Chapter 11: Problem 63
For each positive integer \(n\), let \(a_{n}\) be the largest integral multiple of \(1 / 10^{n}\) such that \(a_{n}^{2} \leqq 2 .\) (a) Prove that \(\left\\{a_{n}\right\\}\) is a bounded increasing sequence, so \(A=\lim a_{n}\) exists. (b) Prove that if \(A^{2}>2\), then \(a_{n}^{2}>2\) for \(n \rightarrow x\) sufficiently large. (c) Prove that if \(A^{2}<2\), then \(a_{n}^{2}
Short Answer
Step by step solution
Understanding Sequence Definition
Prove \( \{a_n\} \) is Increasing
Show \( \{a_n\} \) is Bounded
Prove: If \( A^2 > 2 \), then \( a_n^2 > 2 \) for large \( n \)
Prove: If \( A^2 < 2 \), then \( a_n^2 < B < 2 \) for large \( n \)
Conclusion: Show \( A^2 = 2 \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Bounded Increasing Sequence
Moreover, the sequence is bounded because each \( a_n \) is always less than or equal to \( \sqrt{2} \). This constraint ensures that \( \{a_n\} \) does not exceed \( \sqrt{2} \).
- Increasing: \( a_n \leq a_{n+1} \)
- Bounded: \( a_n \leq \sqrt{2} \)
Sequence Convergence
As each \( a_n \) becomes larger, it moves incrementally closer to \( \sqrt{2} \) while adhering to the condition \( a_n^2 \leq 2 \). Therefore, by the definition of convergence, \( \lim\limits_{n \to \infty} a_n = A \).
- Every increasing and bounded sequence eventually converges to a limit.
- Our sequence meets these conditions and hence converges to \( A \).
Limit of a Sequence
If \( A^2 > 2 \), then eventually \( a_n^2 \) would also be greater than 2, but this contradicts \( a_n^2 \leq 2 \). Similarly, if \( A^2 < 2 \), \( a_n \) would stabilize below a value that cannot touch \( \sqrt{2} \), again leading to inconsistencies given the initial conditions.
- The limit is the value approached as the sequence extends indefinitely.
- The conditions of boundedness and increasing lead to \( A^2 = 2 \).
Mathematical Proofs
Multiple cases are presented using proofs:
- **If \( A^2 > 2 \):** We assume that \( A^2 > 2 \), implying \( a_n^2 \) would also breach this boundary. However, since \( a_n^2 \leq 2 \), this leads to a contradiction.
- **If \( A^2 < 2 \):** Here, the sequence supposedly dips below an upper bound \( B \) where \( B < 2 \). As \( a_n \) approaches \( A \), it should stabilize, which opposes the fact that \( a_n^2 \leq 2 \).