Problem 78 - 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. $$ a_{n}=n\left(1-\cos \frac{1}{n}\right) $$

Step-by-Step Solution

Answer

The sequence converges to 0.

Step 1: Understand the Sequence

The sequence given is \( a_n = n \left( 1 - \cos \frac{1}{n} \right) \). We need to determine whether this sequence converges or diverges, and if it converges, find its limit.

Step 2: Analyze the Inner Expression

The function \( \cos x \) can be expanded using the Taylor series around 0: \( \cos x \approx 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots \). For small \( x \), we can approximate \( \cos \frac{1}{n} \approx 1 - \frac{1}{2n^2} \) because higher order terms are negligible as \( n \to \infty \).

Step 3: Simplify the Expression

Now, substitute the approximation back into the sequence: \[ a_n = n \left( 1 - \left( 1 - \frac{1}{2n^2} \right) \right) = n \cdot \frac{1}{2n^2} = \frac{1}{2n} \].

Step 4: Determine Convergence

The expression \( \frac{1}{2n} \) as \( n \to \infty \) obviously tends to \( 0 \). Therefore, the sequence \( a_n = \frac{1}{2n} \) converges to 0.

Key Concepts

Limits of Sequences

Sequences often play a vital role in mathematical analysis and calculus. A sequence can be understood as a list of numbers that follow a particular rule. Evaluating the limit of a sequence involves identifying the value that the sequence approaches as the number of terms increases indefinitely. In this exercise, our focus is on determining the behavior of the sequence \( a_n = n \left( 1 - \cos \frac{1}{n} \right) \). To explore whether a sequence converges, we must ascertain if its terms approach a fixed, finite number as \( n \to \infty \). If they do, the sequence is said to converge to that number, known as the limit. If not, it is divergent.

• A sequence \( \{a_n\} \) converges to a limit \( L \) if for every \( \epsilon > 0 \), there exists a natural number \( N \) such that for all \( n > N \), \( |a_n - L| < \epsilon \).

In the given example, step-by-step analysis revealed that the sequence term \( \frac{1}{2n} \) approaches zero as \( n \to \infty \). Hence, the sequence converges to 0.

Taylor Series Expansion

The Taylor series is an essential concept for approximating functions that are too complex for straightforward calculation. It breaks a function into an infinite sum of polynomials, which makes complex functions easier to simplify and analyze. In the context of the sequence \( a_n = n \left( 1 - \cos \frac{1}{n} \right) \), the Taylor series allows us to approximate \( \cos \frac{1}{n} \) when \( n \) is large.The Taylor series expansion for the cosine function around 0 is given by:

• \( \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots \)

For very small values of \( x \), higher-order terms diminish so quickly that we can often ignore them for simplification. Hence for \( x = \frac{1}{n} \), we use:\[ \cos \frac{1}{n} \approx 1 - \frac{1}{2n^2} \]This approximation significantly simplifies the expression, enabling us to deduce the sequence's behavior as \( n \to \infty \). This technique shows how powerful and versatile the Taylor series can be in tackling sequences involving trigonometric functions.

Trigonometric Sequences

Trigonometric sequences, like the one in our exercise, often involve sequences whose behavior is dictated by a trigonometric function such as sine, cosine, or tangent. They are quite common in both pure and applied mathematics, especially when dealing with oscillatory or wave-like phenomena.The sequence in question, \( a_n = n \left( 1 - \cos \frac{1}{n} \right) \), makes use of the cosine function, a fundamental trigonometric function. Analyzing such sequences often starts with transforming them using known mathematical tools, such as the Taylor series, to approximate or simplify.

• Trigonometric functions like cosines can be quite complex and their behavior at extremes can be counterintuitive without approximations.
• The cosine function, when expanded using Taylor series, simplifies the investigation of limits or convergence properties of related sequences.

Understanding trigonometric functions' incremental behavior, especially at infinity or zero, is crucial as it aids significantly in determining whether sequences converge to a certain limit. In this example, the cosine sequence converges to zero due to effective approximation of \( \cos \frac{1}{n} \) using its Taylor series expansion.