Math Problem Statement

Let a₁ = 2, b₁ = 1 and for n ≥ 1, aₙ₊₁ = (aₙ + bₙ)/2, bₙ₊₁ = (2aₙbₙ)/(aₙ + bₙ). Show that (i) bₙ ≤ aₙ for all n, (ii) bₙ₊₁ ≥ bₙ for all n, (iii) the sequences {aₙ} and {bₙ} converge to the same limit √2.

Solution

To solve the given problem, we have sequences {an}\{a_n\} and {bn}\{b_n\} defined recursively as:

  • Initial values: a1=2,b1=1a_1 = 2, b_1 = 1
  • For n1n \geq 1: an+1=an+bn2,bn+1=2anbnan+bna_{n+1} = \frac{a_n + b_n}{2}, \quad b_{n+1} = \frac{2a_n b_n}{a_n + b_n}

We need to show three things:

  1. bnanb_n \leq a_n for all nn.
  2. bn+1bnb_{n+1} \geq b_n for all nn.
  3. Both sequences {an}\{a_n\} and {bn}\{b_n\} converge to the same limit, 2\sqrt{2}.

Proof:

(i) Show that bnanb_n \leq a_n for all nn:

We start by induction.

  • Base case: For n=1n = 1, we have a1=2a_1 = 2 and b1=1b_1 = 1. Clearly, b1a1b_1 \leq a_1.

  • Inductive step: Suppose bnanb_n \leq a_n for some nn. Now, we need to show that bn+1an+1b_{n+1} \leq a_{n+1}.

From the recursive definitions:

an+1=an+bn2,bn+1=2anbnan+bna_{n+1} = \frac{a_n + b_n}{2}, \quad b_{n+1} = \frac{2a_n b_n}{a_n + b_n}

Since bnanb_n \leq a_n, an+bn2bna_n + b_n \geq 2b_n. This implies:

an+bn2bn\frac{a_n + b_n}{2} \geq b_n

Also, for bn+1b_{n+1}:

bn+1=2anbnan+bn2anbn2bn=anbn+1an+1b_{n+1} = \frac{2a_n b_n}{a_n + b_n} \leq \frac{2a_n b_n}{2b_n} = a_n \quad \Rightarrow \quad b_{n+1} \leq a_{n+1}

Thus, by induction, bnanb_n \leq a_n for all nn.

(ii) Show that bn+1bnb_{n+1} \geq b_n for all nn:

We need to show that bn+1bn0b_{n+1} - b_n \geq 0.

bn+1=2anbnan+bnb_{n+1} = \frac{2a_n b_n}{a_n + b_n}

Consider the difference:

bn+1bn=2anbnan+bnbn=2anbnbn(an+bn)an+bnb_{n+1} - b_n = \frac{2a_n b_n}{a_n + b_n} - b_n = \frac{2a_n b_n - b_n(a_n + b_n)}{a_n + b_n} bn+1bn=2anbnanbnbn2an+bn=anbnbn2an+bnb_{n+1} - b_n = \frac{2a_n b_n - a_n b_n - b_n^2}{a_n + b_n} = \frac{a_n b_n - b_n^2}{a_n + b_n} bn+1bn=bn(anbn)an+bnb_{n+1} - b_n = \frac{b_n(a_n - b_n)}{a_n + b_n}

Since anbna_n \geq b_n, it follows that anbn0a_n - b_n \geq 0, and hence bn+1bnb_{n+1} \geq b_n.

(iii) Show that both sequences {an}\{a_n\} and {bn}\{b_n\} converge to the same limit 2\sqrt{2}:

Let LL be the common limit of {an}\{a_n\} and {bn}\{b_n\}.

  • As nn \to \infty, anan+1La_n \approx a_{n+1} \approx L and bnbn+1Lb_n \approx b_{n+1} \approx L.

From the recursive definitions:

L=L+L2=LandL=2L2L+L=2L22L=2L = \frac{L + L}{2} = L \quad \text{and} \quad L = \frac{2L^2}{L + L} = \frac{2L^2}{2L} = \sqrt{2}

Thus, both sequences converge to 2\sqrt{2}.


If you'd like a more detailed breakdown of any part or have questions, let me know!

Relative Questions:

  1. What other methods can be used to analyze the convergence of recursive sequences?
  2. Can we generalize this approach for other initial values of a1a_1 and b1b_1?
  3. How does the arithmetic mean-geometric mean inequality relate to these sequences?
  4. What happens if we modify the recurrence relation for bn+1b_{n+1}?
  5. How can this convergence be visualized graphically?

Tip:

When analyzing recursive sequences, always start by establishing bounds and then investigate monotonicity and convergence properties.

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Math Problem Analysis

Mathematical Concepts

Sequences
Inequalities
Limits
Convergence

Formulas

aₙ₊₁ = (aₙ + bₙ)/2
bₙ₊₁ = (2aₙbₙ)/(aₙ + bₙ)
Limit: √2

Theorems

Arithmetic-Geometric Mean Inequality
Monotonicity
Convergence Theorem

Suitable Grade Level

University Level (Advanced Calculus or Real Analysis)