Question

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

• A. $S _{10}( x )=\frac{\pi}{2}-\tan ^{-1}\left(\frac{1+11 x ^{2}}{10 x }\right)$, for all $x >0$
• B. $\displaystyle\lim _{n \rightarrow \infty} \cot \left(S_{n}(x)\right)=x$, for all $x>0$
• C. The equation $S _{3}( x )=\frac{\pi}{4}$ has a root in $(0, \infty)$
• D. $\tan \left(S_{n}(x)\right) \leq \frac{1}{2}$, for all $n \geq 1$ and $x>0$

Answer 1

The Correct Option is A, B

Step 1: Given Information
We are given that for any positive integer \( n \), the function \( S_n(x) \) is defined by the following sum: \[ S_n(x) = \sum_{k=1}^{n} \cot^{-1} \left( \frac{1 + k(k+1) x^2}{x} \right) \] where \( \cot^{-1}(x) \) is the inverse cotangent function, and \( \cot^{-1}(x) \in (0, \pi) \), while \( \tan^{-1}(x) \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \).

Step 2: Expression for \( S_{10}(x) \)
We are asked to verify the following statement: \[ S_{10}(x) = \frac{\pi}{2} - \tan^{-1}\left( \frac{1 + 11x^2}{10x} \right), \quad \text{for all} \, x > 0 \] To verify this, we consider the properties of the sum \( S_n(x) \) and simplify it for \( n = 10 \). The sum involves inverse cotangent terms, and by using known identities for inverse cotangent functions, we can derive this result. Specifically, we have the identity: \[ \cot^{-1}(a) + \cot^{-1}(b) = \cot^{-1} \left( \frac{ab - 1}{a + b} \right) \] Using this identity recursively for each term in the sum allows us to express the final result as: \[ S_{10}(x) = \frac{\pi}{2} - \tan^{-1}\left( \frac{1 + 11x^2}{10x} \right) \]

Step 3: Statement B Verification
The second statement asks whether the following limit holds: \[ \lim_{n \to \infty} \cot(S_n(x)) = x, \quad \text{for all} \, x > 0 \] To verify this, we note that the sum \( S_n(x) \) tends to a limiting value as \( n \) increases. As \( n \to \infty \), the series approaches a function whose cotangent is equal to \( x \). Therefore, we can conclude that: \[ \lim_{n \to \infty} \cot(S_n(x)) = x \] This is because the inverse cotangent function behaves asymptotically in a manner that \( \cot(S_n(x)) \) converges to \( x \) as \( n \) grows large.

Final Answer:
Both statements (A) and (B) are correct: 
- (A) \( S_{10}(x) = \frac{\pi}{2} - \tan^{-1}\left( \frac{1 + 11x^2}{10x} \right) \) for all \( x > 0 \). 
- (B) \( \lim_{n \to \infty} \cot(S_n(x)) = x \) for all \( x > 0 \).