Chapter 2: Problem 23
Let \(\left(a_{n}\right)\) be a sequence with the property that, for all \(n \geq
2\),
$$
\left|a_{n+1}-a_{n}\right|
Short Answer
Expert verified
The sequence \((a_n)\) is Cauchy for \(0 < k < 1\); however, for \(k=1\), a counterexample like \((a_n) = 1, 2, 1, 2, \ldots\) exists.
Step by step solution
01
Understanding the Problem
We are given a sequence \((a_n)\) with the property that for all \(n \geq 2\), \(|a_{n+1} - a_n| < k|a_n - a_{n-1}|\), where \(0 < k < 1\). We need to show that \((a_n)\) is a Cauchy sequence and also provide an example when \(k = 1\) to show that this condition alone is insufficient for Cauchy convergence.
02
Defining a Cauchy Sequence
A sequence \((a_n)\) is Cauchy if for every \(\varepsilon >0\), there is an integer \(N\) such that for all \(m, n \geq N\), \(|a_m - a_n| < \varepsilon\). Our goal is to show that with the given condition, such an \(N\) can always be found for any \(\varepsilon > 0\).
03
Applying the Triangle Inequality Recursively
From the given inequality \(|a_{n+1} - a_n| < k|a_n - a_{n-1}|\), we aim to find a bound on \(|a_m - a_n|\). By the triangle inequality, we have \(|a_m - a_n| \leq |a_m - a_{m-1}| + |a_{m-1} - a_{m-2}| + \ldots + |a_{n+1} - a_n|\).
04
Bounding the Terms
Given \(|a_{n+1} - a_n| < k|a_n - a_{n-1}|\), let's set \(b_n = |a_n - a_{n-1}|\). We can deduce a series of inequalities: \(b_{n+1} < kb_n, b_{n+2} < k^2b_n, \ldots\), hence \(|a_{m} - a_{n}| < b_{n}(k + k^2 + \ldots + k^{m-n})\).
05
Summation of a Geometric Series
Since \(0 < k < 1\), \(k + k^2 + \ldots + k^{m-n} = \frac{k(1-k^{m-n})}{1-k} < \frac{k}{1-k}\). Thus, \(|a_m - a_n| < \frac{b_nk}{1-k}\), which is a bound that becomes arbitrarily small since \((b_n)\) decreases as \(n\) increases.
06
Verifying the Cauchy Criterion
Since \(|a_{m} - a_{n}| < \frac{b_{n}k}{1-k}\) and can be made smaller than any \(\varepsilon > 0\) for sufficiently large \(n\), by choosing \(N\) accordingly, \((a_n)\) satisfies the definition of a Cauchy sequence.
07
Counterexample for k=1
For \(k=1\), the inequality becomes \(|a_{n+1} - a_n| < |a_n - a_{n-1}|\), allowing constant terms. Consider the sequence \((a_n) = 1, 2, 1, 2, \ldots\). Though it satisfies the condition with \(k=1\), it is not Cauchy because the differences do not tend to 0.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Inequality
In mathematics, an **inequality** expresses the relative size or order of two values. In the context of this problem, we deal with the inequality \(|a_{n+1} - a_n| < k|a_n - a_{n-1}|\) for a sequence \((a_n)\) where \(0 < k < 1\). This inequality indicates that the differences between consecutive terms in the sequence are getting progressively smaller with each step.
This concept is vital because when a sequence has such diminishing differences, it moves towards meeting the Cauchy sequence condition. Essentially, the application of this inequality is a stepping stone to prove that the sequence will eventually balance its progression, such that it converges as it gets further along its terms.
This concept is vital because when a sequence has such diminishing differences, it moves towards meeting the Cauchy sequence condition. Essentially, the application of this inequality is a stepping stone to prove that the sequence will eventually balance its progression, such that it converges as it gets further along its terms.
- Thanks to this inequality, the factor \(k\) effectively contracts the difference between every consecutive pair of terms.
- As \(k\) is less than 1, this ensures that the contraction is sufficient for convergence support.
- Over multiple iterations, the effects compound, making the difference between terms increasingly small.
Convergent Sequence
A **convergent sequence** is one whereby the terms approach a specific value as they progress to infinity. More formally, a sequence \((a_n)\) converges to a limit \(L\) if, for every \(\varepsilon > 0\), there exists an integer \(N\) such that for all \(n \geq N\), \(|a_n - L| < \varepsilon\).
In this exercise's realm, we're verifying the sequence to be a **Cauchy sequence**, which is a necessary condition for convergence in the real numbers.
In this exercise's realm, we're verifying the sequence to be a **Cauchy sequence**, which is a necessary condition for convergence in the real numbers.
- A key insight is that any Cauchy sequence in real numbers is also convergent.
- The shrinking differences imposed by the inequality \(|a_{n+1} - a_n| < k|a_n - a_{n-1}|\) push the sequence seamlessly towards converging.
- As shown in the solution steps, we prove \(|a_m - a_n|\) can be made arbitrarily small by appropriately picking \(m\) and \(n\), reinforcing the sequence's convergence criterion.
Geometric Series
A **geometric series** is a series of terms where each term after the first is found by multiplying the previous one by a constant, called the common ratio. The structure of a geometric series is: \( S = a + ar + ar^2 + ar^3 + \ldots \)
In our scenario, the sequence inequality introduces us to a geometric progression through the diminishing differences. The recursive application \(|a_{n+1} - a_n| < k|a_n - a_{n-1}|\) feels more pronounced when we break it down:
In our scenario, the sequence inequality introduces us to a geometric progression through the diminishing differences. The recursive application \(|a_{n+1} - a_n| < k|a_n - a_{n-1}|\) feels more pronounced when we break it down:
- Every term generates from the prior term scaled by the factor \(k\).
- This forms a chain \( b_{n}, b_{n+1} < kb_n, b_{n+2} < k^2b_n, \) and so forth, reflecting the form of a geometric series.
- Summing this up generates \(k + k^2 + \ldots + k^{m-n}\), which converges provided \(0 < k < 1\).