Dene the sequence an inductively by a1 1 and ak1 6ak 5 ak

Dene the sequence an inductively by a₁ = 1 and
a_k+1 =(6a_k + 5) /(a_k + 2) Prove that (i) a_n > 0 and (ii) a_n < 5 for all positive integers n.

Solution

a1 = 1

a_k+1 = (6a_k + 5)/(a_k + 2)

a₂ = (6x1+5)/(1+2) = 11/3

a₃= (6 x (11/3) + 5)/((11/3) + 2) = 81/17

a_n = (6a_n-1 + 5)/(a_n-1 + 2)

now a_n-1 is +ve since a₁, a₂,..... are all +ve.

a_n = (6a_n-1 + 5)/(a_n-1 + 2) = (a_n-1 + 2 + 5a_n-1 + 3)/(a_n-1 + 2) = 1 + (5a_n-1 + 3)/(a_n-1 + 2) > 0

therefore a_n >0.

a_n = (6a_n-1 + 5)/(a_n-1 + 2) = (5(a_n-1 + 2) + a_n-1 - 5)/(a_n-1 + 2) = 5 + (a_n-1 - 5)/(a_n-1 + 2)

now if, a_n-1 <5, then a_n<5.