For any positive integer $n$, define $f_n : (0, \infty) \to \mathbb{R}$ as $f_n(x) = \sum_{j=1}^{n} \tan^{-1}\left(\frac{1}{1+(x+j)(x+j-1)}\right)$ for all $x \in (0, \infty)$. Here, the inverse trigonometric function $\tan^{-1} x$ assumes values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then, which of the following statement(s) is (are) TRUE?

For any positive integer n, define fₙ : (0, ∞) → R as fₙ(x) = ∑ⱼ₌₁ⁿ tan⁻¹(1/(1+(x+j)(x+j-1))) for all x ∈ (0, ∞). Here, the inverse trigonometric function tan⁻¹ x assumes values in (-π/2, π/2). Then, which of the following statement(s) is (are) TRUE?

$f_n(x) = \sum_{j=1}^{n}\tan^{-1}\!\left(\dfrac{1}{1+(x+j)(x+j-1)}\right)$

$\tan^{-1}x \in \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$

$\text{To find: which of statements (1)--(4) are TRUE}$

$\tan^{-1}u - \tan^{-1}v = \tan^{-1}\dfrac{u-v}{1+uv}$

$u=x+j,\ v=x+j-1 \;\Rightarrow\; u-v=1,\ 1+uv = 1+(x+j)(x+j-1)$

$\tan^{-1}\!\left(\dfrac{1}{1+(x+j)(x+j-1)}\right) = \tan^{-1}(x+j) - \tan^{-1}(x+j-1)$

$f_n(x) = \tan^{-1}(x+n) - \tan^{-1}(x)$

$f_j(0) = \tan^{-1}(j) \;\Rightarrow\; \tan(f_j(0)) = j$

$\sum_{j=1}^{5} j^2 = 55 \;\Rightarrow\; (1) \text{ is TRUE}$

$f_j'(0) = \dfrac{1}{1+j^2} - 1 \;\Rightarrow\; 1+f_j'(0) = \dfrac{1}{1+j^2}$

$\sec^2(f_j(0)) = 1 + j^2$

$\sum_{j=1}^{10}\dfrac{1+j^2}{1+j^2} = 10 \;\Rightarrow\; (2) \text{ is TRUE}$

$\tan(f_n(x)) = \dfrac{(x+n)-x}{1+x(x+n)} = \dfrac{n}{1+x(x+n)}$

$\lim_{x\to\infty}\tan(f_n(x)) = 0 \neq \dfrac{1}{n} \;\Rightarrow\; (3) \text{ is FALSE}$

$\sec^2(f_n(x)) = 1 + \tan^2(f_n(x)) \to 1 + 0 = 1$

$\Rightarrow (4) \text{ is TRUE}$

$\boxed{\text{TRUE statements: (1), (2), (4)}}$

$\text{Answer: (1), (2), (4)}$