Chapter 10: Problem 72
Which of the sequences \(\left\\{a_{n}\right\\}\) in Exercises \(27-90\) converge, and which diverge? Find the limit of each convergent sequence. $$ a_{n}=\left(1-\frac{1}{n^{2}}\right)^{n} $$
Short Answer
Expert verified
The sequence converges to 1.
Step by step solution
01
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.
02
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 \( \ln(1 - \frac{1}{n^2}) \approx -\frac{1}{n^2} \).
03
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, \( a_n \approx e^{-\frac{1}{n}} \).
04
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 \( n \to \infty \).
05
Concluding Convergence
Since \( a_n \to 1 \), the sequence \( a_n = \left( 1 - \frac{1}{n^2} \right)^n \) converges to 1.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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:
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.
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:
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.