For any positive integer n, let Sn : (0, ∞) → R be defined by Sn(x) = Σ (k=1 to n) cot⁻¹((1 + k(k+1)

Q: Wait, why do we convert $\cot^{-1}$ to $\tan^{-1}$? Can't we just work with $\cot^{-1}$ directly?

While you *could* technically work with $\cot^{-1}$, the standard identity for the difference of inverse trigonometric functions, $\tan^{-1}(A) - \tan^{-1}(B)$, is much more intuitive and widely used in telescoping series problems. Converting to $\tan^{-1}$ aligns the problem with standard algebraic patterns, making the 'telescoping' structure immediately visible.

Q: How do I know if the telescoping series will actually cancel out?

The beauty of a telescoping series lies in the structure $\sum [f(k+1) - f(k)]$. When you write out the terms, you get $(f(2) - f(1)) + (f(3) - f(2)) + \dots + (f(n+1) - f(n))$. Notice how $f(2)$ cancels with $-f(2)$, $f(3)$ with $-f(3)$, and so on. As long as you can express your general term as a difference of two functions where the second part of the $k$-th term matches the first part of the $(k-1)$-th term, the cancellation is guaranteed.

Q: In Option C, why does the discriminant being negative mean there are no roots?

For a quadratic equation $ax^2 + bx + c = 0$, the roots are given by the quadratic formula. The term under the square root, $b^2 - 4ac$, is the discriminant $D$. If $D < 0$, the square root of a negative number is imaginary, meaning there are no real roots. Since the problem asks for a root in $(0, \infty)$, which is a subset of real numbers, a negative discriminant confirms that no such real root exists.

Question

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

Options:

(A) $S_{10}(x) = \frac{\pi}{2} - 	an^{-1}\left(\frac{1 + 11x^2}{10x}ight)$, for all $x > 0$

(B) $\lim_{n 	o \infty} \cot(S_n(x)) = x$, for all $x > 0$

(C) The equation $S_3(x) = \frac{\pi}{4}$ has a root in $(0, \infty)$

(D) $	an(S_n(x)) \le \frac{1}{2}$, for all $n \ge 1$ and $x > 0$

Sigma Percentile · Accepted Answer

### The Art of Telescoping: Unlocking the Series

Welcome, fellow explorer of the mathematical landscape. Today, we are going to dissect a problem that, at first glance, might seem like a daunting mountain of inverse trigonometric functions.

But as we peel back the layers, you will see that it is actually a beautifully orchestrated dance of cancellation. Let us embark on this journey together.

### Phase 1: The Transformation

We are given the series $S_n(x) = \sum_{k=1}^n \cot^{-1}\left(\frac{1 + k(k+1)x^2}{x}ight)$. The first thing that should strike you is the presence of $\cot^{-1}$.

In the world of JEE Advanced, whenever you see inverse cotangent, your first reflex should be to convert it into inverse tangent. Why? Because our algebraic toolkit for $	an^{-1}$ is far more robust.

Using the identity $\cot^{-1}(y) = 	an^{-1}(1/y)$ for $y > 0$, we can rewrite our general term. Since $x > 0$ and $k \ge 1$, the argument is clearly positive. Thus, our series transforms into:

$$S_n(x) = \sum_{k=1}^n 	an^{-1}\left(\frac{x}{1 + k(k+1)x^2}ight)$$

This is the first step of our transformation, and it already feels much more manageable.

### Phase 2: The Telescoping Magic

Now, look closely at the argument: $\frac{x}{1 + k(k+1)x^2}$. Does it remind you of the tangent difference formula?

Recall that $	an^{-1}(A) - 	an^{-1}(B) = 	an^{-1}\left(\frac{A - B}{1 + AB}ight)$. We want to force our expression into this mold.

Look at the denominator: $1 + k(k+1)x^2$. We can factor this as $1 + (kx) \cdot ((k+1)x)$.

Now, check the numerator. We need $A - B$ to equal $x$. If we set $A = (k+1)x$ and $B = kx$, then $A - B = (k+1)x - kx = x$. It is a perfect fit!

Our general term is now:

$$	an^{-1}\left(\frac{(k+1)x - kx}{1 + (k+1)x \cdot kx}ight) = 	an^{-1}((k+1)x) - 	an^{-1}(kx)$$

This is the **'Aha!' moment**. We have successfully turned a complex summation into a difference of two terms.

### Phase 3: The Great Cancellation

Now, let us expand the summation $S_n(x) = \sum_{k=1}^n [	an^{-1}((k+1)x) - 	an^{-1}(kx)]$. Writing out the terms for $k=1, 2, \dots, n$ gives us:

$$S_n(x) = (	an^{-1}(2x) - 	an^{-1}(x)) + (	an^{-1}(3x) - 	an^{-1}(2x)) + \dots + (	an^{-1}((n+1)x) - 	an^{-1}(nx))$$

Observe the pattern. The $	an^{-1}(2x)$ from the first term cancels with the $-	an^{-1}(2x)$ from the second term. This domino effect continues until only the very first negative term and the very last positive term remain.

We are left with:

$$S_n(x) = 	an^{-1}((n+1)x) - 	an^{-1}(x)$$

To make this even more useful, we can combine these back into a single $	an^{-1}$ expression using the difference formula again, resulting in:

$$S_n(x) = 	an^{-1}\left(\frac{nx}{1 + (n+1)x^2}ight)$$

### Phase 4: Evaluating the Options

With our simplified formula $S_n(x) = 	an^{-1}\left(\frac{nx}{1 + (n+1)x^2}ight)$, we can now tackle the options with confidence.

For Option A, we substitute $n=10$ and use the complementary angle property $	an^{-1}(y) = \frac{\pi}{2} - 	an^{-1}(1/y)$ to show it is true.

For Option B, we take the limit of $\cot(S_n(x))$ as $n 	o \infty$, which simplifies to $x$, confirming it is also true.

For Option C, we set $S_3(x) = \pi/4$, which leads to a quadratic equation with a negative discriminant, proving it has no real roots.

Finally, for Option D, we use the AM-GM inequality to show that the maximum value of $	an(S_n(x))$ can exceed $1/2$, proving it false.

Mathematics is not just about solving for $x$; it is about finding the hidden order in chaos. You have just navigated a complex series by finding its underlying structure. Keep this curiosity alive, and you will conquer any problem that comes your way.

Select Answer:

Visualized Solution

Converting to $tan^{- 1}$

Preparing for Telescoping

Telescoping Cancellation

Simplified $S_{n} (x)$

Evaluating Option A

Evaluating Option B

Evaluating Option C

Evaluating Option D

The Sigma Insight: Properties of Inverse Trigonometric Functions

The Art of Telescoping

Phase 1

Phase 2

Phase 3

Phase 4

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