/*! 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 41 Determine whether the sequence \... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine whether the sequence \(\left\\{a_{n}\right\\}\) converges, and find its limit if it does converge. \(a_{n}=\left(\frac{2-n^{2}}{3+n^{2}}\right)^{n}\)

Short Answer

Expert verified
The sequence does not converge.

Step by step solution

01

Analyze the Sequence

Consider the general term of the sequence: \(a_n = \left(\frac{2-n^2}{3+n^2}\right)^n\). We aim to determine the limit of \(a_n\) as \(n\) approaches infinity.
02

Simplify Inside the Power

Evaluate the ratio inside the power: \(\frac{2-n^2}{3+n^2}\). As \(n\) becomes very large, \(n^2\) dominates the constants 2 and 3 in the numerator and denominator, respectively. So, we approximate \(\frac{2-n^2}{3+n^2} \approx \frac{-n^2}{n^2} = -1\).
03

Evaluate the Behavior as n Approaches Infinity

The term \(\left(\frac{2-n^2}{3+n^2}\right)^n\) resembles \((-1)^n\). Since \((-1)^n\) alternates between -1 and 1, the sequence does not converge.

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Key Concepts

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

Limits
Understanding limits is crucial when evaluating whether a sequence converges. The limit of a sequence \(a_n\) as \(n\) approaches infinity is the value that \(a_n\) gets closer to, as \(n\) becomes very large. If the sequence eventually stays arbitrarily close to a particular number, it is said to converge to that limit. If not, it diverges.

In the given problem, we are tasked to find the limit of the sequence \(a_n = \left(\frac{2-n^2}{3+n^2}\right)^n\) as \(n\) grows. To determine convergence, analyzing the limit behavior of expressions within sequences is essential. In this particular sequence, as \(n\) tends toward infinity, the expression within the parenthesis approaches -1, leading to the sequence exhibiting inconsistent behavior, thus not converging to a fixed limit.
Infinite Sequences
Infinite sequences stretch into eternity, meaning they continue to grow indefinitely. The sequence \(a_n\) with the term \((\frac{2-n^2}{3+n^2})^n\) is one such example of an infinite sequence.

Tackling infinite sequences often involves understanding their long-term behavior. This can include determining if they home in on a single value or if they oscillate or grow unboundedly. In our exercise, since the sequence \(a_n\) becomes similar to the alternating sequence \((-1)^n\), it alternates between 1 and -1 endlessly. Hence, it fails to settle down at a single value, reflecting its lack of convergence.
Dominant Terms
Dominant terms simplify analyzing sequences' or functions' behavior, especially for large values of \(n\). When functions are expressed as fractions, like \(\frac{2-n^2}{3+n^2}\), dominant terms are those that have the greatest power or impact and overshadow other terms.

In the expression \(\frac{2-n^2}{3+n^2}\), the term \(n^2\) is more dominant than the constants 2 and 3. As \(n\) grows, \(n^2\)'s significance surmounts that of 2 and 3, making the expression approximate to \(-1\). Recognizing dominant terms helps simplify the sequence analysis and apply limits, providing clarity about the sequence's potential convergence or divergence.
Power Sequences
Power sequences, such as the sequence given by \(a_n = \left(\frac{2-n^2}{3+n^2}\right)^n\), involve raising expressions to a power. These sequences can reveal intriguing behaviors and require careful examination, especially as the exponent increases.

When a term inside a power sequence approaches a constant value as \(n\) increases, the power can dramatically affect the sequence's convergence. In our case, since \(\frac{2-n^2}{3+n^2}\) approaches \(-1\), raising it to power \(n\) results in \((-1)^n\), an alternating sequence. As expected, this behavior leads to the sequence \(a_n\) failing to converge, since the alternating pattern doesn't allow it to approach one definite number. Power sequences like this highlight how both the base and exponent determine the sequence’s nature.

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Most popular questions from this chapter

Sum the indicated number of terms of the given alternating series. Then apply the alternating series remainder estimate to estimate the error in approximating the sum of the series with this partial sum. Finally, approximate the sum of the series, writing precisely the number of decimal places that thereby are guaranteed to be correct (after rounding). \(\sum_{n=1}^{x} \frac{(-1)^{n+1}}{n !}, 6\) terms

Suppose that \(\sum a_{n}\) is a conditionally convergent infinite series. For each \(n\), let $$a_{n}^{+}=\frac{a_{n}+\left|a_{n}\right|}{2} \text { and } a_{n}=\frac{a_{n}-\left|a_{n}\right|}{2} \text { . }$$ (a) Explain why \(\Sigma a_{n}^{+}\) consists of the positive terms of \(\Sigma a_{n}\) and why \(\Sigma a_{n}^{-}\) consists of the negative terms of \(\suma_{n}\). (b) Given a real number \(r\), show that some rearrangement of the conditionally convergent series \(\Sigma a_{n}\) converges to \(r\). (Snggestion: If \(r\) is positive, for instance, begin with the first partial sum of the positive series \(\Sigma a_{n}^{+}\) that exceeds \(r\). Then add just enough terms of the negative series \(\Sigma a_{n}\) so that the cumulative sum is less than \(r\). Next add just enough terms of the positive series that the cumulative sum is greater than \(r\), and continuc in this way to define the desired rearrangement.) Why does it follow that this rearranged infinite series converges to \(r\) ?

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Deduce from the arctangent series (Example 11 ) that $$\pi=\frac{6}{\sqrt{3}} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2 n+1}\left(\frac{1}{3}\right)^{n}$$ Then use this alternating series to show that \(\pi=3.14\) accurate to two decimal places.

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