⁻¹ (2 1^2 )+ ⁻¹ (2 2^2 )+ ⁻¹ (2 3^2 )+ =

Step-by-step solution

The infinite series S is defined as S = _ n=1 ^ ⁻¹(2n^2) . For positive arguments, ⁻¹(x) = ⁻¹ ( 1 x ) , so the general term becomes T_n = ⁻¹ ( 1 2n^2 ) . Multiplying numerator and denominator by 2 yields T_n = ⁻¹ ( 2 4n^2 ) . Using the identity ⁻¹(A) - ⁻¹(B) = ⁻¹ ( A - B 1 + AB ) with A = 2n + 1 and B = 2n - 1 gives T_n = ⁻¹(2n + 1) - ⁻¹(2n - 1) , since A - B = 2 and 1 + AB = 4n^2 . The partial sum telescopes when expanded: S_k = _ n=1 ^ k [ ⁻¹(2n + 1) - ⁻¹(2n - 1) ] = ⁻¹(2k + 1) - ⁻¹(1) . Taking the limit as k , ⁻