lim_{n\to\infty} \frac{s_n}{n} = \frac{1}{2\pi} \int_{-\pi}^{\pi} \mathrm{sgn}(\sin x) \, dx = 0. $$
This shows that $s_n = o(n)$. But this alone is insufficient to prove or disprove that the given series converges or not. Numerical calculations of $(s_n)$ up to $n = 10^7$

shows that the growth of $(s_n)$ is highly suppressed, indicating the possibility that the series converges.
