If $\sum_{n=0}^{\infty} 2\cot^{-1}\left(\frac{n^2 + n + 4}{2}\right) = k\pi$, then find the value of $k$.
Step-by-Step Solution
Key Concept: Recognize that the expression can be decomposed into a telescoping series using the tangent subtraction formula.
The sum $\sum_{n=0}^{\infty} 2\tan^{-1}\left(\frac{2}{n^2+n+4}\right)$ telescopes using the identity $2\tan^{-1}\left(\frac{2}{n^2+n+4}\right) = \tan^{-1}\left(\frac{n+1}{2}\right) - \tan^{-1}\left(\frac{n}{2}\right)$. The telescoping series yields $\sum_{n=0}^{\infty}2\left[\tan^{-1}\left(\frac{n+1}{2}\right) - \tan^{-1}\left(\frac{n}{2}\right)\right] = 2\tan^{-1}(\infty) - 2\tan^{-1}(0) = \pi$.
Correct Answer: 1