Chapter 10: Problem 74
Which of the sequences \(\left\\{a_{n}\right\\}\) in Exercises \(27-90\) converge, and which diverge? Find the limit of each convergent sequence. $$ a_{n}=\frac{(10 / 11)^{n}}{(9 / 10)^{n}+(11 / 12)^{n}} $$
Short Answer
Expert verified
The sequence converges and its limit is 0.
Step by step solution
01
Express the Sequence Terms
The sequence is given by \( a_n = \frac{(10/11)^n}{(9/10)^n + (11/12)^n} \). We are tasked with finding whether this sequence converges or diverges and determining the limit if it converges.
02
Simplify the Denominator
The denominator \( (9/10)^n + (11/12)^n \) consists of exponential terms. As \( n \to \infty \), compare each term's growth rate. Observe that \( (11/12) > (9/10) \), thus \( (11/12)^n \) dominates as \( n \to \infty \).
03
Analyze the Limit of the Sequence
Focus on simplifying \( a_n \) as \( n \to \infty \):\[a_n = \frac{(10/11)^n}{(9/10)^n + (11/12)^n} \approx \frac{(10/11)^n}{(11/12)^n}\]For large \( n \), \((9/10)^n\) becomes negligible.
04
Simplify Using Exponents
Simplify the expression:\[a_n \approx \left( \frac{10}{11} \right)^n \times \left( \frac{12}{11} \right)^n = \left( \frac{120}{121} \right)^n\]Notice that \( \frac{120}{121} < 1 \).
05
Conclude the Convergence and Find the Limit
Since \( \left( \frac{120}{121} \right)^n < 1 \) and converges to 0 as \( n \to \infty \), the sequence \( a_n \) converges. The limit of the sequence is 0.
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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 fundamental concept in calculus and real analysis. It refers to the value that the terms of a sequence approach as the sequence progresses to infinity. In simpler terms, as the index number of a sequence grows larger, the sequence will get closer and closer to its limit, if such a limit exists.
For the sequence given by \( a_n = \frac{(10/11)^n}{(9/10)^n + (11/12)^n} \), we aim to find whether it converges. Convergence means the sequence approaches a finite number. By simplifying and analyzing the terms, it is established that \((11/12)^n\) dominates, and thus allows us to simplify the sequence further.
By observing that as \( n \to \infty \), the dominant term \((11/12)^n\) makes \((9/10)^n\) negligible, the sequence simplifies to \(\left(\frac{120}{121}\right)^n\), indicating that the sequence converges to 0 because \(\left(\frac{120}{121}\right) < 1\). Hence, as \( n \to \infty \), \( a_n \to 0 \).
For the sequence given by \( a_n = \frac{(10/11)^n}{(9/10)^n + (11/12)^n} \), we aim to find whether it converges. Convergence means the sequence approaches a finite number. By simplifying and analyzing the terms, it is established that \((11/12)^n\) dominates, and thus allows us to simplify the sequence further.
By observing that as \( n \to \infty \), the dominant term \((11/12)^n\) makes \((9/10)^n\) negligible, the sequence simplifies to \(\left(\frac{120}{121}\right)^n\), indicating that the sequence converges to 0 because \(\left(\frac{120}{121}\right) < 1\). Hence, as \( n \to \infty \), \( a_n \to 0 \).
Exponential Functions
Exponential functions play an important role in many mathematical contexts, including sequence convergence. An exponential function is typically in the form \( b^x \) where \( b \) is a constant base, and \( x \) is the exponent. When used in sequences, exponential terms determine the growth or decay of the sequence.
In this exercise, both the numerator \((10/11)^n\) and the two terms in the denominator \((9/10)^n + (11/12)^n\) are exponential functions. What makes them interesting is their bases:
In this exercise, both the numerator \((10/11)^n\) and the two terms in the denominator \((9/10)^n + (11/12)^n\) are exponential functions. What makes them interesting is their bases:
- \((10/11) < 1\), indicating decay as \( n \) increases.
- \((9/10) < (11/12)\), showing comparative rates of decay with \( (11/12) \) decaying slower.
Dominance of Terms
Determining the dominance of terms within a sequence is crucial for understanding its behavior for large \( n \). When dealing with a sequence that has multiple terms in its expression, especially exponential ones, we look at which term grows or shrinks the slowest.
In the given sequence, comparing \((9/10)^n\) and \((11/12)^n\), we observe that \((11/12)^n\) is the dominant term because it shrinks slower than \((9/10)^n\). Thus, as \( n \) becomes larger, the influence of \((9/10)^n\) diminishes, allowing \((11/12)^n\) to effectively dictate the behavior of the denominator.
By focusing on dominant terms, we can simplify sequences significantly. In our sequence \( a_n \), the dominance of \((11/12)^n\) is why we approximated the denominator by \( (11/12)^n \), leading to an easier computation of \( a_n \) approaching 0 as \( n \to \infty \). This analysis of dominance is a powerful tool for confirming convergence.
In the given sequence, comparing \((9/10)^n\) and \((11/12)^n\), we observe that \((11/12)^n\) is the dominant term because it shrinks slower than \((9/10)^n\). Thus, as \( n \) becomes larger, the influence of \((9/10)^n\) diminishes, allowing \((11/12)^n\) to effectively dictate the behavior of the denominator.
By focusing on dominant terms, we can simplify sequences significantly. In our sequence \( a_n \), the dominance of \((11/12)^n\) is why we approximated the denominator by \( (11/12)^n \), leading to an easier computation of \( a_n \) approaching 0 as \( n \to \infty \). This analysis of dominance is a powerful tool for confirming convergence.