# Question 5 [JEE adv, 2021]

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

1. $$S_{10}(x) = \dfrac{\pi}{2}-\tan^{-1}\Big(\dfrac{1 + 11x^2}{10x}\Big)$$, for all $$x>0$$
2. $$\lim_{n\to \infty}\cot(S_n(x)) = x$$, for all $$x>0$$
3. The equation $$S_3(x)=\dfrac{\pi}{4}$$ has a root in $$(0,\infty)$$
4. $$\tan(S_n(x))\leq \dfrac{1}{2}$$, for all $$n\geq 1$$ and $$x > 0$$

## Concepts

Seems like an ugly sum. Try to see any pattern that gives difference of terms. Then, in-between terms cancel out.

## Solution

#### Step 1

Simplify the sum. $\begin{eqnarray} \cot^{-1}\Big(\frac{1 + k(k + 1)x^2}{x}\Big) &=& \tan^{-1}\Big(\frac{x}{1 + k(k + 1)x^2}\Big)\\ &=& \tan^{-1}\Big(\frac{(k+1)x - kx}{1 + kx\cdot (k + 1)x}\Big)\\ &=& \tan^{-1}(k + 1)x - \tan^{-1}kx \end{eqnarray}$ Nice! This is promising. \begin{aligned} S_n(x) &= \sum_{k = 1}^n\cot^{-1}\Big(\frac{1 + k(k + 1)x^2}{x}\Big)\\ &= \sum_{k = 1}^n\Big(\tan^{-1}(k + 1)x - \tan^{-1}kx\Big)\\ &= \tan^{-1}(n+1)x - \tan^{-1}x\\ &= \tan^{-1}\Big(\frac{nx}{1 + (n+1)x^2}\Big) \end{aligned}

#### Step 3: 1.

\begin{aligned} S_{10}(x) &= \tan^{-1}\Big(\frac{10x}{1 + 11x^2}\Big)\\ &= \frac{\pi}{2} - \cot^{-1}\Big(\frac{10x}{1 + 11x^2}\Big)\\ &= \frac{\pi}{2} - \tan^{-1}\Big(\frac{1 + 11x^2}{10x}\Big) \end{aligned}

#### Step 3: 2.

\begin{aligned} \lim_{n\to \infty}\cot(S_n(x)) &= \lim_{n\to \infty}\cot\Big(\tan^{-1}\Big(\frac{nx}{1 + (n+1)x^2}\Big)\Big)\\ &= \lim_{n\to \infty} \frac{1 + (n + 1)x^2}{nx}\\ &= x \end{aligned}

#### Step 3: 3.

\begin{aligned} S_3(x) &= \frac{\pi}{4}\\ \tan^{-1}\Big(\frac{3x}{1 + 4x^2}\Big) &= \frac{\pi}{4}\\ \frac{3x}{1 + 4x^2} &= 1\\ 4x^2 - 3x + 1 &= 0 \end{aligned} has no real roots.

#### Step 3: 4

\begin{aligned} \tan(S_n(x)) &= \frac{nx}{1 + (n + 1)x^2}\\ \tan(S_n(1)) &= \frac{n}{n + 2}\\ &= \frac{1}{1 + 2/n}\\ &\geq\frac{1}{2}, \text{ for } n\geq 2 \end{aligned}

Only options 1 and 2 are correct.