/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 123 Which of the sequences converge,... [FREE SOLUTION] | 91Ó°ÊÓ

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Which of the sequences converge, and which diverge? Give reasons for your answers. \(a_{n}=\frac{4^{n+1}+3^{n}}{4^{n}}\)

Short Answer

Expert verified
The sequence converges to 4.

Step by step solution

01

Simplify the Sequence Expression

To begin, simplify the sequence expression: \(a_n = \frac{4^{n+1} + 3^n}{4^n}\). First, notice that \(4^{n+1} = 4 \cdot 4^n\). Substitute to get:\[ a_n = \frac{4 \cdot 4^n + 3^n}{4^n} \].
02

Separate the Fraction

Now, separate the fraction into two parts:\[ a_n = \frac{4 \cdot 4^n}{4^n} + \frac{3^n}{4^n} \].This simplifies to:\[ a_n = 4 + \left(\frac{3}{4}\right)^n \].
03

Analyze Each Term

Let's examine the behavior of each term as \( n \to \infty \):1. The first term is 4, which remains constant as \( n \) increases.2. For the second term, \( \left(\frac{3}{4}\right)^n \), note that \(\frac{3}{4} < 1\). As \(n \to \infty\), \( \left(\frac{3}{4}\right)^n \to 0\) because raising a number less than 1 to an increasingly large power makes it approach 0.
04

Determine Convergence

Since the first term is constant (4), and the second term approaches 0, the combined sequence \( a_n = 4 + \left(\frac{3}{4}\right)^n \) approaches 4 as \( n \to \infty \). Thus, the sequence converges to 4.

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Key Concepts

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

Convergent Sequences
A convergent sequence is a sequence of numbers that approaches a specific value, called the limit, as the index number becomes infinitely large. Identifying whether a sequence converges can sometimes be a tricky task, but there are a few key indicators to help determine convergence.

- A sequence converges if, for any small distance, there is a point beyond which all subsequent terms of the sequence are within that distance from the limit.
- In simple terms, the further along you go in the sequence, the closer the terms get to a certain number.

Let's consider the sequence from the original problem: \[a_n = \frac{4^{n+1} + 3^n}{4^n}.\]
By simplifying, we recognized the pattern:
\[a_n = 4 + \left(\frac{3}{4}\right)^n.\] Here, the term \(4\) remains constant, and the term \(\left(\frac{3}{4}\right)^n\) becomes smaller and smaller, getting closer to zero as \(n\) increases.

Thus, the sequence approaches the value 4. This is the hallmark of a convergent sequence.
Limit of a Sequence
The limit of a sequence is the value that the terms of a sequence approach as the index (usually denoted as \(n\)) goes to infinity. It's like predicting the behavior of the sequence far off into the future.

For our sequence:
\[a_n = 4 + \left(\frac{3}{4}\right)^n,\]
we found that as \(n\) becomes very large, \(\left(\frac{3}{4}\right)^n\) tends towards zero. This is because any power of a number less than 1 will diminish toward zero when the exponent becomes very large. Therefore, the limit of \(a_n\) as \(n\) approaches infinity is 4.
  • This tells us that the sequence will eventually get arbitrarily close to 4, no matter how small a neighborhood around 4 you choose.
  • In mathematical notation, this is written as:\[\lim_{{n \to \infty}} a_n = 4.\]
Understanding the limit helps us predict long-term behavior of sequences without calculating each term individually.
Algebraic Manipulation in Sequences
Algebraic manipulation involves simplifying or transforming sequences to make them easier to analyze or understand. It's especially useful in determining whether a sequence converges and finding its limit.

Let's revisit the original sequence expression: \[a_n = \frac{4^{n+1} + 3^n}{4^n}.\]
Algebraic steps helped us rewrite the expression:
  • First, recognize that \( 4^{n+1} = 4 \cdot 4^n \) to enable factorization.
  • Separate the expression: \[a_n = \frac{4 \cdot 4^n}{4^n} + \frac{3^n}{4^n} = 4 + \left(\frac{3}{4}\right)^n.\]
Through algebraic simplification, we separated terms in such a way that made convergence more obvious.
This demonstrates how transforming expressions can provide clarity on the behavior of sequences, aiding in the analysis of convergence and the determination of the sequence's limit.

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Most popular questions from this chapter

To construct this set, we begin with the closed interval \([0,1] .\) From that interval, remove the middle open interval (1/3, 2/3). leaving the two closed intervals [ 0, 1/3 ] and [2/3, 1 ]. At the second step we remove the open middle third interval from each of those remaining. From \([0.1 / 3]\) we remove the open interval \((1 / 9,2 / 9),\) and from \([2 / 3,1]\) we remove \((7 / 9,8 / 9),\) leaving behind the four closed intervals \([0,1 / 9]\) \([2 / 9,1 / 3],[2 / 3,7 / 9],\) and \([8 / 9,1] .\) At the next step, we remove the middle open third interval from each closed interval left behind, so \((1 / 27,2 / 27)\) is removed from \([0,1 / 9],\) leaving the closed intervals \([0,1 / 27]\) and \([2 / 27,1 / 9] ;(7 / 27,8 / 27)\) is removed from \([2 / 9,1 / 3]\), leaving behind \([2 / 9,7 / 27]\) and \([8 / 27,1 / 3],\) and so forth. We continue this process repeatedly without stopping, at each step removing the open third interval from every closed interval remaining behind from the preceding step. The numbers remaining in the interval \([0,1],\) after all open middle third intervals have been removed, are the points in the Cantor set (named after Georg Cantor, \(1845-1918\) ). The set has some interesting properties. a. The Cantor set contains infinitely many numbers in [0,1] List 12 numbers that belong to the Cantor set. b. Show, by summing an appropriate geometric series, that the total length of all the open middle third intervals that have been removed from [0,1] is equal to 1

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