Among S 1 : lim n → ∞ 1 n 2 ( 2 + 4 + 6 + … + 2 n ) = 1 S 2 : lim n → ∞ 1 n 16 1 15 + 2 15 + 3 15 + … + n 15 = 1 16

Step-by-step solution

S 1 : lim n → ∞ 1 n 2 [ 2 + 4 + 6 + … + 2 n ] lim n → ∞ 2 n ( n + 1 ) 2 n 2 = 1 S 2 : lim n → ∞ 1 n 16 1 15 + 2 15 + 3 15 + … + n 15 This is limit of sum. We know that lim n → ∞ ∑ r = 1 n f r n 1 n = ∫ 0 1 f x d x ∵ lim n → ∞ ∑ r = 1 n r k n k + 1 = lim n → ∞ ∑ r = 1 n r n k n = ∫ 0 1 x k d x = 1 k + 1 Here k = 15 ∴ lim n → ∞ ∑ r = 1 n r 15 n 16 = 1 16 ∴ Both S 1 and