The series,

\(\sum_{n=1}^\infty(\cos^{-1}(\frac{1}{n+1})-\cos^{-1}(\frac{1}{n+2}))\)

Consider a general term,

\(a_n=\cos^{-1}(\frac{1}{n+1})-\cos^{-1}(\frac{1}{n+2})\)

Consider n-th partial sum:

\(S_n=\sum_{n=1}^\infty(\cos^{-1}(\frac{1}{n+1})-\cos^{-1}(\frac{1}{n+2}))\)

\(=(\cos^{-1}(\frac{1}{2})-\cos^{-1}(\frac{1}{3}))+(\cos^{-1}(\frac{1}{3})-\cos^{-1}(\frac{1}{4})...+(\cos^{-1}(\frac{1}{n})-\cos^{-1}(\frac{1}{n+1}))+(\cos^{-1}(\frac{1}{n+1})-\cos^{-1}(\frac{1}{n+2}))\)

\(=\cos^{-1}(\frac{1}{2})-\cos^{-1}(\frac{1}{3})+\cos^{-1}(\frac{1}{4})...\cos^{-1}(\frac{1}{n})-\cos^{-1}(\frac{1}{n+1})+\cos^{-1}(\frac{1}{n+1})-\cos^{-1}(\frac{1}{n+2})\)

By cancelling finite number of terms only first term and last term remains

\(=\cos^{-1}(\frac{1}{2})-\cos^{-1}(\frac{1}{n+2})\)

Therefore,

\(S_n=\cos^{-1}(\frac{1}{2})-\cos^{-1}(\frac{1}{n+2})\)

Sequence of partial sum:

\(\left\{S_n\right\}=\cos^{-1}(\frac{1}{2})-\cos^{-1}(\frac{1}{n+2})\)

Applying limit

\(\lim_{n\to\infty}S_n=\lim_{n\to\infty}(\cos^{-1}(\frac{1}{2})-\cos^{-1}(\frac{1}{n+2}))\)

\(=\lim_{n\to\infty}(\cos^{-1}(\frac{1}{2}))-\lim_{n\to\infty}(\cos^{-1}(\frac{1}{n+2}))\)

\(=\frac{\pi}{3}-\lim_{n\to\infty}(\cos^{-1}(\frac{1}{n+2}))\)

Using limit chain rule to solve

\(\lim_{n\to\infty}(\cos^{-1}(\frac{1}{n+2}))\)

\(\lim_{n\to\infty}(\frac{1}{n+2})=\frac{1}{\infty}=0\)

\(\lim_{n\to0}(\cos^{-1}(\frac{1}{n+2}))=\cos^{-1}(0)=\frac{\pi}{2}\)

\(\Rightarrow\lim_{n\to\infty}S_n=\frac{\pi}{3}-\frac{\pi}{2}\)

\(=-\frac{\pi}{6}\)

Therefore, the given series is convergent to \(\frac{-\pi}{6}\)

\(\sum_{n=1}^\infty(\cos^{-1}(\frac{1}{n+1})-\cos^{-1}(\frac{1}{n+2}))\)

Consider a general term,

\(a_n=\cos^{-1}(\frac{1}{n+1})-\cos^{-1}(\frac{1}{n+2})\)

Consider n-th partial sum:

\(S_n=\sum_{n=1}^\infty(\cos^{-1}(\frac{1}{n+1})-\cos^{-1}(\frac{1}{n+2}))\)

\(=(\cos^{-1}(\frac{1}{2})-\cos^{-1}(\frac{1}{3}))+(\cos^{-1}(\frac{1}{3})-\cos^{-1}(\frac{1}{4})...+(\cos^{-1}(\frac{1}{n})-\cos^{-1}(\frac{1}{n+1}))+(\cos^{-1}(\frac{1}{n+1})-\cos^{-1}(\frac{1}{n+2}))\)

\(=\cos^{-1}(\frac{1}{2})-\cos^{-1}(\frac{1}{3})+\cos^{-1}(\frac{1}{4})...\cos^{-1}(\frac{1}{n})-\cos^{-1}(\frac{1}{n+1})+\cos^{-1}(\frac{1}{n+1})-\cos^{-1}(\frac{1}{n+2})\)

By cancelling finite number of terms only first term and last term remains

\(=\cos^{-1}(\frac{1}{2})-\cos^{-1}(\frac{1}{n+2})\)

Therefore,

\(S_n=\cos^{-1}(\frac{1}{2})-\cos^{-1}(\frac{1}{n+2})\)

Sequence of partial sum:

\(\left\{S_n\right\}=\cos^{-1}(\frac{1}{2})-\cos^{-1}(\frac{1}{n+2})\)

Applying limit

\(\lim_{n\to\infty}S_n=\lim_{n\to\infty}(\cos^{-1}(\frac{1}{2})-\cos^{-1}(\frac{1}{n+2}))\)

\(=\lim_{n\to\infty}(\cos^{-1}(\frac{1}{2}))-\lim_{n\to\infty}(\cos^{-1}(\frac{1}{n+2}))\)

\(=\frac{\pi}{3}-\lim_{n\to\infty}(\cos^{-1}(\frac{1}{n+2}))\)

Using limit chain rule to solve

\(\lim_{n\to\infty}(\cos^{-1}(\frac{1}{n+2}))\)

\(\lim_{n\to\infty}(\frac{1}{n+2})=\frac{1}{\infty}=0\)

\(\lim_{n\to0}(\cos^{-1}(\frac{1}{n+2}))=\cos^{-1}(0)=\frac{\pi}{2}\)

\(\Rightarrow\lim_{n\to\infty}S_n=\frac{\pi}{3}-\frac{\pi}{2}\)

\(=-\frac{\pi}{6}\)

Therefore, the given series is convergent to \(\frac{-\pi}{6}\)