Convergence of an alternating series : $ \sum_{n\geq 1} \frac{(-1)^n|\sin n|}{n}$ 
Study the convergence of $$\displaystyle \sum_{n\geq 1} \frac{(-1)^n|\sin n|}{n}.$$

I am stuck with this series, we need probably some measure of irrationally of $\pi$, unfortunately I am unfamiliar with this. So here is my attempt :
Let $f(x) = \sum \frac{|\sin{n}|}{n} x^n, |x| < 1$
It's not difficult to compute the Fourier series of $|\sin(x)|$ :
$$
\displaystyle|\sin(x)|=\frac{2}{\pi}-\frac{4}{\pi}\sum_{n=1}^{+\infty}\frac{\cos(2nx)}{4n^2-1}
$$
Then Fubini's theorem (Series Version) works very well (because the previous series converges absolutely at $x$ fixed ) and all calculations made, we find that for all $x\in( -1,1)$:
$$
\displaystyle f(x)=\frac{2}{\pi}\sum_{n=1}^{+\infty}\frac{x^n}{n}-\frac{4}{\pi}\sum_{p=1}^{+\infty}\frac{x^2-2x\cos(p)}{(4p^2-1)(x^2-2x\cos(p)+1)}
$$
However, the second sum I have not been able to show the convergence. I feel the series diverge because the following series
$$
\displaystyle\sum\frac{1}{p^2\sin^2\left(\frac{p}{2}\right)}
$$
diverge because  $0$ is an accumulation point of $\displaystyle (n\sin(n))$ sequence.
Any ideas (for the original series) ?
 A: $\color{red}{\text{Not an answer, just an idea needing more work}}$ :

I'd say $\sum_n \frac{(-1)^n}{n} b_n$ converges whenever $\Delta^k b(n) = O(w^{k}),w< 2$  where $\Delta^k b(n)$ is the $k$th forward difference, here $b(n) = |\sin n|$
If you sum by parts $k$ times, using that $\sum_{n=1}^N \frac{(-1)^n}{n} = \frac{(-1)^N}{2 N}+ O(\frac{1}{2 N^2})$ you'll get a main term $2^{-k} \sum_{n=1}^{N-k} \frac{(-1)^n}{n} \Delta^k b(n)$

use that $A(2M)=\sum_{n=M}^\infty \frac{1}{(2n+1)(2n+2)} = \sum_{n=m}^\infty \frac{1}{(2n+1)^2}-\frac{1}{(2n+1)^2(2n+2)}$ and approximate with $\int_{2M}^\infty \frac{dx}{x^2}-\int_{2M}^\infty \frac{dx}{x^3}$ to obtain
$$A(N) = \ln 2+\sum_{n=1}^N \frac{(-1)^n}{n}=\frac{(-1)^N}{2N}+O(\frac{1}{2 N^2})$$
Summing by parts
$$\sum_{n=1}^N \frac{(-1)^n}{n} |\sin n| = A(N)|\sin(N)|+\sum_{n=1}^{N-1} A(n) (|\sin n| - |\sin (n +1)|)$$
The problematic term is $\sum_{n=1}^{N-1} \frac{(-1)^N}{2N} (|\sin n| - |\sin (n +1)|) $ 
that we can sum by parts again to get a new problematic term
$$\sum_{n=1}^{N-2} \frac{(-1)^n}{4n} \Delta^2 b(n)$$
where $\Delta^2 b(n)=(|\sin n| - |\sin( n +1)|)-(|\sin( n+1)| - |\sin( n +2)|)$
summing by parts $k$ times we'll have
$$\frac{1}{2^k}\sum_{n=1}^{N-k} \frac{(-1)^n}{n} \Delta^k b(n)$$
Where $\Delta^k b(n)$ is the $k$th forward difference of $b(n) = |\sin n|$
A: By Dirichlet's convergence test, this series will converge if we can show that there exists a constant $C$ such that $$\left|\sum_{n\leq x}(-1)^{n}|\sin(n)|\right|\leq C$$ for all $x$.
Lets write $$\sum_{n\leq x}(-1)^{n}|\sin(n)|=\sum_{n\leq\frac{x}{2}}|\sin(2n)|-\sum_{n\leq\frac{x+1}{2}}|\sin(2n-1)|.$$ Then for $x=2N$, an even number, Euler Maclaurin summation yields $$\sum_{n\leq N}|\sin(2n)|=\int_{1}^{N}|\sin(2t)|dt+\sum_{k=1}^{K}\frac{(-1)^{k}}{k!}B_{k}\left(\frac{d^{k-1}}{dt^{k-1}}|\sin(2t)|\biggr|_{t=1}^{t=N}\right)$$ $$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\frac{(-1)^{K}}{K!}\int_{1}^{N}B_{K}(\{t\})\left(\frac{d^{k}}{dt^{k}}|\sin(2t)|\right)dt.$$ Note that $|\sin(x)|$ has infinitely many derivatives everywhere except at integer multiples of $\pi$, and so the above holds for any $K>0$. Since $$|B_{k}(\{x\})|\leq k!2^{1-k}\pi^{-k}\zeta(k),$$ and since the derivatives of $|\sin(t)|$ are bounded in absolute value by $1$, it follows that $$\left|\sum_{n\leq N}|\sin(2n)|-\int_{1}^{N}|\sin(2t)|dt\right|\leq4\sum_{k=1}^{K}\frac{\zeta(k)}{(2\pi)^{k}}+\frac{2\zeta(K)N}{(2\pi)^{K}}.$$ The series $\sum_{k=1}^{\infty}\frac{\zeta(k)}{(2\pi)^{k}}$  converges absolutely, so by taking $K=N$ we see that there exists a constant $C_{1}$ such that $$\left|\sum_{n\leq N}|\sin(2n)|-\int_{1}^{N}|\sin(2t)|dt\right|\leq C_{1}$$ for all $N$. Similarly, there exists a constant $C_{2}$ such that $$\left|\sum_{n\leq N}|\sin(2n-1)|-\int_{1}^{N}|\sin(2t-1)|dt\right|\leq C_{2}.$$ Thus by the triangle inequality, $$\left|\sum_{n\leq x}(-1)^{n}|\sin(n)|\right|\leq C_{1}+C_{2}+\left|\int_{1}^{N}|\sin(2t)|dt-\int_{1}^{N}|\sin(2t-1)|dt\right|$$ 
$$\leq C_{1}+C_{2}+\int_{N-1/2}^{N}|\sin(2t)|dt+\int_{1}^{3/2}|\sin(2t-1)|dt$$ 
$$\leq C_{3}$$ for some constant $C_{3}$. This implies the desired result.
A: By using the Fourier series of $\left|\sin(n)\right|$ the problem of showing that
$$ \sum_{n\geq 1}\frac{(-1)^n \left|\sin n\right|}{n} $$
is convergent boils down to the problem of showing that
$$ \sum_{m\geq 1}\frac{\log\left|\cos m\right|}{4m^2-1}$$
is convergent. The only issue is given by the values of $m$ such that $m$ is close to an odd multiple of $\frac{\pi}{2}$.
On the other hand, since the irrationality measure of $\pi$ is finite, even if $\cos m$ is close to zero it cannot be closer than $\frac{1}{m^{10}}$ for any $m$ large enough. Since
$$ \sum_{m\geq 1}\frac{10\log m}{4m^2-1} $$
is absolutely convergent, the original series is convergent as well.
