Convergence of the series $\sum\limits_{n=0}^{\infty }\left ( \tan\frac{\pi n+1}{4n+2}-2\sin\frac{\pi n+1}{6n+1} \right )$ 
Does the series: $\sum\limits_{n=0}^{\infty }\left ( \tan\frac{\pi n+1}{4n+2}-2\sin\frac{\pi n+1}{6n+1} \right )$ converge or diverge?

I've tried to apply Taylor expansion, but I wasn't able to find a proper way to apply it.

We have
$$ \tan\left(\frac{\pi}{4}+x\right)=1+2x+O(x^2),\qquad 2\sin\left(\frac{\pi}{6}+x\right)= 1+\sqrt{3}\,x+O(x^2)\tag{1}$$
$$ \frac{\pi n+1}{4n+2}=\frac{\pi}{4}+\left(\frac{1}{4}-\frac{\pi}{8}\right)\frac{1}{n}+O\left(\frac{1}{n^2}\right)\tag{2}$$
$$ \frac{\pi n+1}{6n+1}=\frac{\pi}{6}+\left(\frac{1}{6}-\frac{\pi}{36}\right)\frac{1}{n}+O\left(\frac{1}{n^2}\right)\tag{3}$$
hence by combining $(1),(2)$ and $(3)$ it follows that:
$$ \tan\left(\frac{\pi n+1}{4n+2}\right)-2\sin\left(\frac{\pi n+1}{6n+1}\right) = \frac{C}{n}+O\left(\frac{1}{n^2}\right)\tag{4} $$
with $C\approx -0.42292335$, so the given series is divergent by comparison with the harmonic series.
