Arctangent Summation

Similar solution with Ikkyu San 's, different presentation.

$\begin{aligned} S & = \lim_{m \to \infty} \sum_{n=1}^m \tan^{-1} \left(\frac{2n}{n^4+n^2+2} \right) \\ & = \lim_{m \to \infty} \sum_{n=1}^m \tan^{-1} \left(\frac{\color{#3D99F6}{(n^2+n+1)} - \color{#D61F06}{(n^2-n+1)}}{1+\color{#3D99F6}{(n^2+n+1)} \color{#D61F06}{(n^2-n+1)}} \right) \\ & = \lim_{m \to \infty} \sum_{n=1}^m \left(\tan^{-1} \color{#3D99F6}{(n^2+n+1)} - \tan^{-1} \color{#D61F06}{(n^2-n+1)} \right) \\ & = \lim_{m \to \infty} \left( \sum_{n=1}^m \tan^{-1} (n^2+n+1) - \sum_{n=1}^m \tan^{-1} (n^2-n+1) \right) \\ & = \lim_{m \to \infty} \left( \sum_{n=1}^m \tan^{-1} (n^2+n+1) - \sum_\color{#D61F06}{n=0}^\color{#D61F06}{m-1} \tan^{-1} [(\color{#D61F06}{n}+1)^2-(\color{#D61F06}{n}+1)+1] \right) \\ & = \lim_{m \to \infty} \left( \sum_{n=1}^m \tan^{-1} (n^2+n+1) - \sum_{n=0}^{m-1} \tan^{-1} (n^2+n+1) \right) \\ & = \lim_{m \to \infty} \left(\tan^{-1} (m^2+m+1) - \tan^{-1} (0^2+0+1) \right) \\ & = \tan^{-1} (\infty) - \tan^{-1} (1) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \end{aligned}$

$\Rightarrow a + b = 1 + 4 = \boxed{5}$

$\begin{aligned}\displaystyle\sum_{n=1}^{\infty}\tan^{-1}\left(\dfrac{2n}{n^4+n^2+2}\right)=&\displaystyle\sum_{n=1}^{\infty}\tan^{-1}\left(\dfrac{(n^2+n+1)-(n^2-n+1)}{1+(n^2+n+1)(n^2-n+1)}\right)\\=&\displaystyle\lim_{m\to\infty}\sum_{n=1}^m\left[\tan^{-1}(n^2+n+1)-\tan^{-1}(n^2-n+1)\right]\\=&\displaystyle\lim_{m\to\infty}\bigg[\tan^{-1}(3)-\tan^{-1}(1)+\tan^{-1}(7)-\tan^{-1}(3)+\tan^{-1}(13)-\tan^{-1}(7)+\cdots+\\&\tan^{-1}(m^2+m+1)-\tan^{-1}(m^2-m+1)\bigg]\\=&\displaystyle\lim_{m\to\infty}\bigg[\tan^{-1}(m^2+m+1)-\tan^{-1}(1)\bigg]\\=&\dfrac{\pi}2-\dfrac{\pi}4=\dfrac{\pi}4\end{aligned}$

Thus, $a+b=1+4=\boxed5$

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Notes:

$\tan^{-1}(x)-\tan^{-1}(y)=\tan^{-1}\left(\dfrac{x-y}{1+xy}\right)$