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 given sequence A_(n+1) = sqrt(b*a_n + c) is bounded and convergent, we need to show two things:
1. Boundedness: We need to demonstrate that the sequence does not grow infinitely large.
2. Convergence: We need to show that the sequence approaches a certain value as n approaches infinity.
To prove boundedness, we can start by assuming that the terms of the sequence are bounded, which means they do not exceed a certain value M for all n. We need to find an appropriate M such that it will hold true for all terms.
Let's assume that M is an upper bound for the sequence A_n, meaning that A_n ≤ M for all n. Now, let's consider the term A_(n+1):
A_(n+1) = sqrt(b*a_n + c)
Since a, b, and c are all greater than 0, we can say that √(b*a_n + c) ≤ √(b*M + c). This means that if M is an upper bound for A_n, then √(b*M + c) will be an upper bound for A_(n+1) as well.
Assuming that M is indeed an upper bound for the sequence, we can conclude that each term in the sequence is less than or equal to M, proving boundedness.
To prove convergence, we need to demonstrate that the sequence approaches a specific value as n approaches infinity. We can do this by establishing that the sequence is both increasing and bounded above.
To show that the sequence is increasing, we need to prove that A_(n+1) ≥ A_n for all n. Let's substitute the expressions for A_n and A_(n+1):
√(b*a_(n-1) + c) ≤ √(b*a_n + c)
Now, let's square both sides to remove the square root:
b*a_(n-1) + c ≤ b*a_n + c
As we can see, the c terms cancel out, leaving us with:
b*a_(n-1) ≤ b*a_n
Since a and b are both greater than 0, we can divide both sides by b to obtain:
a_(n-1) ≤ a_n
Therefore, we have shown that the given sequence is increasing.
Lastly, we have already proven boundedness. By combining the fact that the sequence is both increasing and bounded, we can conclude that it is convergent.