Trigonometry & Inverse Trigonometry
General
Grade 12
Question:
<p><span class="math-inline">\(\displaystyle\sum_{n=1}^{\infty}\tan^{-1}\!\frac{1}{2n^2}=\)</span></p>
Step-by-Step Solution
Key Concept: General
<div class="solution"><p>Rewrite as <span class="math-inline">$\tan^{-1}(n+1)-\tan^{-1}(n-1)$</span>... after telescoping: <span class="math-inline">$\tan^{-1}(\infty)-\tan^{-1}(1)=\pi/2-\pi/4=\pi/4$</span>.</p><div class="key-concept"><strong>Key Concept:</strong> Infinite tan⁻¹ sums — always try telescoping before evaluating directly</div></div>
Correct Answer: π/4