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Chapter 9: Problem 3

Give an example of a bounded sequence that has a limit.

Short Answer

Expert verified
Question: Determine whether the following sequence is bounded, and if so, find its limit: \(a_n = \frac{1}{2^n}\) for all natural numbers \(n\). Answer: The sequence \(\left(a_n\right)\) is bounded with an upper bound of \(\frac{1}{2}\) since all of its elements are non-negative and less than or equal to \(\frac{1}{2}\). The limit of the sequence is \(0\), as the terms tend to \(0\) when the powers of \(2\) in the denominator increase.

Step by step solution

01

1. Definition of the sequence

Let's define the sequence \((a_n)\) as \(a_n = \frac{1}{2^n}\) for all natural numbers \(n\). So, the sequence is: \(\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \dots\).
02

2. Show the sequence is bounded

To show that the sequence is bounded, we need to find \(M > 0\) such that \(|a_n| \leq M\) for all \(n \in \mathbb{N}\). For our sequence, since all the elements are positive, we can simply find an upper bound. Notice that all the terms in the sequence are non-negative and less than or equal to \(\frac{1}{2}\). Therefore, we have an upper bound of \(M = \frac{1}{2}\). So, the sequence is bounded.
03

3. Determine the limit

Now, we need to find the limit of the sequence \((a_n)\). As the powers of \(2\) in the denominator increase, the fractions become smaller and smaller, tending to \(0\). Thus, the limit of the sequence is \(0\).
04

4. Prove the limit is correct

To prove that the limit of \((a_n)\) is indeed \(0\), we can use the definition of a limit. For any \(\epsilon > 0\), we need to find a natural number \(N\) such that \(n \geq N\) implies \(|a_n - 0| < \epsilon\). In our case, \(|a_n - 0| = |a_n| = \frac{1}{2^n}\). To ensure that \(\frac{1}{2^n} < \epsilon\), we can use the Archimedean property and find a natural number \(N\) such that \(\frac{1}{2^N} < \epsilon\). Then, for any \(n \geq N\), we have \(\frac{1}{2^n} \leq \frac{1}{2^N} < \epsilon\), which proves that the limit of \((a_n)\) is, indeed, \(0\).

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Most popular questions from this chapter

Determine whether the following series converge absolutely or conditionally, or diverge. $$\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k^{2 / 3}}$$

Determine whether the following series converge or diverge. $$\sum_{k=2}^{\infty} \frac{4}{k \ln ^{2} k}$$

An early limit Working in the early 1600 s, the mathematicians Wallis, Pascal, and Fermat were attempting to determine the area of the region under the curve \(y=x^{p}\) between \(x=0\) and \(x=1\) where \(p\) is a positive integer. Using arguments that predated the Fundamental Theorem of Calculus, they were able to prove that $$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n-1}\left(\frac{k}{n}\right)^{p}=\frac{1}{p+1}$$ Use what you know about Riemann sums and integrals to verify this limit.

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