The series converges. It is enough to show that the sequence of the following partial sums converges:
\begin{align}
s_N &= \sum_{n=1}^{N} \left(\frac{ |\sin 2n|}{2n}-\frac{|\sin (2n+1)|}{2n+1}\right)=\sum_{n=1}^N \left(\frac{(2n+1)|\sin 2n|- 2n|\sin (2n+1)| }{2n(2n+1)} \right)\\
&=\sum_{n=1}^N \left( \frac{|\sin 2n|-|\sin (2n+1)|}{2n+1}+\frac{|\sin 2n|}{2n(2n+1)}\right).
\end{align}
Thus, it is enough to show that the following converges:
$$
S_N = \sum_{n=1}^N \frac{|\sin 2n|-|\sin (2n+1)|}{2n+1}.
$$
We consider a partition of $\mathbb{N}$ into four disjoint sets $A_{1}$, $A_{2}$, $A_{3}$, $A_{4}$ defined by:
$$
A_{1}=\{n\in\mathbb{N}: \sin 2n >0, \sin (2n+1)>0\}, \ \ A_{2}=\{n\in\mathbb{N}: \sin 2n >0, \sin (2n+1)<0\},
$$
$$
A_{3}=\{n\in\mathbb{N}: \sin 2n <0, \sin (2n+1)>0\}, \ \ A_{4}=\{n\in\mathbb{N}: \sin 2n <0, \sin (2n+1)<0\}.
$$
Note that
$$
A_{1}=\{n\in\mathbb{N}: 2n \ \mathrm{mod} \ 2\pi \in (0,\pi-1)\}, \ \ A_{2}=\{n\in\mathbb{N}: 2n \ \mathrm{mod} \ 2\pi \in (\pi-1,\pi)\},$$
$$
A_{3}=\{n\in\mathbb{N}: 2n \ \mathrm{mod} \ 2\pi \in (-1,0)\}, \ \ A_{4}=\{n\in\mathbb{N}: 2n \ \mathrm{mod} \ 2\pi \in (-\pi, -1)\}.$$
By trigonometric identities,
$$
n\in A_{1} \Longrightarrow |\sin 2n|-|\sin (2n+1)| = \sin 2n - \sin(2n+1) = -2\cos(2n+\frac12)\sin \frac12, $$
$$
n\in A_{2} \Longrightarrow |\sin 2n|-|\sin (2n+1)| = \sin 2n + \sin(2n+1) = 2\sin(2n+\frac12)\cos \frac12, $$
$$
n\in A_{3} \Longrightarrow |\sin 2n|-|\sin (2n+1)| = -\sin 2n - \sin(2n+1) = -2\sin(2n+\frac12)\cos\frac12, $$
$$
n\in A_{4} \Longrightarrow |\sin 2n|-|\sin (2n+1)| = -\sin 2n + \sin(2n+1) = 2\cos(2n+\frac12)\sin \frac12.$$
We define
$$
f_1(x)=-I_{(0,\pi-1)}(x)2\cos(x+\frac12)\sin\frac12, \ \ f_2(x)=I_{(\pi-1,\pi)}(x)2\sin(x+\frac12)\cos\frac12,$$
$$
f_3(x)=-I_{(-1,0)}(x)2\sin(x+\frac12)\cos\frac12, \ \ f_4(x)=I_{(-\pi,-1)}(x)2\cos(x+\frac12)\sin\frac12 $$
where $I_A$ is the characteristic function of $A$. Note that these functions $f_i(x)$ are of bounded variation in $[-\pi,\pi]$. Thus, $f=f_1+f_2+f_3+f_4$ is of a bounded variation.
We need Koksma's inequality p. 143, Theorem 5.1 of 'Uniform Distribution of Sequences' by Kuipers and Niederreiter:
Theorem [Koksma]
Let $f$ be a function on $I=[0,1]$ of bounded variation $V(f)$, and suppose we are given $N$ points $x_1, \ldots , x_N$ in $I$ with discrepancy
$$
D_N:=\sup_{0\leq a\leq b\leq 1} \left|\frac1N \#\{1\leq n\leq N: x_n \in (a,b) \} -(b-a)\right|.
$$
Then
$$
\left|\frac1N \sum_{n\leq N} f(x_n) - \int_I f(x)dx \right|\leq V(f)D_N.
$$
To control the discrepancy, we apply Erdos-Turan inequality p. 112, Theorem 2.5 of Kuipers and Niederreiter's book.
