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Question #41196

Difficulty: 8/10 | 🟢 Free

The Question

Evaluate the sum of the infinite series $\sum_{n=3}^{\infty} \tan^{-1}\left( \frac{5}{n^2 - n + 25} \right)$

Options

A.
$\tan^{-1}\left(\frac{5}{2}\right)$
B.
$\tan^{-1}\left(\frac{5}{2}\right) - \pi$
C.
$\pi - \tan^{-1}\left(\frac{2}{5}\right)$
D.
$\tan^{-1}\left(\frac{2}{5}\right)$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Infinite Series involving Inverse Tangent|General Term Formulation|Even Multiplicity Roots|Domain Constraint Intersections

Why This Question Is Tricky

Common Mistakes

This question tests the 'Domain Constraint Intersection' trap within telescoping series.

Students commonly apply the method of differences (V-method) blindly, assuming the sum is simply $V_\infty - V_{initial}$ .

However, the sequence $V_n = -\tan^{-1}\left(\frac{n+5}{n-5}\right)$ has a critical jump discontinuity at $n=5$ , where the argument passes through a pole.

This causes the arctangent function to jump from $\frac{\pi}{2}$ to $-\frac{\pi}{2}$ .

Ignoring this discontinuity leads to the extraneous negative sum $\tan^{-1}\left(\frac{5}{2}\right) - \pi$ .

The problem heavily punishes rote application of the telescoping formula without verifying the domain continuity of the general term across the summation index.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Inverse Trigonometric Functions - Class 12
Concept	Infinite Series involving Inverse Tangent
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free