Problem 74 - Thomas Calculus

Question

Which of the sequences \(\left\\{a_{n}\right\\}\) in Exercises \(27-90\) converge, and which diverge? Find the limit of each convergent sequence. $$ a_{n}=\frac{(10 / 11)^{n}}{(9 / 10)^{n}+(11 / 12)^{n}} $$

Step-by-Step Solution

Answer

The sequence converges and its limit is 0.

Step 1: 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.

Step 2: 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 \( n \to \infty \).

Step 3: 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.

Step 4: 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 \( \frac{120}{121} < 1 \).

Step 5: 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.

Key Concepts

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 \( n \to \infty \), \( a_n \to 0 \).

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:

• \((10/11) < 1\), indicating decay as \( n \) increases.
• \((9/10) < (11/12)\), showing comparative rates of decay with \( (11/12) \) decaying slower.

As \( n \) goes to infinity, terms with bases between 0 and 1 will shrink towards zero. Such behavior is key to simplifying sequences and finding their limits. Observing which term dominates further influences the simplification process.

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.