/*! 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 52 Determine whether the following ... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine whether the following sequences converge or diverge and describe whether they do so monotonically or by oscillation. Give the limit when the sequence converges. $$\left\\{(-0.003)^{n}\right\\}$$

Short Answer

Expert verified
Answer: The sequence $\left\\{(-0.003)^{n}\right\\}$ converges oscillatingly to the limit 0 as $n$ approaches infinity.

Step by step solution

01

Identify the Pattern of the Sequence

The given sequence is: $$\left\\{(-0.003)^{n}\right\\}$$ Where \(n\) is a natural number. We can simply write the sequence as \(a_n = (-0.003)^n\).
02

Check for Monotonicity or Oscillation

As we know, a sequence is called monotonic if it is either increasing or decreasing, and oscillating if it keeps changing its signs from positive to negative repeatedly. In our given sequence, \((-0.003)^n\), the sign of \(a_n\) changes in every term. When \(n\) is even, \(a_n\) is positive; and when \(n\) is odd, \(a_n\) is negative. This sign changing pattern indicates that the given sequence is oscillating.
03

Checking for Convergence or Divergence

To check for convergence or divergence, we need to find the limit of the sequence as \(n\) approaches infinity. We have, $$\lim_{n\to\infty} (-0.003)^n$$ Since the absolute base value (\(0.003\)) is less than 1, the limit of the sequence will be zero.
04

Determining the Convergence or Divergence of the Sequence

Since the limit of the sequence as \(n\) approaches infinity is 0, the given sequence converges.
05

Combining the Results

Therefore, the given sequence, \(\left\\{(-0.003)^{n}\right\\}\) converges oscillatingly to the limit 0 as \(n\) approaches infinity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Monotonicity
Monotonicity is a term used to describe sequences that consistently move in one direction. These sequences can either be increasing or decreasing. For a sequence to be increasing, each successive term should be greater than or equal to the previous term. Similarly, a decreasing sequence is one where each term is less than or equal to the term before it. In our exercise, we are looking at the sequence \((-0.003)^n\). This sequence is **not monotonic** because it doesn't consistently increase or decrease. Instead, the sign of its terms alternates between positive and negative. This behavior of changing signs is called oscillation, which we'll explore in more detail next.
Oscillation
Oscillation in sequences occurs when the terms of a sequence switch back and forth between positive and negative. This is different from monotonic sequences, where terms move consistently in one direction.
When analyzing the sequence \((-0.003)^n\), we notice a clear pattern in its behavior. Every time we increase \(n\) by 1, the sign of the sequence term changes:
  • When \(n\) is even, \((-0.003)^n\) results in a positive value.
  • When \(n\) is odd, \((-0.003)^n\) results in a negative value.
This alternating sign pattern is a distinctive feature of oscillating sequences. Despite this back-and-forth change, the absolute value of each term is getting smaller as \(n\) increases, which affects its convergence.
Limit of a Sequence
The limit of a sequence is a core concept that helps determine if a sequence converges or diverges. When we say a sequence converges, it means that as \(n\) becomes very large, the terms of the sequence approach a specific value, which we call the limit. For the sequence \((-0.003)^n\), we calculate the limit as \(n\) approaches infinity. Because the absolute value of \(-0.003\) is less than 1, each term becomes smaller, approaching zero. Thus, we have:\[ \lim_{n\to\infty} (-0.003)^n = 0 \]Given this result, the sequence converges to a limit of zero. Despite the oscillation in sign, the magnitude of each term diminishes towards zero as \(n\) increases, confirming that the sequence converges.

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Most popular questions from this chapter

Consider the following infinite series. a. Write out the first four terms of the sequence of partial sums. b. Estimate the limit of \(\left\\{S_{n}\right\\}\) or state that it does not exist. $$\sum_{k=1}^{\infty} 3^{-k}$$

Infinite products An infinite product \(P=a_{1} a_{2} a_{3} \ldots,\) which is denoted \(\prod_{k=1}^{\infty} a_{k}\) is the limit of the sequence of partial products \(\left\\{a_{1}, a_{1} a_{2}, a_{1} a_{2} a_{3}, \dots\right\\}\) a. Show that the infinite product converges (which means its sequence of partial products converges) provided the series \(\sum_{k=1}^{\infty} \ln a_{k}\) converges. b. Consider the infinite product $$P=\prod_{k=2}^{\infty}\left(1-\frac{1}{k^{2}}\right)=\frac{3}{4} \cdot \frac{8}{9} \cdot \frac{15}{16} \cdot \frac{24}{25} \cdots$$ Write out the first few terms of the sequence of partial products, $$P_{n}=\prod_{k=2}^{n}\left(1-\frac{1}{k^{2}}\right)$$ (for example, \(P_{2}=\frac{3}{4}, P_{3}=\frac{2}{3}\) ). Write out enough terms to determine the value of the product, which is \(\lim _{n \rightarrow \infty} P_{n}\). c. Use the results of parts (a) and (b) to evaluate the series $$\sum_{k=2}^{\infty} \ln \left(1-\frac{1}{k^{2}}\right)$$

Consider the sequence \(\left\\{F_{n}\right\\}\) defined by $$F_{n}=\sum_{k=1}^{\infty} \frac{1}{k(k+n)^{\prime}}$$ for \(n=0,1,2, \ldots . .\) When \(n=0,\) the series is a \(p\) -series, and we have \(F_{0}=\pi^{2} / 6\) (Exercises 65 and 66 ). for \(n=0,1,2, \ldots . .\) When \(n=0,\) the series is a \(p\) -series, and we have \(F_{0}=\pi^{2} / 6\) (Exercises 65 and 66 ).

Evaluate the limit of the following sequences. $$a_{n}=\frac{n^{8}+n^{7}}{n^{7}+n^{8} \ln n}$$

Pick two positive numbers \(a_{0}\) and \(b_{0}\) with \(a_{0}>b_{0}\) and write out the first few terms of the two sequences \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}:\) $$a_{n+1}=\frac{a_{n}+b_{n}}{2}, \quad b_{n+1}=\sqrt{a_{n} b_{n}}, \quad \text { for } n=0,1,2 \dots$$ (Recall that the arithmetic mean \(A=(p+q) / 2\) and the geometric mean \(G=\sqrt{p q}\) of two positive numbers \(p\) and \(q\) satisfy \(A \geq G\). a. Show that \(a_{n}>b_{n}\) for all \(n\). b. Show that \(\left\\{a_{n}\right\\}\) is a decreasing sequence and \(\left\\{b_{n}\right\\}\) is an increasing sequence. c. Conclude that \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}\) converge. d. Show that \(a_{n+1}-b_{n+1}<\left(a_{n}-b_{n}\right) / 2\) and conclude that \(\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} b_{n} .\) The common value of these limits is called the arithmetic-geometric mean of \(a_{0}\) and \(b_{0},\) denoted \(\mathrm{AGM}\left(a_{0}, b_{0}\right)\). e. Estimate AGM(12,20). Estimate Gauss' constant \(1 / \mathrm{AGM}(1, \sqrt{2})\).

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