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Question #40699

Difficulty: 8/10 | 🟢 Free

The Question

Let $H_1, H_2, H_3, \dots$ be a sequence of positive real numbers such that $H_{n+1}$ is the harmonic mean of $H_n$ and $H_{n+2}$ for all $n \ge 1$ . If $H_1 = \frac{1}{5}$ and $H_2 = \frac{1}{9}$ , evaluate the infinite series: $S = \sum_{n=1}^{\infty} \tan^{-1}\left( \frac{8 H_n H_{n+2}}{1 + H_n H_{n+2}} \right)$

Options

A.
$\tan^{-1}\left( \frac{7}{22} \right)$
B.
$\pi + \tan^{-1}\left( \frac{7}{22} \right)$
C.
$\tan^{-1}\left( \frac{14}{45} \right)$
D.
$\frac{\pi}{2} - \tan^{-1}\left( \frac{7}{22} \right)$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Harmonic Mean Constraints|Infinite Series involving Inverse Tangent|Algebraic Simplification

Why This Question Is Tricky

Common Mistakes

This question tests the ability to recognize harmonic mean constraints within inverse trigonometric series.

The primary trap lies in the algebraic simplification phase: students who substitute $H_n = \frac{1}{4n+1}$ will obtain the argument $\frac{4}{8n^2+20n+5}$ .

Attempting to factor this quadratic fails as its roots are irrational.

If they instead rewrite it as $\frac{8}{1+(4n+1)(4n+9)}$ to use the AP terms, they telescope the series to $\pi - (\tan^{-1}(5) + \tan^{-1}(9))$ .

Here, the secondary trap triggers: calculating $\tan^{-1}(5) + \tan^{-1}(9)$ requires the $\pi + \tan^{-1}(x)$ branch cut since $5 \times 9 > 1$ .

Missing this periodicity yields the incorrect option B.

The elegant solution uses the HP property $H_n - H_{n+2} = 8H_n H_{n+2}$ directly, avoiding both the unfactorable quadratic and the branch cut anomaly.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Sequences and Series - Class 11
Concept	Harmonic Mean Constraints
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free