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Chapter 10: Problem 90

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 converges, with a limit of 5.

Step by step solution

01

Understanding the Sequence

The sequence given is \( a_n = \left(3^n + 5^n\right)^{1/n} \). We need to analyze this sequence to determine if it converges or diverges and, if it converges, find the limit.
02

Simplifying the Expression

For large \( n \), \( 5^n \) dominates \( 3^n \) because 5 is larger than 3. Hence, we can approximate \( 3^n + 5^n \approx 5^n \).
03

Applying the Power Law

Using the approximation, we have \( a_n \approx \left(5^n\right)^{1/n} \). This further simplifies to \( a_n \approx 5^{n/n} = 5 \).
04

Analyzing Convergence

As \( n \to \infty \), the effect of \( 3^n \) becomes negligible compared to \( 5^n \), thus \( a_n \to 5 \). This implies the sequence \( a_n \) converges to 5.
05

Confirming the Limit

Since \( a_n = \left(3^n + 5^n\right)^{1/n} \approx 5 \) as \( n \) grows, the sequence does converge, and its limit is 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
When we talk about the 'limit of a sequence,' we're looking at understanding what value a sequence approaches as we continue to generate more terms. In mathematical terms, the limit of a sequence \( \{a_n\} \), as \( n \) approaches infinity, is a single value that the terms \( a_n \) get closer to the greater \( n \) becomes. In other words, if we keep adding more and more terms, they should all start clustering around a specific number. For our particular sequence \( a_n = (3^n + 5^n)^{1/n} \), through our analysis, we determined that the sequence converges, meaning it does indeed approach a specific value, which in this case is 5.
Dominance of Terms
Within a sequence, the 'dominance of terms' refers to identifying which parts of the expression grow faster as \( n \) becomes large. This step is crucial to simplifying and understanding complex sequences. For our sequence, we need to compare \( 3^n \) and \( 5^n \). Since 5 is greater than 3, \( 5^n \) grows much faster than \( 3^n \) as \( n \) increases. This means that for large values of \( n \), the term \( 3^n \) becomes insignificant next to \( 5^n \). This dominant term essentially dictates the behavior of the entire expression.
Sequence Simplification
Once we know which term dominates, we can perform 'sequence simplification' to make our problem more manageable. In our sequence, we see that for large \( n \), \( 3^n + 5^n \approx 5^n \). Simplifying the given sequence \( a_n = (3^n + 5^n)^{1/n} \) to \( a_n \approx (5^n)^{1/n} \) reduces the complexity. This simplification reveals the sequence's limit more directly by removing lesser influential components, making it easier to see the sequence's convergence and determine the limit.
Power Law in Sequences
A useful tool in the analysis of sequences is the 'power law,' which aids in understanding the simplification of sequence expressions. The power law states that if you raise a power to another power, you multiply the exponents. In our example, after simplification, we use the power law: \( a_n \approx 5^{n/n} = 5^1 = 5 \). This application of the power law shows that not only is the sequence simplified, but also provides a clear pathway to realizing that the limit of our sequence is 5. It highlights the beauty of mathematical laws in making complex expressions much simpler to handle.

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Most popular questions from this chapter

Prove that $$ \lim _{n \rightarrow \infty} \sqrt[n]{n}=1 $$

Uniqueness of convergent power series $$ \begin{array}{l}{\text { a. Show that if two power series } \sum_{n=0}^{\infty} a_{n} x^{n} \text { and } \sum_{n=0}^{\infty} b_{n} x^{n}} \\\ {\text { are convergent and equal for all values of } x \text { in an open }} \\ {\text { interval }(-c, c), \text { then } a_{n}=b_{n} \text { for every } n . \text { Hint: Let }} \\ {f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}=\sum_{n=0}^{\infty} b_{n} x^{n} . \text { Differentiate term by term to }} \\ {\text { show that } a_{n} \text { and } b_{n} \text { both equal } f^{(n)}(0) /(n !) . )}\\\\{\text { b. Show that if } \sum_{n=0}^{\infty} a_{n} x^{n}=0 \text { for all } x \text { in an open interval }} \\ {(-c, c), \text { then } a_{n}=0 \text { for every } n .}\end{array} $$

Prove that if \(\left\\{a_{n}\right\\}\) is a convergent sequence, then to every positive number \(\varepsilon\) there corresponds an integer \(N\) such that $$ \left|a_{m}-a_{n}\right|<\varepsilon \quad \text { whenever } \quad m>N \quad \text { and } \quad n>N $$

How close is the approximation \(\sin x=x\) when \(|x|<10^{-3} ?\) For which of these values of \(x\) is \(x<\sin x ?\)

Use a CAS to perform the following steps for the sequences. a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L\) ? b. If the sequence converges, find an integer \(N\) such that \(\quad\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$ a_{n}=(0.9999)^{n} $$
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