allery$\displaystyle \sum_{n=1}^{\infty} \tan^{-1}\left(\frac{1}{n^2 + n + 1}\right)$ equals
A$\tfrac{\pi}{2}$
B$\tfrac{\pi}{3}$
C$\tan^{-1}(2)$
D$\tfrac{\pi}{4}$
Answer & Solution
Correct answer: D. $\tfrac{\pi}{4}$
Observe that $\tan^{-1}\tfrac{1}{1 + n(n+1)} = \tan^{-1}\tfrac{(n+1)-n}{1+(n+1)n} = \tan^{-1}(n+1) - \tan^{-1}(n)$ (difference formula).
**So the partial sum telescopes:** $\sum_{n=1}^{N} = \tan^{-1}(N+1) - \tan^{-1}(1)$.
As $N \to \infty$, $\tan^{-1}(N+1) \to \tfrac{\pi}{2}$, giving $\tfrac{\pi}{2} - \tfrac{\pi}{4} = \tfrac{\pi}{4}$.
Keep practising Unit 1A2: Trigonometric Functions & Inverse Trig
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