Question

# Find a formula for the nth partial sum of the series and use it to determine whether the series converges or diverges. If a series converges, find its sum. sum_{n=1}^infty(cos^{-1}(frac{1}{n+1})-cos^{-1}(frac{1}{n+2}))

Series
Find a formula for the nth partial sum of the series and use it to determine whether the series converges or diverges. If a series converges, find its sum.
$$\sum_{n=1}^\infty(\cos^{-1}(\frac{1}{n+1})-\cos^{-1}(\frac{1}{n+2}))$$

2021-01-07
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 $$\frac{-\pi}{6}$$