For any positive integer $n$, let $S_n:(0,\infty)\to\mathbb{R}$ be defined by $S_n(x)=\sum_{k=1}^{n}\cot^{-1}\left(\frac{1+k(k+1)x^{2}}{x}\right)$ where for any $x\in\mathbb{R}$, $\cot^{-1}x\in(0,\pi)$ and $\tan^{-1}(x)\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. Then which of the following statements is (are) TRUE?

For any positive integer n, let Sₙ:(0,∞)→R be defined by Sₙ(x)=∑ₖ₌₁ⁿcot⁻¹((1+k(k+1)x²)/x) where for any x∈R, cot⁻¹x∈(0,π) and tan⁻¹(x)∈(-π/2,π/2). Then which of the following statements is (are) TRUE? $S_n(x) = \sum_{k=1}^{n} \cot^{-1}\!\left(\frac{1 + k(k+1)x^{2}}{x}\right)$ $\cot^{-1}x \in (0,\pi), \qquad \tan^{-1}x \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$ $\text{Find a closed form of } S_n(x), \text{ then test (1)–(4)}$ $x>0 \Rightarrow \frac{1+k(k+1)x^2}{x} > 0, \quad \cot^{-1}y = \tan^{-1}\frac{1}{y}$ $\text{term}_k = \tan^{-1}\dfrac{x}{1 + k(k+1)x^2}$ $x = (k+1)x - kx, \qquad k(k+1)x^2 = (kx)\big((k+1)x\big)$ $\tan^{-1}\dfrac{A-B}{1+AB} = \tan^{-1}A - \tan^{-1}B$ $\Rightarrow \text{term}_k = \tan^{-1}((k+1)x) - \tan^{-1}(kx)$ $S_n = \sum_{k=1}^{n} \left[\tan^{-1}((k+1)x) - \tan^{-1}(kx)\right]$ $S_n(x) = \tan^{-1}((n+1)x) - \tan^{-1}(x)$ $S_{10} = \tan^{-1}(11x) - \tan^{-1}(x) = \tan^{-1}\dfrac{10x}{1+11x^2}$ $= \dfrac{\pi}{2} - \tan^{-1}\dfrac{1+11x^2}{10x} \ \Rightarrow (1) \text{ TRUE}$ $\lim_{n\to\infty} S_n = \dfrac{\pi}{2} - \tan^{-1}(x)$ $\lim_{n\to\infty} \cot(S_n) = \tan\!\big(\tan^{-1}x\big) = x \ \Rightarrow (2) \text{ TRUE}$ $S_3 = \tan^{-1}\!\left(\dfrac{3x}{1+4x^2}\right) \leq \tan^{-1}\!\left(\dfrac{3}{4}\right) < \dfrac{\pi}{4}$ $\Rightarrow S_3 = \dfrac{\pi}{4} \text{ has no root} \ \Rightarrow (3) \text{ FALSE}$ $\tan(S_n) = \dfrac{nx}{1+(n+1)x^2}$ $n=2,\ x=\dfrac{1}{\sqrt{3}}:\ \tan(S_2) = \dfrac{1}{\sqrt{3}} \approx 0.577 > \dfrac{1}{2}$ $\Rightarrow (4) \text{ FALSE}$ $\boxed{\text{TRUE statements: } (1) \text{ and } (2)}$ $S_n(x) = \tan^{-1}((n+1)x) - \tan^{-1}(x) \ \Rightarrow \text{TRUE: } (1), (2)$