Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

Write the first four terms of the sequence \(\left\\{a_{n}\right\\}\). Then use Theorem \(D\) to show that the sequence converges. $$ a_{n}=\frac{4 n-3}{2^{n}} $$

Step-by-Step Solution

Answer

First four terms: \( \frac{1}{2}, \frac{5}{4}, \frac{9}{8}, \frac{13}{16} \). Sequence converges to 0.

Step 1: Calculate the First Term

To find the first term of the sequence, substitute \(n = 1\) into the sequence formula: \[a_1 = \frac{4(1) - 3}{2^1} = \frac{1}{2}\]The first term is \(\frac{1}{2}\).

Step 2: Calculate the Second Term

Substitute \(n = 2\) into the sequence formula:\[a_2 = \frac{4(2) - 3}{2^2} = \frac{5}{4}\]The second term is \(\frac{5}{4}\).

Step 3: Calculate the Third Term

Substitute \(n = 3\) into the sequence formula:\[a_3 = \frac{4(3) - 3}{2^3} = \frac{9}{8}\]The third term is \(\frac{9}{8}\).

Step 4: Calculate the Fourth Term

Substitute \(n = 4\) into the sequence formula:\[a_4 = \frac{4(4) - 3}{2^4} = \frac{13}{16}\]The fourth term is \(\frac{13}{16}\).

Step 5: Applying Theorem D for Convergence

Theorem \(D\) often refers to the criterion that \(\lim_{n \to \infty} a_n = L\) implies the sequence converges to \(L\). Calculate the limit of \(a_n\):Consider: \[\lim_{n \to \infty} \frac{4n - 3}{2^n}\]The numerator grows linearly, and the denominator grows exponentially, meaning the denominator grows faster. Thus,\[\lim_{n \to \infty} \frac{4n - 3}{2^n} = 0\]Therefore, according to Theorem \(D\), the sequence converges to 0.

Key Concepts

Limits of sequences

In mathematical terms, finding the limit of a sequence is akin to deciphering the eventual destiny of the sequence as it extends towards infinity. This narrative is crucial as it helps us identify whether a sequence converges to a specific value, known as the limit. For our sequence \(a_n = \frac{4n - 3}{2^n}\), the critical task is to determine its behavior as \(n\) approaches infinity.
When we observe the sequence, it becomes evident that the integer denominator \(2^n\) grows exponentially. Meanwhile, the numerator \(4n - 3\) expands linearly at a slower pace. Hence, as \(n\) becomes sufficiently large, the denominator vastly overpowers the numerator, pushing the fractions infinitesimally closer to zero. Therefore, the limit of \(a_n\) as \(n\) approaches infinity is \(0\). This observation underpins the sequence's convergence to this limit.

Sequence terms

Before delving into the convergence aspect, understanding and calculating the terms of a sequence is like laying down foundational bricks. Each term of the sequence \(a_n = \frac{4n - 3}{2^n}\) quantifies a specific element in the sequence's progression. Let's break it down for the initial terms:

• The first term \(a_1 = \frac{1}{2}\)
• The second term \(a_2 = \frac{5}{4}\)
• The third term \(a_3 = \frac{9}{8}\)
• The fourth term \(a_4 = \frac{13}{16}\)

These calculations reveal the sequence's nature, reflecting how each fraction hones in on a specific value with increasing \(n\). Notably, while there is an initial increase in magnitude, the presence of the exponentially growing denominator suggests that terms will shrink over time, affirming the sequence's eventual convergence to a limit.

Exponential growth

Exponential growth may sound daunting, but it's merely a description of rapid escalation as demonstrated by the powers of 2 in this sequence. Consider the \(2^n\) in the denominator of \(a_n = \frac{4n - 3}{2^n}\). As \(n\) increases, the value of \(2^n\) skyrockets, dramatically overshadowing the linearly expanding numerator \(4n - 3\).
This relationship is essential for understanding the convergence of sequences with exponential components. The rapidly expanding denominators quash the overall value of each term towards zero as \(n\) grows, which in turn indicates the sequence's convergence to the limit zero despite the linear growth of the numerator. It’s this overpowering exponential growth of the denominator that is the critical characteristic pushing the sequence toward its limit.

Theorem D

Theorem D is a linchpin in the study of sequence convergence, and it asserts a simple yet profound concept: if the sequence \(a_n\) approaches a single value \(L\) as \(n\) becomes infinitely large, then \(a_n\) converges to \(L\). In other words, the distance between the sequence's terms and the value \(L\) diminishes with the increase of \(n\).
For the sequence \(a_n = \frac{4n - 3}{2^n}\), we've shown that as \(n\) tensely approaches infinity, \(\lim_{n \to \infty} a_n = 0\). This result, according to Theorem D, confirms that the sequence converges, settling at a zero limit.
The remarkable power of Theorem D lies in providing a mechanism to test and affirm sequences' convergence, offering a decisive conclusion through the observation of trends as \(n\) stretches toward infinity.