Problem 72 - 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. $$ a_{n}=\left(1-\frac{1}{n^{2}}\right)^{n} $$

Step-by-Step Solution

Answer

The sequence converges to 1.

Step 1: Understanding the Sequence

We have the sequence \( a_n = \left( 1 - \frac{1}{n^2} \right)^n \). We need to determine if this sequence converges or diverges, and if it converges, find its limit.

Step 2: Transforming the Sequence for Analysis

Consider the expression \( a_n = \left( 1 - \frac{1}{n^2} \right)^n \). By taking the natural logarithm, we get \( \ln(a_n) = n \ln(1 - \frac{1}{n^2}) \). To simplify, we can use the Taylor series expansion \( \ln(1 + x) \approx x - \frac{x^2}{2} + \ldots \) for small \(x\), where \( x = -\frac{1}{n^2} \). This gives \( \ln(1 - \frac{1}{n^2}) \approx -\frac{1}{n^2} \).

Step 3: Simplifying the Expression

Substituting back into our expression, we have \( \ln(a_n) \approx n \left(-\frac{1}{n^2}\right) = -\frac{1}{n} \). Therefore, \( a_n \approx e^{-\frac{1}{n}} \).

Step 4: Finding the Limit

As \( n \to \infty \), \(-\frac{1}{n} \to 0\). Thus, \( e^{-\frac{1}{n}} \to e^0 = 1 \). This means \( a_n \to 1 \) as \( n \to \infty \).

Step 5: Concluding Convergence

Since \( a_n \to 1 \), the sequence \( a_n = \left( 1 - \frac{1}{n^2} \right)^n \) converges to 1.

Key Concepts

Taylor Series Expansion

A Taylor series expansion breaks down a complex function into an infinite polynomial around a specific point, usually to make calculations easier.

When we look at our sequence \( a_n = \left( 1 - \frac{1}{n^2} \right)^n \), it involves a logarithmic part that can be hard to simplify directly. This is where the Taylor series comes in handy.

The expression \( \ln(1 + x) \approx x - \frac{x^2}{2} + \ldots \) helps us to approximate functions like \( \ln(1 - \frac{1}{n^2}) \) when \( x \) is small. Here, \( x = -\frac{1}{n^2} \), which is indeed small for large \( n \). This simplifies the function to \( -\frac{1}{n^2} \), giving a big simplification step for further calculations.

• The smaller \( x \), the closer the approximation. This is great for sequences with large \( n \).
• The Taylor series essentially transforms a non-linear term to a linear approximation, making it more digestible.

Limit of a Sequence

The limit of a sequence is a fundamental concept in calculus, which helps us understand the behavior of a sequence as it approaches a specific value as \( n \) increases indefinitely.

In our given sequence, \( a_n = \left(1 - \frac{1}{n^2}\right)^n \), we determined that its limit is 1 as \( n \to \infty \).

Here's how we calculate it:

• We rewrote \( a_n \) using natural logarithms and Taylor series approximations, simplifying it to \( a_n \approx e^{-\frac{1}{n}} \).
• Observing \( e^{-\frac{1}{n}} \) as \( n \) grows, we see that \( -\frac{1}{n} \to 0 \), so \( e^0 = 1 \).
• This means that eventually, as \( n \to \infty \), \( a_n \) converges on 1.
• A convergent sequence approaches a particular number, unlike a divergent one, which does not settle.

Understanding these values helps grasp the behavioral pattern of infinite sequences, crucial for deeper calculus topics.

Logarithmic Transformation

Logarithmic transformation is a technique used in various mathematical fields to transform balanced exponential processes into more manageable forms.

Consider our sequence \( a_n = \left(1 - \frac{1}{n^2}\right)^n \). Direct analysis is complex, so we transform it:

• Taking the natural logarithm, we have \( \ln(a_n) = n \ln(1 - \frac{1}{n^2}) \).
• The logarithm simplifies multiplication operations to addition, making them easier to handle.
• This step is crucial before approximation because expressions become more straightforward to work with.

Therefore, applying a logarithmic transformation assists us in simplifying the sequence to find limits, particularly using Taylor series for further simplification. Overall, this aids in concluding about the sequence's convergence or divergence.