If lim n → ∞ ⁡ 1 a + 2 a + … + n a n + 1 a - 1 n a + 1 + n a + 2 + … + n a + n = 1 60 for some positive real number a , then a is equal to

Step-by-step solution

lim n → ∞ n a ( ∑ r = 1 n ( r n ) a ) ( n + 1 ) a − 1 . n 2 ( a + 1 + 1 n 2 ) = 1 60 ⇒ lim n → ∞ n a ∑ r = 1 r n a n a + 1 1 + 1 n a - 1 a + 1 + 1 n 2 = 1 60 ⇒ lim n → ∞ 1 n ∑ r = 1 n ( r n ) a ( 1 + 1 n ) a -1 a + 1 + 1 n 2 = 1 60 ⇒ ∫ 0 1 x a d x ( a + 1 2 ) = 1 60 ⇒   1 a + 1 a + 1 2 = 1 60 ⇒   a + 1   2 a + 1 = 120 ⇒ 2 a 2 + 3 a - 119 = 0 ⇒ 2 a 2 + 17 a - 14 a - 119 = 0 ⇒   a - 7