Get ready Part 3

Let $\displaystyle S_n = \sum_{k=0}^n \cot^{-1} \left(k^2+k+1\right)$ . The first few $S_n$ appears to claim that $S_n = \cot^{-1} \left(\dfrac 1{n+1}\right)$ . Let us prove by induction that the claim is true for all $n \ge 0$ .

Proof: For $n=0$ , $S_0 = \cot^{-1} 1 = \cot^{-1} \left(\dfrac 1{0+1}\right)$ . The claim is true for $n=0$ . Assuming the claim is true for $n$ , then:

$\begin{aligned} S_{n+1} & = S_n + \cot^{-1} \left((n+1)^2 + (n+1) + 1\right) \\ & = \cot^{-1} \left(\dfrac 1{n+1}\right) + \cot^{-1} \left(n^2+3n+3\right) \\ & = \cot^{-1} \left(\frac {\frac {n^2+3n+3}{n+1}-1}{\frac 1{n+1} + n^2+3n+3} \right) \\ & = \cot^{-1} \left(\frac {n^2+2n+2}{n^3+4n^2+6n+4} \right) \\ & = \cot^{-1} \left(\frac {n^2+2n+2}{(n+2)(n^2+2n+2)} \right) \\ & = \cot^{-1} \left(\frac 1{n+2} \right) \end{aligned}$

Therefore, the claim is true for $n+1$ and true for all $n \ge 0$ .

Then $\displaystyle \lim_{n \to \infty} S_n = \lim_{n \to \infty} \cot^{-1} \left(\frac 1{n+1}\right) = \cot^{-1}(0) = \frac \pi 2 \approx \boxed{1.5708}$ .