Question

\(\lim_{n \to \infty} \tan \left\{ \sum_{r=1}^{n} \tan^{-1} \left( \frac{1}{1 + r + r^2} \right) \right\}\) is equal to __________.

Hint:

The expression $1+r+r^2$ is a classic setup for a telescoping series in $\tan^{-1}$. Always try to rewrite the numerator as the difference of terms that form the product in the denominator.

Answer 1

Correct Answer: 1

Step 1: \(\frac{1}{1 + r + r^2} = \frac{(r+1) - r}{1 + r(r+1)}\).
Step 2: \(\tan^{-1}(.......) = \tan^{-1}(r+1) - \tan^{-1} r\).
Step 3: Sum \(S_n = \sum_{r=1}^n (\tan^{-1}(r+1) - \tan^{-1} r) = \tan^{-1}(n+1) - \tan^{-1} 1\).
Step 4: \(\lim_{n \to \infty} S_n = \pi/2 - \pi/4 = \pi/4\).
Step 5: \(\tan(\pi/4) = 1\).