TrigonometrySuperchallengeInverse Trigonometric Functions ShareInverse Trigonometric Functions · Superchallenge problemUsing the identity arctan12k2=arctan12k−1−arctan12k+1\arctan\dfrac{1}{2k^{2}}=\arctan\dfrac{1}{2k-1}-\arctan\dfrac{1}{2k+1}arctan2k21=arctan2k−11−arctan2k+11, evaluate the infinite telescoping sumarctan12+arctan18+arctan118+arctan132+⋯=∑k=1∞arctan12k2.\arctan\dfrac12+\arctan\dfrac18+\arctan\dfrac{1}{18}+\arctan\dfrac{1}{32}+\cdots=\sum_{k=1}^{\infty}\arctan\dfrac{1}{2k^{2}}.arctan21+arctan81+arctan181+arctan321+⋯=k=1∑∞arctan2k21.Check Show answer Walk through it step by step Practice more like this Review this concept