Theorem[Erdos-Turan]
Let $x_1, \ldots, x_N$ be $N$ points in $I=[0,1]$. Then there is an absolute constant $C>0$ such that for any positive integer $m$,
$$
D_N\leq C \left( \frac1m+ \sum_{h=1}^m \frac1h \left| \frac1N\sum_{n=1}^N e^{2\pi i h x_n}\right|\right).
$$
The sequence of our interest is $x_n = 2n$ mod $2\pi$. The above two inequalities together applied to $f=f_1+f_2+f_3+f_4$ yields the following: There is an absolute constant $C>0$ such that for any positive integer $m$,
$$
\left|\frac1N \sum_{n\leq N} f(x_n) - \frac1{2\pi} \int_{-\pi}^{\pi} f(x)dx \right|\leq C \left( \frac1m+ \frac1N\sum_{h=1}^m \frac1{h\langle \frac h{\pi} \rangle}\right).
$$
A result on the irrationality measure of $\pi$ by Salikhov implies that
$$
\left| \frac1{\pi} - \frac pq \right| \geq \frac 1{q^{\mu+\epsilon}}
$$
for all integers $p, q$ and $q$ is sufficiently large, and $\mu=7.60631$, $\epsilon>0$. This implies
$$
h\left\langle \frac h{\pi} \right\rangle \geq h^{2-\mu-\epsilon}
$$
for sufficiently large $h$. Then for some absolute constant $C>0$,
$$
\left|\frac1N \sum_{n\leq N} f(x_n) - \frac1{2\pi} \int_{-\pi}^{\pi} f(x)dx \right|\leq C \left( \frac1m+ \frac1N m^{\mu-1+\epsilon}\right).
$$
Taking $m=\lfloor N^{1/\mu}\rfloor$, we obtain
$$
\left|\frac1N \sum_{n\leq N} f(x_n) - \frac1{2\pi} \int_{-\pi}^{\pi} f(x)dx \right|\leq C N^{-\frac1{\mu}+\epsilon}.
$$
It is easy to see that $\int_{-\pi}^{\pi} f(x) dx = 0$. Therefore,
$$
\left|\sum_{n\leq N}f(x_n)\right|\leq CN^{1-\frac1{\mu}+\epsilon}.
$$
The convergence of the series now follows from Abel's summation formula.
Added on 12/1/2018
By the bound for the discrepancy here: https://en.wikipedia.org/wiki/Low-discrepancy_sequence#Additive_recurrence
We may have a better error term:
$$
\left|\sum_{n\leq N}f(x_n)\right|\leq CN^{1-\frac1{\mu-1}+\epsilon}.
$$
Best Answer
Note that $$1+z^2+z^4+\cdots = \frac{1}{1-z^2}.$$ Integrating both sides, $$z+\frac{z^3}{3}+\frac{z^5}{5}+\cdots=\frac 12 (\log(1+z)-\log(1-z)) = \tanh^{-1}(z).$$ Observe that multiplying both sides by $z$ gives $$\sum_{n=1}^\infty \frac{z^{2n}}{2n-1} = z\tanh^{-1}(z)$$ and multiplying the left hand side by $\frac 1z$ and subtracting the $1$ term gives $$\sum_{n=1}^\infty \frac{z^{2n}}{2n+1} = -1 + \frac{\tanh^{-1}(z)}{z}.$$ Rewriting $\tanh^{-1}(z)$ as $\ln\left(\frac{1-z}{1+z}\right)$, remembering that $z$ is on the unit circle, we can draw vectors $1+z$ and $1-z$ in the complex plane. Doing some basic geometry, we can see that the angle between these is $\frac \pi 2$, and that the lengths of $1+z$ and $1-z$ are $2 \cos \left(\frac \theta 2 \right)$ and $2 \cos \left( \frac \pi 2 - \frac \theta 2 \right) = 2 \sin \left( \frac \theta 2 \right)$.
So $\frac{1-z}{1+z} = -i \cdot \tan \left(\frac x2 \right)$, and so the $\log$ of this is $$-\frac{i \pi}{2} + \ln\left(\tan \left(\frac x2 \right)\right)$$ (since $\log(-i) = -\frac{i \pi}{2}$). From here, everything is easily calculatable.