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