Answer
$-\frac{\pi}{6}$
Work Step by Step
Given :
\[ \sum_{n=1}^{\infty} \left( \cos^{-1}\left(\frac{1}{n+1}\right) - \cos^{-1}\left(\frac{1}{n+2}\right) \right) \]
Consider the series:
\[ S_n = \left( \cos^{-1}\left(\frac{1}{2}\right) - \cos^{-1}\left(\frac{1}{3}\right) \right) + \left( \cos^{-1}\left(\frac{1}{3}\right) - \cos^{-1}\left(\frac{1}{4}\right) \right) + \ldots + \left( \cos^{-1}\left(\frac{1}{n+1}\right) - \cos^{-1}\left(\frac{1}{n+2}\right) \right) \]
Combine the like terms, noticing cancellations:
\[ S_n = \cos^{-1}\left(\frac{1}{2}\right) - \cos^{-1}\left(\frac{1}{n+2}\right) \]
Now, find the limit as \( n \) approaches infinity:
\[ \lim_{n\to\infty} S_n = \cos^{-1}\left(\frac{1}{2}\right) - \lim_{n\to\infty} \cos^{-1}\left(\frac{1}{n+2}\right) \]
As \( n \) approaches infinity, \( \frac{1}{n+2} \) approaches zero, and \( \cos^{-1}(0) = \frac{\pi}{2} \):
\[ \lim_{n\to\infty} S_n = \cos^{-1}\left(\frac{1}{2}\right) - \frac{\pi}{2} \]
Evaluate \(\cos^{-1}\left(\frac{1}{2}\right)\):
\[ \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \]
Substitute back:
\[ \lim_{n\to\infty} S_n = \frac{\pi}{3} - \frac{\pi}{2} = -\frac{\pi}{6} \]
Therefore, the series converges to \(-\frac{\pi}{6}\).
