Trigo(sum)mation

$\begin{aligned} S & = \sum_{n=0}^\infty \tan^{-1} \left(\frac {\cot^{-1}(n^2+3n+3)}{1+\cot^{-1}(n+1)\cot^{-1}(n+2)} \right) & \small \color{#3D99F6} \text{Let }\alpha = \cot^{-1} (n+1) \text{ and } \beta = \cot^{-1} (n+2) \\ & = \sum_{n=0} ^\infty \tan^{-1} \left(\frac {\color{#3D99F6}\alpha - \beta}{1+\alpha \beta} \right) & \small \color{#3D99F6} \text{See note below.} \\ & = \sum_{n=0} ^\infty \tan^{-1} \left(\frac {\tan x - \tan y}{1+\tan x \tan y} \right) & \small \color{#3D99F6} \text{Let }\tan x = \alpha \text{ and } \tan y = \beta \\ & = \sum_{n=0} ^\infty \tan^{-1} \left(\tan (x -y) \right) \\ & = \sum_{n=0} ^\infty (x -y) \\ & = \sum_{n=0} ^\infty \left( \tan^{-1} \alpha -\tan^{-1} \beta \right) \\ & = \sum_{n=0} ^\infty \left( \tan^{-1} \left(\cot^{-1}(n+1)\right) -\tan^{-1} \left(\cot^{-1}(n+2)\right) \right) \\ & = \tan^{-1} \left(\cot^{-1}1\right) \\ & = \tan^{-1} \frac \pi 4 \end{aligned}$

$\implies p = \pi \implies \dfrac {2p}\pi = \boxed{2}$

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Note:

$\small \begin{aligned} \cot (\alpha - \beta) & = \frac {1+\tan \alpha \tan \beta}{\tan \alpha -\tan \beta} = \frac {1+\frac 1{(n+1)(n+2)}}{\frac 1{n+1}-\frac 1{n+2}} = n^2 + 3n + 3 \\ \implies \alpha - \beta & = \cot^{-1}(n^2+3n+3) \end{aligned}$