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Chapter 8: Problem 29

\(9-32\) n Determine whether the sequence converges or diverges. If it converges, find the limit. $$\\{0,1,0,0,1,0,0,0,1, \ldots\\}$$

Short Answer

Expert verified
The sequence diverges.

Step by step solution

01

Identify the Pattern

Observe the given sequence: \(\{0,1,0,0,1,0,0,0,1, \ldots\}\). Notice that there is a repeating pattern in the sequence: every fourth term is '1', while all other terms are '0'. Thus, the pattern in the sequence is 0, 1, 0, 0, repeating indefinitely.
02

Determine the Sequence Limit

In a convergent sequence, the terms approach a specific value as the sequence progresses. For the given sequence, as \(n\) becomes very large, the terms will not approach any single number because every fourth term deviates to '1' and the others are '0'. Thus, the sequence does not stabilize at any particular value.
03

Conclude on Convergence or Divergence

Since the sequence does not settle to any one particular value, and it continues to oscillate between '0' and '1', the sequence neither converges nor approaches a limit. Therefore, the sequence diverges.

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Key Concepts

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

Divergence of Sequences
When we talk about whether a sequence converges or diverges, we're basically asking if the sequence settles towards a particular value or not. A diverging sequence, on the other hand, never reaches or approaches a specific number and often fluctuates or grows without bound. In the sequence \[\{0, 1, 0, 0, 1, 0, 0, 0, 1, \ldots\} \]we observe that it does not get closer to any particular number as it progresses. This is because the sequence repeatedly returns to '1' after a few '0鈥檚, leading to no permanent stabilization.
  • Non-convergence: Even as we extend the length of the sequence, we cannot zero in on a single value that it tries to reach or settle towards.
  • Fluctuation: The mix of '0's and '1's exemplifies divergence through oscillation, as it never sticks to one value.
Sequences like this are prime examples of divergence due to their lack of pattern indicating progression towards a point.
Sequence Patterns
Understanding the pattern of a sequence is crucial for analyzing its behavior over time. Patterns give us hints about how the sequence can be categorized and studied further.
The given sequence \[\{0, 1, 0, 0, 1, 0, 0, 0, 1, \ldots\} \]clearly shows a repeating pattern.
  • Repetitive Nature: There鈥檚 a clear pattern where every fourth term is '1', with the rest being '0'.
  • Predictability: This periodic repetition makes it predictable but does not lead to convergence.
Patterns like these are essential for determining the characteristics of the sequence, such as whether or not it might eventually simplify to a limit.
Limit of a Sequence
The limit of a sequence is a concept used to describe what value the terms of a sequence get closer to as the sequence progresses without end.
A convergent sequence has a distinct number, known as the limit, that it approaches. However, our sequence: \[\{0, 1, 0, 0, 1, 0, 0, 0, 1, \ldots\} \]never stabilizes at any single value as we make the sequence longer.
  • Absence of Convergence: The terms fluctuate between '0' and '1' repeatedly, illustrating no consistent progression towards a fixed number.
  • Irregular Stabilization: If a sequence does not stabilize, a limit does not exist. Therefore, this sequence has no limit.
Determining the limit is crucial for understanding the end behavior of sequence, but divergence indicates the absence of such limits.

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Most popular questions from this chapter

Find a power series representation for the function and determine the interval of convergence. $$ f(x)=\frac{x}{2 x^{2}+1} $$

Find a power series representation for the function and determine the radius of convergence. $$ f(x)=\frac{x}{(1+4 x)^{2}} $$

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In Section 4.6 we considered Newton's method for approxi- mating a root \(r\) of the equation \(f(x)=0,\) and from an initial approximation \(x_{1}\) we obtained successive approximations \(x_{2}, x_{3}, \ldots,\) where $$x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$$ Use Taylor's Formula with \(n=1, a=x_{n},\) and \(x=r\) to show that if \(f^{\prime \prime}(x)\) exists on an interval \(I\) containing \(r, x_{n}\) and \(x_{n+1},\) and \(\left|f^{\prime \prime}(x)\right| \leqslant M,\left|f^{\prime}(x)\right| \geqslant K\) for all \(x \in I,\) then $$ \left|x_{n+1}-r\right| \leqslant \frac{M}{2 K}\left|x_{n}-r\right|^{2}$$ [This means that if \(x_{n}\) is accurate to \(d\) decimal places, then \(x_{n+1}\) is accurate to about 2\(d\) decimal places. More precisely,if the error at stage \(n\) is at most \(10^{-m}\) , then the error at stage \(n+1\) is at most \((M / 2 K) 10^{-2 m} . ]\)

(a) Show that the function defined by $$f(x)=\left\\{\begin{array}{ll}{e^{-1 / x^{2}}} & {\text { if } x \neq 0} \\\ {0} & {\text { if } x=0}\end{array}\right.$$ is not equal to its Maclaurin series. (b) Graph the function in part (a) and comment on its behavior near the origin.
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