Math Problem Statement

Überprüfe die gegebenen Ergebnisse

Überprüfe die gegebenen Ergebnisse

Solution

To verify the given results from the image, let us carefully analyze the provided mathematical problem and the accompanying work:

Problem Overview:

The task involves analyzing the convergence of a recursively defined sequence ana_n given by: a1=3,an+1=7an+8,for all nN.a_1 = 3, \quad a_{n+1} = \sqrt{7a_n + 8}, \quad \text{for all } n \in \mathbb{N}.

You are required to:

  1. Determine the candidate for the potential limit of the sequence.
  2. Prove boundedness and monotonicity, and ultimately establish whether the sequence converges.

Step-by-Step Review

  1. Candidate for the Limit (aa)

    • The potential limit aa satisfies: a=7a+8.a = \sqrt{7a + 8}. Squaring both sides: a2=7a+8    a27a8=0.a^2 = 7a + 8 \implies a^2 - 7a - 8 = 0. Factoring or solving the quadratic equation gives: a=7±49+322=7±92.a = \frac{7 \pm \sqrt{49 + 32}}{2} = \frac{7 \pm 9}{2}. Hence, the solutions are: a1=8anda2=1.a_1 = 8 \quad \text{and} \quad a_2 = -1. Since the sequence ana_n is positive, the only viable candidate is: a=8.a = 8.

    ✅ The candidate for the limit aa is correctly identified as 88.


  1. Boundedness

    • Initial analysis shows an>0a_n > 0 for all nn, so the sequence is bounded below by 00.

    • To prove the sequence is bounded above by b=8b = 8, mathematical induction is used:

      Inductive Step: Assume an8a_n \leq 8. Then: an+1=7an+878+8=64=8.a_{n+1} = \sqrt{7a_n + 8} \leq \sqrt{7 \cdot 8 + 8} = \sqrt{64} = 8. Hence, an+18a_{n+1} \leq 8, proving the sequence is bounded above.

    ✅ The boundedness proof appears correct.


  1. Monotonicity

    • To check if the sequence is monotonic, evaluate whether an+1ana_{n+1} \geq a_n (monotonically increasing): an+1an=7an+8an.a_{n+1} - a_n = \sqrt{7a_n + 8} - a_n. For an<8a_n < 8, note that 7an+8>an\sqrt{7a_n + 8} > a_n, which confirms the sequence is increasing.

    ✅ The sequence is correctly identified as monotonically increasing.


  1. Convergence

    • Since the sequence is bounded and monotonically increasing, it converges by the Monotone Convergence Theorem. The limit is the candidate a=8a = 8, as verified earlier.

    ✅ The convergence and limit are correctly determined.


If you would like, I can provide additional clarification or rewrite the solution with detailed explanations.

Related Questions

  1. What other methods can you use to check boundedness?
  2. How do you prove monotonicity in recursive sequences in general?
  3. Can this approach be generalized for other quadratic recurrence relations?
  4. What is the role of the Monotone Convergence Theorem in such problems?
  5. How would the results differ if the initial condition a1a_1 were negative?

Tip: When solving recursive sequences, always verify initial terms to ensure the sequence remains within the derived bounds.

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Math Problem Analysis

Mathematical Concepts

Recursive Sequences
Convergence
Monotonicity
Boundedness

Formulas

Recursive formula: a_{n+1} = sqrt(7a_n + 8)
Quadratic equation: a^2 = 7a + 8

Theorems

Monotone Convergence Theorem
Mathematical Induction

Suitable Grade Level

Grade 11-12