Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find an explicit formula a \(a_{n}=\) ____ for each sequence, determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\). $$ \sin 1,2 \sin \frac{1}{2}, 3 \sin \frac{1}{3}, 4 \sin \frac{1}{4}, \ldots $$

Step-by-Step Solution

Answer

The sequence converges, and the limit is 1.

Step 1: Identify the Pattern in the Sequence

The sequence is given as \( \sin 1, 2 \sin \frac{1}{2}, 3 \sin \frac{1}{3}, 4 \sin \frac{1}{4}, \ldots \). By looking at the general term, we can see that each term can be represented by the formula \( a_n = n \sin \frac{1}{n} \).

Step 2: Develop the General Term

From the pattern observed, we can express the general term of the sequence as \( a_n = n \sin \frac{1}{n} \).

Step 3: Determine Convergence or Divergence

To determine whether the sequence converges or diverges, we need to examine the limit of the sequence as \( n \to \infty \). We evaluate \( \lim_{n \to \infty} n \sin \frac{1}{n} \).

Step 4: Evaluate the Limit

As \( n \to \infty \), \( \frac{1}{n} \to 0 \), and since \( \sin x \approx x \) for small \( x \), \( \sin \frac{1}{n} \approx \frac{1}{n} \). Therefore, \( n \sin \frac{1}{n} \approx n \cdot \frac{1}{n} = 1 \). Hence, \( \lim_{n \to \infty} n \sin \frac{1}{n} = 1 \).

Step 5: Conclusion

The sequence converges, and the limit of the sequence \( \lim_{n \to \infty} a_n = 1 \).

Key Concepts

Explicit Formula

An explicit formula provides a direct way to find the nth term of a sequence without needing to calculate all previous terms. It is akin to having a shortcut for revealing the sequence's pattern. In the given exercise, observe how each term in the sequence follows a pattern of the form: \( \sin 1, 2 \sin \frac{1}{2}, 3 \sin \frac{1}{3}, 4 \sin \frac{1}{4}, \ldots \). You may notice a recurring structure where each term can be represented as \( a_n = n \sin \frac{1}{n} \).

• Here, \( n \) is the sequence index, and
• \( \sin \frac{1}{n} \) is a sine function applied to the reciprocal of the index.

This explicit formula allows us to easily calculate any term in the sequence efficiently without recalculation of prior terms. Understanding this general formula is essential for further analysis, like determining convergence and limits.

Convergence and Divergence

Convergence and divergence are crucial concepts in understanding the long-term behavior of sequences. Simply put, a sequence converges if it approaches a particular value as its index, \( n \), grows to infinity. If it doesn't approach a specific value, it is said to diverge.

In this exercise, you want to determine whether the sequence \( a_n = n \sin \frac{1}{n} \) converges. As \( n \) goes to infinity, we analyze the formula to see the general direction of the sequence.

• When \( n \) increases, \( \frac{1}{n} \to 0 \), causing \( \sin \frac{1}{n} \approx \frac{1}{n} \).
• Thus, the term \( a_n = n \cdot \sin \frac{1}{n} \approx n \cdot \frac{1}{n} = 1 \).

Since the sequence approaches the value of 1, it converges. Understanding convergence allows us to predict that no matter how large \( n \) gets, the terms of this sequence will not sprawl wildly but rather settle towards one value.

Limit Evaluation

Calculating limits is fundamental in evaluating the end behavior of sequences. For this sequence, \( a_n = n \sin \frac{1}{n} \), to find \( \lim_{n \to \infty} a_n \), we consider the asymptotic behavior of the formula.

As \( n \to \infty \):

• \( \frac{1}{n} \to 0 \), leading to \( \sin \frac{1}{n} \to \frac{1}{n} \) for small values, based on the approximation \( \sin x \approx x \) when \( x \) is near zero.
• This gives us \( a_n = n \sin \frac{1}{n} \approx n \cdot \frac{1}{n} = 1 \).

Therefore, the limit is \( \lim_{n \to \infty} a_n = 1 \). Such evaluations confirm the sequence converges to a finite limit, enhancing our understanding of its long-term outcome. Mastering limit evaluation empowers students to analyze sequences efficiently.