Convergence of the series $\sum_{n=1}^{\infty }\frac{1}{n}\left ( \arctan n-2\arctan\frac{n-1}{n} \right )$ 
Does the series: $\sum_{n=1}^{\infty }\frac{1}{n}\left ( \arctan n-2\arctan\frac{n-1}{n} \right )$ 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
$$\arctan n-2\arctan\frac{n-1}{n} = \arctan\frac{n}{2n^3-2n^2+2n-1} \tag{1}$$
due to $\arctan(x)+\arctan(y)=\arctan\frac{x+y}{1-xy}$, and the RHS of $(1)$ behaves like $\arctan\frac{1}{2n^2}\sim\frac{1}{2n^2}$ for large values of $n$, hence the given series is convergent by asymptotic comparison and the $p$-test.
An alternative approach is to notice that, for any $n\geq 2$:
$$\begin{eqnarray*}a_n=\arctan n-2\arctan\frac{n-1}{n}&=&\arctan\left(\frac{1}{2n}+\frac{1}{2n-2}\right)-\arctan\left(\frac{1}{2n}+\frac{1}{2n}\right)\\&=&\int_{\frac{1}{2n-2}}^{\frac{1}{2n}}\frac{dx}{1+\left(x+\frac{1}{2n}\right)^2}\\&\leq&\int_{\frac{1}{2n-2}}^{\frac{1}{2n}}\,dx = \frac{1}{2n(n-1)}.\tag{2}\end{eqnarray*}$$
