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.
 A: 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 $n\geq 2$:
$$\begin{eqnarray*}a_n=\arctan n-2\arctan\frac{n-1}{n}&=&\arctan\left(\frac{1}{2n}+\frac{1}{2n-2}\right)-\arctan\left(\frac{1}{2n}+\frac{1}{2n}\right)\\&=&\int_{\frac{1}{2n-2}}^{\frac{1}{2n}}\frac{dx}{1+\left(x+\frac{1}{2n}\right)^2}\\&\leq&\int_{\frac{1}{2n-2}}^{\frac{1}{2n}}\,dx = \frac{1}{2n(n-1)}.\tag{2}\end{eqnarray*}$$
A: Using the formula for the arctan of a sum and $\arctan(x)\le x$,
$$
\begin{align}
\arctan(n)-2\arctan\left(\frac{n-1}n\right)
&=\arctan(n)-\arctan\left(\frac{2n^2-2n}{2n-1}\right)\\
&=\arctan\left(\frac{n}{2n^3-2n^2+2n-1}\right)\\
&\le\frac1{2n^2-2n}\\
&=\frac12\left(\frac1{n-1}-\frac1n\right)\tag{1}
\end{align}
$$
Thus,
$$
\begin{align}
\sum_{n=1}^\infty\left(\arctan(n)-2\arctan\left(\frac{n-1}n\right)\right)
&=\frac\pi4+\sum_{n=2}^\infty\left(\arctan(n)-2\arctan\left(\frac{n-1}n\right)\right)\\
&\le\frac\pi4+\sum_{n=2}^\infty\frac12\left(\frac1{n-1}-\frac1n\right)\\
&=\frac{\pi+2}4\tag{2}
\end{align}
$$
Since the series in $(2)$ converges,
$$
\sum_{n=1}^\infty\frac1n\left(\arctan(n)-2\arctan\left(\frac{n-1}n\right)\right)\tag{3}
$$
converges by comparison.
A: $$
\arctan n \to \frac \pi 2 \quad \text{as } n\to\infty.
$$
$$
2\arctan \frac{n-1} n \to \frac \pi 2 \quad \text{as } n\to\infty.
$$
Since they're both approaching $\pi/2$, the situation is more delicate than what it would be if they had different limits. (In that case we'd have $\displaystyle \sum_n \frac{\text{something}} n$ with "something" approaching some non-zero number, and the sum would diverge.)
$$
\arctan a + \arctan b = \arctan \frac{a+b}{1-ab}.
$$
Consequently
\begin{align}
& \arctan a + \arctan b + \arctan c = \arctan \frac{a+b}{1-ab} + \arctan c \\[10pt]
= {} & \arctan \frac{\frac{a+b}{1-ab} + c}{1 - \frac{a+b}{1-ab} c} = \arctan \frac{a+b+c-abc}{1 - ab - ac - bc}.
\end{align}
Apply those with $a= n$, $b=(n-1)/n=c$.
Then use the fact that for $x>0$ near $0$, you have $\dfrac x 2 < \arctan x < x.$
A: Another approach:
$$\lim_{x\to\infty}\frac{\arctan x-2\arctan\frac{x-1}x}{\frac1x}\stackrel{\text{l'Hospital}}=\lim_{x\to\infty}\frac{\frac1{1+x^2}-\frac2{x^2}\frac1{1+\frac{(x-1)^2}{x^2}}}{-\frac1{x^2}}=$$$${}$$
$$=\lim_{x\to\infty}\left(\frac{2x^2}{2x^2-2x+1}-\frac{x^2}{1+x^2}\right)=0$$
Thus, we can use the generalized limit comparison test (as the series is positive):
$$\frac{\frac1n\left(\arctan n-2\arctan\frac{n-1}n\right)}{\frac1{n^2}}=\frac{\left(\arctan n-2\arctan\frac{n-1}n\right)}{\frac1n}\xrightarrow[n\to\infty]{}0$$
and thus our series converges since $\;\sum\frac1{n^2}\;$ does.
