/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 72 Which of the sequences \(\left\\... [FREE SOLUTION] | 91Ó°ÊÓ

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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.
  • 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.

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Most popular questions from this chapter

By multiplying the Taylor series for \(e^{x}\) and \(\sin x,\) find the terms through \(x^{5}\) of the Taylor series for \(e^{x} \sin x .\) This series is the imaginary part of the series for \begin{equation} e^{x} \cdot e^{i x}=e^{(1+i) x} \end{equation} Use this fact to check your answer. For what values of \(x\) should the series for \(e^{x}\) sin \(x\) converge?

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \frac{1}{1+2+3+\cdots+n}\end{equation}

Approximation properties of Taylor polynomials Suppose that \(f(x)\) is differentiable on an interval centered at \(x=a\) and that \(g(x)=b_{0}+b_{1}(x-a)+\cdots+b_{n}(x-a)^{n}\) is a polynomial of degree \(n\) with constant coefficients \(b_{0}, \ldots, b_{n}\) . Let \(E(x)=\) \(f(x)-g(x) .\) Show that if we impose on \(g\) the conditions i) \(E(a)=0\) ii) $$\lim _{x \rightarrow a} \frac{E(x)}{(x-a)^{n}}=0$$ then $$\begin{array}{r}{g(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\cdots} \\ {+\frac{f^{(n)}(a)}{n !}(x-a)^{n}}\end{array}$$ Thus, the Taylor polynomial \(P_{n}(x)\) is the only polynomial of degree less than or equal to \(n\) whose error is both zero at \(x=a\) and negligible when compared with \((x-a)^{n}.\)

Computer Explorations Taylor's formula with \(n=1\) and \(a=0\) gives the linearization of a function at \(x=0 .\) With \(n=2\) and \(n=3\) we obtain the standard quadratic and cubic approximations. In these exercises we explore the errors associated with these approximations. We seek answers to two questions: \begin{equation} \begin{array}{l}{\text { a. For what values of } x \text { can the function be replaced by each }} \\ {\text { approximation with an error less than } 10^{-2} \text { ? }} \\ {\text { b. What is the maximum error we could expect if we replace the }} \\ {\text { function by each approximation over the specified interval? }}\end{array} \end{equation} Using a CAS, perform the following steps to aid in answering questions (a) and (b) for the functions and intervals in Exercises \(53-58 .\) \begin{equation} \begin{array}{l}{\text { Step } 1 : \text { Plot the function over the specified interval. }} \\ {\text { Step } 2 : \text { Find the Taylor polynomials } P_{1}(x), P_{2}(x), \text { and } P_{3}(x) \text { at }} \\\ {x=0 .}\\\\{\text { Step } 3 : \text { Calculate the }(n+1) \text { st derivative } f^{(n+1)}(c) \text { associated }} \\ {\text { with the remainder term for each Taylor polynomial. }} \\ {\text { Plot the derivative as a function of } c \text { over the specified interval }} \\ {\text { and estimate its maximum absolute value, } M .}\\\\{\text { Step } 4 : \text { Calculate the remainder } R_{n}(x) \text { for each polynomial. }} \\ {\text { Using the estimate } M \text { from Step } 3 \text { in place of } f^{(n+1)}(c), \text { plot }} \\ {R_{n}(x) \text { over the specified interval. Then estimate the values of }} \\ {x \text { that answer question (a). }}\\\\{\text { Step } 5 : \text { Compare your estimated error with the actual error }} \\ {E_{n}(x)=\left|f(x)-P_{n}(x)\right| \text { by plotting } E_{n}(x) \text { over the specified }} \\ {\text { interval. This will help answer question (b). }} \\ {\text { Step } 6 : \text { Graph the function and its three Taylor approximations }} \\ {\text { together. Discuss the graphs in relation to the information }} \\ {\text { discovered in Steps } 4 \text { and } 5 .}\end{array} \end{equation} $$f(x)=e^{-x} \cos 2 x, \quad|x| \leq 1$$

Use series to evaluate the limits. \begin{equation} \lim _{x \rightarrow 0} \frac{\sin 3 x^{2}}{1-\cos 2 x} \end{equation}

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