Final answer:
To prove the convergence of the sequence defined by an = (2n+1)/(3n * an-1), we need to show that the sequence is bounded and monotonic. The sequence is bounded below by 1 and is monotonically increasing, so it is convergent. The limit of the sequence is sqrt(2/3).
Step-by-step explanation:
To prove the convergence of the sequence defined by an = (2n+1)/(3n * an-1), we need to show that the sequence is bounded and monotonic.
First, we can prove by induction that an > n for all n ≥ 2. Assuming that an-1 > (n-1), we can see that an = (2n+1)/(3n * an-1) > (2n+1)/(3n * (n-1)) > (n+1)/n > 1. Since an > 1, the sequence is bounded below.
Furthermore, we can show that the sequence is monotonic. By taking the ratio of consecutive terms, we have an/an-1 = (2n+1)/(2n-1). If we simplify this ratio, we get (2n+1)/(2n-1) > 1 for all n ≥ 2.
Therefore, the sequence {an} is bounded below by 1 and is also monotonically increasing. By the Monotone Convergence Theorem, the sequence is convergent.
To find the limit of the sequence, we can take the limit as n approaches infinity:
lim(n -> ∞) an = lim(n -> ∞) ((2n+1)/(3n * an-1)) = 2/3 * lim(n -> ∞) (1/an-1) = 2/3 * (1/lim(n -> ∞) (an-1)).
Since the sequence is convergent, we can substitute L for lim(n -> ∞) (an-1), giving us:
L = 2/3 * (1/L).
This equation can be rearranged to: L^2 = 2/3, which gives two possible solutions: L = sqrt(2/3) or L = -sqrt(2/3). However, since all terms in the sequence are positive, we can conclude that the limit L = sqrt(2/3).