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Chapter 9: Problem 18

Do the sequences, converge or diverge? If a sequence converges, find its limit. $$\frac{2^{n}}{3^{n}}$$

Short Answer

Expert verified
The sequence converges, and its limit is 0.

Step by step solution

01

Understand the Sequence

The given sequence is \( \left( \frac{2^{n}}{3^{n}} \right) \). We need to determine whether this sequence converges or diverges. If it converges, we will find its limit.
02

Simplify the Sequence

Observe that the term \( \frac{2^{n}}{3^{n}} \) can be rewritten as \( \left( \frac{2}{3} \right)^{n} \). This step helps us understand the structure of the sequence.
03

Determine Limit Behavior

Since \( \left( \frac{2}{3} \right) \) is a positive fraction less than one, as \( n \) becomes very large, \( \left( \frac{2}{3} \right)^{n} \) approaches zero. This is because repeatedly multiplying by a number smaller than one results in smaller and smaller values.
04

Conclusion on Convergence

We conclude that the sequence \( \left( \frac{2^{n}}{3^{n}} \right) = \left( \frac{2}{3} \right)^{n} \) converges. The limit of the sequence as \( n \to \infty \) is 0.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

The Limit of a Sequence
When working with sequences, one of the main questions is whether a sequence reaches a specific value as it progresses. This specific value is known as the "limit" of the sequence. For any sequence \(a_n\), if there exists a real number \(L\) such that as \(n\) increases towards infinity, \(a_n\) gets arbitrarily close to \(L\), then \(L\) is called the limit of the sequence.
  • If \(a_n\) actually approaches \(L\) in this way, we write \( \lim_{{n \to \infty}} a_n = L \).
  • In simple terms, the terms of the sequence "settle down" at \(L\).
In the given sequence \(\frac{2^n}{3^n}\), also expressed as \(\left( \frac{2}{3} \right)^n\), calculations show that as \(n\) increases, the terms approach zero. Therefore, the limit of this sequence is 0.
Exponential Sequences
Exponential sequences are sequences where each term can be expressed as a constant base raised to an increasing power, typically written as \(a^n\). They exhibit behavior that drastically depends on the base:
  • If the base \(a\) is greater than 1, the sequence grows rapidly without bound.
  • If the base \(a\) is exactly 1, the sequence remains constant at 1 for all terms.
  • If the base \(a\) is between 0 and 1, the sequence shrinks and heads towards zero as \(n\) increases.
  • If the base \(a\) is negative, the sequence will oscillate, making its behavior more complex.
For our sequence \(\left( \frac{2}{3} \right)^n\), the base \(\frac{2}{3}\) is between 0 and 1, hence the sequence tends towards zero as \(n\) becomes very large.
Convergence and Divergence
In the study of sequences, determining whether a sequence converges or diverges is crucial. Convergence means that the terms of the sequence are getting closer together around a certain value, while divergence means they are not settling at any particular value.
To determine if a sequence converges or diverges:
  • Examine if the sequence's terms approach a fixed number as the sequence progresses. If they do, it converges.
  • If the terms keep increasing or decreasing without approaching a specific number, or if they oscillate indefinitely, the sequence diverges.
In practice, look at the terms' behaviors as \(n\) grows very large. For \(\left( \frac{2}{3} \right)^n\), because the sequence approaches zero, we say it converges to 0. Such clarity helps in easily distinguishing converging sequences from those that diverge.

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Most popular questions from this chapter

Decide if the statements are true or false. Give an explanation for your answer. You can tell if a sequence converges by looking at the first 1000 terms.

Determine whether the series converges. $$\sum_{n=1}^{\infty} \frac{n 2^{n}}{3^{n}}$$

What values of \(a\) does the series converge? $$\sum_{n=1}^{\infty}(-1)^{n} \arctan \left(\frac{a}{n}\right), a>0$$

Decide if the statements are true or false. Give an explanation for your answer. If a monotone sequence of positive terms does not converge, then it has a term greater than a million.

Which of the sequences I-IV is monotone and bounded for \(n \geq 1 ?\) I. \(s_{n}=10-\frac{1}{n}\) II. \(s_{n}=\frac{10 n+1}{n}\) III. \(s_{n}=\cos n\) IV. \(s_{n}=\ln n\) (a) 1 (b) I and II (c) II and IV (d) I, II, and III
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