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}))

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asked 2021-01-06
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.
\(\sum_{n=1}^\infty(\cos^{-1}(\frac{1}{n+1})-\cos^{-1}(\frac{1}{n+2}))\)

Answers (1)

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