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Chapter 9: Problem 80

Which of the sequences \(\left\\{a_{n}\right\\}\) converge, and which diverge? Find the limit of each convergent sequence. $$a_{n}=\left(3^{n}+5^{n}\right)^{1 / n}$$

Short Answer

Expert verified
The sequence \(a_n = (3^n + 5^n)^{1/n}\) converges to 5.

Step by step solution

01

Understand the Form of the Sequence

The sequence given is \(a_n = (3^n + 5^n)^{1/n}\). It is in the form of \((b_n + c_n)^{1/n}\). We'll analyze how each term behaves as \(n\) approaches infinity.
02

Dominant Term Analysis

In the expression \(3^n + 5^n\), notice that as \(n\) becomes very large, \(5^n\) grows much faster than \(3^n\). This makes \(5^n\) the dominant term. Thus, \(3^n + 5^n \approx 5^n\) as \(n\) approaches infinity.
03

Simplify Dominant Term Expression

Using the dominant term approximation, we write:\[ a_n = (3^n + 5^n)^{1/n} \approx (5^n)^{1/n} = 5^{n/n} = 5. \]
04

Conclusion About Convergence

Since \((3^n + 5^n)^{1/n} \approx 5\), the sequence \(a_n\) converges to 5. This is due to the fact that as \(n\rightarrow\infty\), the influence of \(3^n\) diminishes relative to \(5^n\).
05

Verification of the Limit

To further verify, consider the limit:\[\lim_{{n \to \infty}} (3^n + 5^n)^{1/n}\approx \lim_{{n \to \infty}} (5^n)^{1/n} = 5,\]confirming our analysis that the sequence \(a_n\) converges to 5.

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Key Concepts

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

Limit of a Sequence
The limit of a sequence is a value that the terms of the sequence approach as the index (often denoted as \( n \)) goes to infinity. In simpler terms, it is the number that the terms get closer to as the sequence progresses. Observing a sequence's behavior as \( n \) grows larger helps us determine whether it converges (reaches a specific value) or diverges (does not settle on any one value).

To find the limit of a sequence like \( a_n = (3^n + 5^n)^{1/n} \), we study how the terms change with increasing \( n \). Through step-by-step analysis, if we find a finite number that represents the terms' destination, this number is the sequence's limit.
  • If the terms reach a specific number, the sequence converges to that number, which is its limit.
  • If the terms do not settle to a specific point, then the sequence diverges.
Dominant Term Analysis
Dominant term analysis is a technique used in mathematics to simplify expressions, especially when analyzing limits of sequences or functions. When dealing with expressions that involve a sum or a combination of terms, like \( 3^n + 5^n \), not all terms contribute equally as \( n \) increases.

In this specific sequence, \( 5^n \) is said to "dominate" because it increases much faster than \( 3^n \) as \( n \) becomes very large. This means that for large \( n \), \( 5^n + 3^n \) is approximately equal to \( 5^n \).
  • This simplifies the analysis because we can approximate the sequence \( a_n = (3^n + 5^n)^{1/n} \) as \( (5^n)^{1/n} \) when \( n \) is large.
  • By focusing on the dominant term, complicated expressions become easier to handle, allowing us to assess convergence/divergence quickly.
Sequence Divergence
Sometimes sequences do not approach any single number as their terms progress. These sequences are termed as 'divergent'. A sequence that diverges does not have a limit, which means the terms continue to grow without stabilizing to a particular value.

However, for our sequence \( a_n = (3^n + 5^n)^{1/n} \), the dominant term analysis showed that it converges to 5, rather than diverging. This is because the influence of \( 3^n \) becomes negligible compared to \( 5^n \), thus driving the sequence toward a definite limit.
  • To determine divergence, one needs to show that no single point can be approached indefinitely by the sequence鈥檚 terms.
  • In situations where divergence is suspected, checking by comparing it with simple diverging sequences (like linear or exponential growth) can be insightful.

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Most popular questions from this chapter

For approximately what values of \(x\) can you replace \(\sin x\) by \(x-\left(x^{3} / 6\right)\) with an error of magnitude no greater than \(5 \times 10^{-4} ?\) Give reasons for your answer.

Use the definition of convergence to prove the given limit. $$\lim _{n \rightarrow \infty} \frac{\sin n}{n}=0$$

Prove that a sequence \(\left\\{a_{n}\right\\}\) converges to 0 if and only if the sequence of absolute values \(\left\\{\left|a_{n}\right|\right\\}\) converges to 0.

Use the identity \(\sin ^{2} x=(1-\cos 2 x) / 2\) to obtain the Maclaurin series for \(\sin ^{2} x .\) Then differentiate this series to obtain the Maclaurin series for \(2 \sin x \cos x .\) Check that this is the series for \(\sin 2 x\).

The zipper theorem \(\quad\) Prove the "zipper theorem" for sequences: If \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}\) both converge to \(L\), then the sequence $$a_{1}, b_{1}, a_{2}, b_{2}, \ldots, a_{n}, b_{n}, \ldots$$ converges to \(L\).
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