Problem 77 - 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. $$ a_{n}=\frac{n^{2}}{2 n-1} \sin \frac{1}{n} $$

Step-by-Step Solution

Answer

The sequence converges, with a limit of \(\frac{1}{2}\).

Step 1: Analyze the Sinusoidal Part

The sequence is defined as \( a_n = \frac{n^2}{2n-1} \sin \frac{1}{n} \). First, observe \( \sin \frac{1}{n} \). As \( n \to \infty \), \( \frac{1}{n} \to 0 \), and we know that \( \sin(0) = 0 \). Hence, \( \sin \frac{1}{n} \to 0 \) as \( n \to \infty \).

Step 2: Analyze the Rational Expression

Next, consider the rational expression \( \frac{n^2}{2n-1} \). The degree of the polynomial in the numerator is 2, and the degree in the denominator is 1. For large \( n \), this behaves like \( \frac{n^2}{2n} = \frac{n}{2} \). Therefore, as \( n \to \infty \), \( \frac{n^2}{2n-1} \to \infty \).

Step 3: Combine the Components for Convergence Determination

The sequence is the product of an expression that tends to \( 0 \), \( \sin \frac{1}{n} \), and an expression that tends to \( \infty \), \( \frac{n^2}{2n-1} \). To determine if the product converges, we compute the limit: \[ \lim_{n \to \infty} \frac{n^2}{2n-1} \sin \frac{1}{n}. \]

Step 4: Use an Approximation for Small Angles

For small \( x \), \( \sin x \approx x \). Hence, \( \sin \frac{1}{n} \approx \frac{1}{n} \) when \( n \) is large. Substitute this approximation into the sequence:\[ a_n \approx \frac{n^2}{2n-1} \cdot \frac{1}{n} = \frac{n}{2n-1}. \]

Step 5: Compute the Limit of the Approximated Sequence

Now find the limit:\[ \lim_{n \to \infty} \frac{n}{2n-1} = \lim_{n \to \infty} \frac{1}{2 - \frac{1}{n}} = \frac{1}{2}, \]as the term \( \frac{1}{n} \to 0 \) as \( n \to \infty \). Thus, the sequence converges to \( \frac{1}{2} \).

Key Concepts

Sequence Limit

When we talk about sequence limits, we're exploring what happens to the terms of a sequence as the index, usually denoted by \( n \), gets larger and larger without bound. In simple terms, we're asking where the sequence is headed as you go forever. The limit of a sequence \( \{a_n\} \) is the value that the terms \( a_n \) approach as \( n \) approaches infinity.For instance, in our example, we have the sequence \( a_n = \frac{n^2}{2n-1} \sin \frac{1}{n} \). The sequence limit is determined by examining the behavior of each component:

• \( \sin \frac{1}{n} \) approaches \( 0 \) as \( n \to \infty \) because \( \frac{1}{n} \to 0 \) and \( \sin(x) \) is \( 0 \) when \( x \) is \( 0 \).
• \( \frac{n^2}{2n-1} \) approaches infinity as the degree of the numerator is higher than the denominator, behaving like \( \frac{n}{2} \).

The sequence converges if the product of these two components results in a finite number. Here, the sequence limit is found to be \( \frac{1}{2} \). This means each term of the sequence gets closer and closer to \( \frac{1}{2} \) as \( n \) becomes very large.

Trigonometric Limits

Trigonometric limits often come into play when dealing with sequences or functions that include sine, cosine, or other trigonometric expressions. A basic trigonometric limit that you might encounter involves the sine function: \( \lim_{x \to 0} \sin x = 0 \).In the given exercise, the function \( \sin \frac{1}{n} \) is a trigonometric component. As \( n \) tends towards infinity, \( \frac{1}{n} \to 0 \), causing \( \sin \frac{1}{n} \to 0 \). Understanding these simple trigonometric limits can be crucial in evaluating more complex sequences.Moreover, for very small angles, \( \sin x \approx x \) holds true. This linear approximation is incredibly helpful. It allows us to transform problems involving trigonometric functions into more manageable forms. Applying this trick to our sequence, \( \sin \frac{1}{n} \approx \frac{1}{n} \), simplifies our calculation of the sequence limit by making it easier to handle algebraically.

Rational Expressions

Rational expressions are fractions where the numerator and the denominator are polynomials. Analyzing the behavior of such expressions as \( n \to \infty \) is key in understanding sequences and limits.The main tool here is assessing the degree of the polynomials:

• If the numerator's degree is higher, the expression tends to infinity.
• If the degrees are equal, the expression approaches the ratio of their leading coefficients.
• If the denominator's degree is higher, the expression tends to zero.

In our sequence, \( \frac{n^2}{2n-1} \), the numerator has a higher degree than the denominator. This means it behaves similarly to \( \frac{n^2}{2n} = \frac{n}{2} \), heading towards infinity. Rational expressions might seem tricky at first, but recognizing the polynomial degrees quickly shows which way the expression might swing as \( n \) becomes very large. This knowledge, combined with other components like the sine function in this example, helps determine whether a sequence converges or diverges.