Problem 82 - 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. $$ a_{n}=\frac{1}{\sqrt{n}} \tan ^{-1} n $$

Step-by-Step Solution

Answer

The sequence converges to 0.

Step 1: Analyze the Function Inside the Sequence

Observe that the sequence given is \(a_n = \frac{1}{\sqrt{n}} \tan^{-1}(n)\). We need to understand the behavior of both \(\frac{1}{\sqrt{n}}\) and \(\tan^{-1}(n)\) separately as \(n\) approaches infinity.

Step 2: Determine the Limit of \(\frac{1}{\sqrt{n}}\)

As \( n \to \infty \), \( \frac{1}{\sqrt{n}} \to 0 \). This occurs because the square root of \(n\) grows without bound, making the entire fraction tend towards zero.

Step 3: Determine the Limit of \(\tan^{-1}(n)\)

The function \(\tan^{-1}(n)\) or the inverse tangent function has a horizontal asymptote at \(\frac{\pi}{2}\) as \(n\) approaches infinity. Thus, \(\tan^{-1}(n) \to \frac{\pi}{2}\) as \(n \to \infty\).

Step 4: Analyze the Product of the Limits

Since the sequence is \(a_n = \frac{1}{\sqrt{n}} \tan^{-1}(n)\), its behavior at infinity is determined by multiplying two limits. The term \(\frac{1}{\sqrt{n}}\) tends to 0, while \(\tan^{-1}(n)\) tends to \(\frac{\pi}{2}\). Multiplying gives \(\frac{1}{\sqrt{n}} \cdot \frac{\pi}{2} \to 0\).

Step 5: Conclude the Behavior of the Sequence

Based on the product of these limits, \(a_n\) converges to 0.

Key Concepts

Convergence of Sequences

When discussing sequences in calculus, it is crucial to understand the concept of convergence. A sequence converges if its terms approach a specific number as the sequence progresses toward infinity. This specific number is called the limit of the sequence. When the terms do not approach any particular number, the sequence is said to diverge.
To determine convergence, a common method is to analyze the behavior of the sequence's terms as the index goes to infinity. You identify if the terms get arbitrarily close to a certain number. If they do, the sequence is convergent.

• Convergence indicates a predictable end point for a sequence.
• Divergence shows the sequence lacks a definitive approach as it extends.

In the original exercise, we considered a sequence involving inverse trigonometric functions and fractional terms: \( a_n = \frac{1}{\sqrt{n}} \tan^{-1}(n) \) . We analyzed each component, checked their limits, and found that the sequence converges to 0 since the fraction dominates and approaches zero.

Limit of a Sequence

The limit of a sequence helps in understanding the long-term behavior of a sequence's terms. When a sequence has a limit, it means that as you progress along the sequence, the terms get closer and closer to a specific value. This value is the limit.
Mathematically, we denote this by saying that the limit of the sequence \( a_n \to L \) as \( n \to \infty \) .
For evaluating limits, each component of a sequence’s formula can be analyzed:

• Consider the arithmetic or algebraic operation affecting the progression of the terms.
• Utilize known limits such as \( \frac{1}{n} \to 0 \) as \( n \to \infty \) .

In the given problem, we found that evaluating \( \frac{1}{\sqrt{n}} \) and \( \tan^{-1}(n) \) separately provided key insights:
The former approached zero, while the latter approached \( \frac{\pi}{2} \). Multiplication of these results led us to conclude that the entire sequence \( a_n \) tends to a limit of zero.

Inverse Trigonometric Functions

Inverse trigonometric functions are crucial when dealing with limits involving angles or angle-like quantities. \( \tan^{-1}(n) \), the inverse tangent function, is one of such functions. It helps find an angle whose tangent is \( n \).
Understanding the asymptotic behavior of these functions as their variable approaches infinity or zero is important.

• As \( n \to \infty \) , \( \tan^{-1}(n) \) approaches \( \frac{\pi}{2} \), providing a horizontal asymptote.
• This behavior affects how products involving inverse tangents like \( \frac{1}{\sqrt{n}} \tan^{-1}(n) \) are evaluated.

In this exercise, recognizing that \( \tan^{-1}(n) \to \frac{\pi}{2} \) as \( n \to \infty \) was essential. This understanding allowed us to predict that \( \tan^{-1}(n) \), despite its growth, does not diverge to infinity. Instead, it converges to a bound, which could then be multiplied by the fraction \( \frac{1}{\sqrt{n}} \) to assess the sequence's overall limit.