Problem 43 - 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. $$ a_{n}=\sin \left(\frac{\pi}{2}+\frac{1}{n}\right) $$

Step-by-Step Solution

Answer

The sequence converges to 1.

Step 1: Analyze the Argument of Sine Function

The sequence is given as \( a_n = \sin \left( \frac{\pi}{2} + \frac{1}{n} \right) \). Note that as \( n \to \infty \), the term \( \frac{1}{n} \to 0 \). Thus, the argument of the sine function, \( \frac{\pi}{2} + \frac{1}{n} \), approaches \( \frac{\pi}{2} \).

Step 2: Evaluate Limit of Sine Function

Since \( \sin(\theta) \) is continuous, we can express the limit of the sequence as:\[ \lim_{{n \to \infty}} \sin \left( \frac{\pi}{2} + \frac{1}{n} \right) = \sin \left( \lim_{{n \to \infty}} \left( \frac{\pi}{2} + \frac{1}{n} \right) \right). \]We have already established that \( \lim_{n \to \infty} \left( \frac{\pi}{2} + \frac{1}{n} \right) = \frac{\pi}{2} \).

Step 3: Calculate Limit

Thus, the sequence limit simplifies to:\[ \lim_{{n \to \infty}} a_n = \sin \left( \frac{\pi}{2} \right). \]The sine of \( \frac{\pi}{2} \) is known to be 1, so:\[ \lim_{{n \to \infty}} a_n = 1. \]

Step 4: Conclusion on Convergence

Since the sequence approaches a finite number (1), the sequence \( a_n = \sin \left( \frac{\pi}{2} + \frac{1}{n} \right) \) converges.

Key Concepts

Limit of a Sequence

Understanding the limit of a sequence is crucial in analyzing whether a sequence converges or diverges. In simple terms, the limit of a sequence is the value that the terms of the sequence approach as the index (usually represented by \( n \)) goes to infinity. To determine if a sequence converges, you need to identify if such a finite limit exists. If the sequence approaches a specific number as \( n \) increases indefinitely, it converges; otherwise, it diverges.

For example, in the sequence given in the exercise, \( a_n = \sin \left( \frac{\pi}{2} + \frac{1}{n} \right) \), we assessed the behavior of the expression as \( n \to \infty \). The term \( \frac{1}{n} \) gets closer to zero, affecting the whole expression: \( \frac{\pi}{2} + \frac{1}{n} \to \frac{\pi}{2} \).

This analysis helps us establish that the sequence approaches a limit that is finite, namely 1. Hence, the convergence is easier to understand when we recognize that \( \lim_{n \to \infty} a_n = 1 \). A sequence like this one that approaches a single number clearly converges.

Continuous Functions

Continuous functions play a significant role in determining the limit of sequences, especially when these sequences are based on the continuous functions themselves. A function is continuous if, intuitively, you can draw its graph without lifting your pencil off the paper.

This characteristic becomes important when dealing with sequences because if a function \( f(x) \) is continuous at a point \( c \), then the limit of the sequence formed by \( f(a_n) \), where \( a_n \to c \), will be \( f(c) \).

In the given example, \( \sin(x) \) is a continuous function. This means that if \( \lim_{n \to \infty} \left( \frac{\pi}{2} + \frac{1}{n} \right) = \frac{\pi}{2} \), then:

• \( \lim_{n \to \infty} \sin \left( \frac{\pi}{2} + \frac{1}{n} \right) = \sin \left( \lim_{n \to \infty} \left( \frac{\pi}{2} + \frac{1}{n} \right) \right) \)
• Which simplifies to \( \sin \left( \frac{\pi}{2} \right) \)
• And since \( \sin \left( \frac{\pi}{2} \right) = 1 \), \( \lim_{n \to \infty} a_n = 1 \)

The continuity of the function allows us to "bring the limit inside," simplifying the process of determining the limit of a sequence.

Trigonometric Limits

Trigonometric limits involve finding the limits of functions that include trigonometric expressions. These are important because many sequences, like the one in this exercise, involve trigonometric terms.

To evaluate a trigonometric limit, understand that the behavior of the trig function on its argument influences the outcome. For example, \( \sin(x) \) and \( \cos(x) \) are continuous, periodic, and bounded functions. The bounded nature (i.e., always between -1 and 1) ensures they have specific limits at respective points.

Considering our example, as \( n \) goes to infinity, the inner expression \( \frac{\pi}{2} + \frac{1}{n} \) approaches a simple angle \( \frac{\pi}{2} \). Understanding that the sine of \( \frac{\pi}{2} \) equals 1 helps in computing:

• The straightforward trigonometric limit:
• \( \lim_{n \to \infty} \sin \left( \frac{\pi}{2} + \frac{1}{n} \right) = \sin \left( \frac{\pi}{2} \right) \)

It simplifies to \( 1 \), showcasing how trigonometric limits can effectively simplify finding the convergence behavior of sequences.