Following sequence is given.
A_(n+1)=sqrt(b*a_n+c)
a_1=a a,b,c>0
How do you prove it is bounded and convergent?
To prove that the sequence A_n is bounded, we need to show that it does not exceed a certain upper limit or fall below a certain lower limit for all n.
First, let's assume that there exists an upper limit M such that A_n ≤ M for all n.
We can re-write the recurrence relation as follows:
A_(n+1) = √(b * A_n + c)
Since a, b, and c are all positive, we can conclude that for any value of A_n,
0 ≤ b * A_n + c
Taking the square root of both sides, we have:
0 ≤ √(b * A_n + c)
Now, if we assume A_n ≤ M, we can substitute it into the inequality:
0 ≤ √(b * M + c)
Since b and c are positive, b * M + c is a positive constant. Let's denote it as X:
0 ≤ √X
The square root of any positive constant is always greater than or equal to zero. Therefore, for any value of A_n ≤ M, A_(n+1) will also be ≤ M.
By using the same logic, we can prove that the sequence A_n is also bounded below by a certain lower limit L.
Now, to prove convergence, we need to show that the sequence A_n approaches a specific value as n goes to infinity.
Let's assume A_n → L as n → ∞. Substituting this assumption into the recurrence relation, we get:
L = √(b * L + c)
Now, we can solve this equation to find the value of L:
L^2 - b * L - c = 0
Using the quadratic formula, we can find the two possible values of L:
L = (b ± √(b^2 + 4c)) / 2
Since a, b, and c are positive, L will be positive as well. Therefore, the sequence A_n converges to a positive value.
In conclusion, the sequence A_n is both bounded and convergent.