Find lim n → ∞ x n if x n = 1 + 1 2 + 1 4 + … . + 1 2 n 1 + 1 3 + 1 9 + … + 1 3 n

We have x n = 1 + 1 2 + 1 4 + … . + 1 2 n 1 + 1 3 + 1 9 + … + 1 3 n The numerator is a GP with first term a 1 = 1 , common ratio r 1 = 1 2 , and number of terms n ; and denominator is another GP with first term a 2 = 1 , common ratio r 2 = 1 3 and number of terms n . As n → ∞ , both become infinite GPs. Using a + a r + a r 2 + . . . . = a 1 - r , r < 1 , we get lim n → ∞ x n = 1 1 - 1 2 1 1 - 1 3 = 4 3

Find lim n → ∞ x n if x n = 1 + 1 2 + 1 4 + … . + 1 2 n 1 + 1 3 + 1 9 + … + 1 3 n

We have x n = 1 + 1 2 + 1 4 + … . + 1 2 n 1 + 1 3 + 1 9 + … + 1 3 n The numerator is a GP with first term a 1 = 1 , common ratio r 1 = 1 2 , and number of terms n ; and denominator is another GP with first term a 2 = 1 , common ratio r 2 = 1 3 and number of terms n . As n → ∞ , both become infinite GPs. Using a + a r + a r 2 + . . . . = a 1 - r , r < 1 , we get lim n → ∞ x n = 1 1 - 1 2 1 1 - 1 3 = 4 3

Prove the existence of the limit of the sequence y n = a 1 2 n a > 1 and calculate it.

Given that :- y n = a 1 2 n a > 1 We have to calculate that y n = a 1 2 n . Applying the limits on both sides we get ⇒ lim n → ∞ y n = lim n → ∞ a 1 2 n We know that if it is given that lim x → 0 x 2 then its value can be found by putting value of as ⇒ lim x → 0 x 2 = 0 2 = 0 So, applying this we get ⇒ lim n → ∞ y n = a 1 2 n We know that ∞ can be written as 1 0 so, applying this we get ⇒ lim n → ∞ y n = a 1 2 ∞ ⇒ lim n → ∞ y n = a 0 ⇒ lim n → ∞ y n = 1 Hence, the value of lim n → ∞ y n = 1 .

Prove that the sequence x 1 = a ; x 2 = a + a , x 3 = a + a + a ; … ; x n = a + a + … . + a ⏟ n radicals has the limits b = 4 a + 1 + 1 2 .

According to question x 1 2 = a , Taking Square on both side x 2 2 = a + a and x n = a + a + a ⋯ ⋯ ⋯ ⋯ up to n term Put x n = b when n → ∞ ⇒ b = a + b ⇒ b 2 = a + b ⇒ b 2 - b - a = 0 ⇒ b = 1 ± 1 + 4 a 2 Hence, it is proved that b has a limit. 4 a + 1 + 1 2 .

Prove that the sequence with the general term x n = 1 3 + 1 + 1 3 2 + 2 + … + 1 3 n + n has a finite limit.

We know that 0 < 1 3 + 1 + 1 3 2 + 1 ⋯ ⋯ ⋯ ⋯ + 1 3 n < 1 3 + 1 3 2 + ⋯ ⋯ ⋯ + 1 3 n ⇒ 0 < x n < 1 3 1 - 1 3 ⇒ 0 < x n < 1 2 Hence, given condition prove that . Given sequence has finite limit.