$\displaystyle \sum^{\infty}_{n=0} \tan^{-1} \left( \frac{ \cot^{-1}(n^2+3n+3) }{ 1+\cot^{-1}(n+1)\cot^{-1}(n+2) } \right)$

If the value of the above expression is in the form $\tan^{-1}\left(\dfrac{p}{4}\right)$, then find the value of $\dfrac{2p}{\pi}$.

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