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. ∑n=1∞(cos−1⁡(1n+1)−cos−1⁡(1n+2))

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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.  ∑n=1∞(cos−1⁡(1n+1)−cos−1⁡(1n+2))

BleabyinfibiaG · Accepted Answer

The series,  ∑n=1∞(cos−1⁡(1n+1)−cos−1⁡(1n+2))  Consider a general term,  an=cos−1⁡(1n+1)−cos−1⁡(1n+2)  Consider n-th partial sum:  Sn=∑n=1∞(cos−1⁡(1n+1)−cos−1⁡(1n+2))  =(cos−1⁡(12)−cos−1⁡(13))+(cos−1⁡(13)−cos−1⁡(14)...+(cos−1⁡(1n)−cos−1⁡(1n+1))+(cos−1⁡(1n+1)−cos−1⁡(1n+2))  =cos−1⁡(12)−cos−1⁡(13)+cos−1⁡(14)...cos−1⁡(1n)−cos−1⁡(1n+1)+cos−1⁡(1n+1)−cos−1⁡(1n+2)  By cancelling finite number of terms only first term and last term remains  =cos−1⁡(12)−cos−1⁡(1n+2)  Therefore,  Sn=cos−1⁡(12)−cos−1⁡(1n+2)  Sequence of partial sum:  {Sn}=cos−1⁡(12)−cos−1⁡(1n+2)  Applying limit  limn→∞Sn=limn→∞(cos−1⁡(12)−cos−1⁡(1n+2))  =limn→∞(cos−1⁡(12))−limn→∞(cos−1⁡(1n+2))  =π3−limn→∞(cos−1⁡(1n+2))  Using limit chain rule to solve  limn→∞(cos−1⁡(1n+2))  limn→∞(1n+2)=1∞=0  limn→0(cos−1⁡(1n+2))=cos−1⁡(0)=π2

Jeffrey Jordon · Accepted Answer

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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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