Question

$x_1 = 1$ and $x_{n+1} = \frac{1}{4}(x_n + 5)$, $n \ge 1$. (a) Prove $1 \le x_n \le \frac{5}{3}$ for all $n \ge 1$ (by induction). (b) Prove $x_n \le x_{n+1}$ for all $n \ge 1$ (using (a)). (c) Use the Monotone Convergence Theorem to say {$x_n$} has a limit, then find the limit.

          $x_1 = 1$ and $x_{n+1} = \frac{1}{4}(x_n + 5)$, $n \ge 1$.
(a) Prove $1 \le x_n \le \frac{5}{3}$ for all $n \ge 1$ (by induction).
(b) Prove $x_n \le x_{n+1}$ for all $n \ge 1$ (using (a)).
(c) Use the Monotone Convergence Theorem to say {$x_n$} has a limit, then find the limit.

x1 = 1 and xn+1 = (1)/(4)(xn + 5), n ≥ 1.
(a) Prove 1 ≤ xn ≤(5)/(3) for all n ≥ 1 (by induction).
(b) Prove xn ≤ xn+1 for all n ≥ 1 (using (a)).
(c) Use the Monotone Convergence Theorem to say xn has a limit, then find the limit.

Added by Jaime J.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert David Mccaslin

Step 1

For n = 1, we have T1 = x1 = 1. Since 1 is between 1 and 5, the base case holds. Now, let's assume that 1 ≤ Tn ≤ 5 for some arbitrary value of n ≥ 1. We will prove that this implies 1 ≤ Tn+1 ≤ 5. Using the recursive formula, we have Tn+1 = √(n+5). Since n ≥ 1, Show more…

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x1 = 1 and xn+1 = âˆš(n+5), n > 1 (a) Prove 1 â‰¤ Tn â‰¤ 5 for all n â‰¥ 1 (by induction). (b) Prove xn < xn+1 for all n > 1 (using (a)). (c) Use the Monotone Convergence Theorem to show that {xn} has a limit, then find the limit.