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If \(\Sigma_{n=0}^{\infty} c_{n} 4^{n}\) is convergent, does it follow that the following series are convergent? (a) \(\sum_{n=0}^{\infty} c_{n}(-2)^{n} \quad\) (b) \(\sum_{n=0}^{\infty} c_{n}(-4)^{n}\)

Short Answer

Expert verified
(a) Convergent. (b) Divergent.

Step by step solution

01

Understand the Convergence of the Original Series

The series \( \Sigma_{n=0}^{\infty} c_{n} 4^{n} \) is given to be convergent. This implies that the coefficients \( c_n \) are such that when multiplied by \( 4^n \), the series converges. This happens typically if the terms of the series become arbitrarily small as \( n \) approaches infinity, given the base \( 4 \) in the series term.
02

Check the Convergence of the First Series

Consider the series \( \sum_{n=0}^{\infty} c_{n}(-2)^{n} \). This series is derived from multiplying \( c_n \) by \( (-2)^n \) instead of \( 4^n \). Check the convergence by considering the ratio of bases \( \left(\frac{-2}{4}\right)^n = \left(-\frac{1}{2}\right)^n \). Since the original series with base \( 4^n \) converges, multiplying terms by \( \left(-\frac{1}{2}\right)^n \) will also result in a convergent series because \( \left(-\frac{1}{2}\right)^n \) converges to zero.
03

Verify the Convergence of the Second Series

Now consider the series \( \sum_{n=0}^{\infty} c_{n}(-4)^{n} \). Here, the terms are multiplied by \( (-4)^n \). Compare the ratios \( \frac{(-4)^n}{4^n} = (-1)^n \). This is a sequence alternating between -1 and 1, which does not converge to zero. Therefore, altering the sign, without controlling magnitude, results in the potential divergence of the series.

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Key Concepts

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

Infinite Series
An infinite series is a sum of infinitely many terms, typically written in the form \( \sum_{n=0}^{\infty} a_n \) where \( a_n \) are the terms of the series. These series can be challenging to work with, as they do not end and the sum is not immediately obvious. There are many fascinating types of series, and depending on their properties, they can converge or diverge.

An important aspect of infinite series is whether they converge—meaning they approach a finite limit as more terms are added—and if they diverge—meaning they do not approach any limit but instead grow indefinitely. Some convergent series have well-known sums, such as the geometric series, which converges when the absolute value of the common ratio is less than one.

Convergence tests are used to determine whether a series converges or diverges. These include the ratio test, root test, and direct comparison test, among others. Using these tests can help us decide if the addition of infinitely many terms results in a finite number, or if the series grows beyond all bounds.
Alternating Series
Alternating series are a special type of series where the signs of the terms alternate between positive and negative. They are typically of the form \( \sum_{n=0}^{\infty} (-1)^n b_n \) where \( b_n \) are positive terms. The alternating sign can sometimes lead to convergence when an otherwise positive series might diverge.

One of the classic examples is the alternating harmonic series \( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \) which converges, even though the corresponding non-alternating series diverges. An important test for determining the convergence of an alternating series is the Alternating Series Test. This test requires:
  • The absolute value of the terms \( b_n \) is decreasing.
  • The limit of \( b_n \) as \( n \rightarrow \infty \) is zero.
When these conditions are met, the series converges. Alternating series can create interesting patterns, sometimes resulting in a better approximation of a number due to the canceling effect of consecutive terms.
Convergent Series
A convergent series is an infinite series where the sum approaches a finite value as more and more terms are added. This may seem counterintuitive given the infinite nature of the series, but many series, despite having infinitely many terms, end in a well-defined sum.

Consider the geometric series: \( \sum_{n=0}^{\infty} ar^n \) converges when \(|r| < 1\). The sum of such a series is \( \frac{a}{1-r} \). Understanding these series is crucial as they appear in many branches of mathematics, including calculus, number theory, and statistical problems.

Series convergence is vital in applications like power series expansions. To determine if a series is convergent, you can use tests such as the ratio test, where you examine the ratio \( |a_{n+1}/a_n| \). If the limit of this ratio as \( n \rightarrow \infty \) is less than 1, the series is convergent. This provides powerful tools in analyzing whether infinite processes can yield finite, meaningful results.

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Most popular questions from this chapter

Evaluate the indefinite integral as an infinite series. \(\int \frac{\cos x-1}{x} d x\)

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If \(\Sigma a_{n}\) is convergent and \(\Sigma b_{n}\) is divergent, show that the series \(\Sigma\left(a_{n}+b_{n}\right)\) is divergent. IHint: Argue by contradiction. \(]\)

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The resistivity \(\rho\) of a conducting wire is the reciprocal of the conductivity and is measured in units of ohm-meters \((\Omega-m) .\) The resistivity of a given metal depends on the temperature according to the equation $$\rho(t)=\rho_{20} e^{\alpha(t-20)}$$ where \(t\) is the temperature in \(^{\circ} \mathrm{C}\) . where \(t\) is the temperature in \(^{\circ} \mathrm{C} .\) There are tables that list the values of \(\alpha\) (called the temperature coefficient) and \(\rho_{20}\) (the resistivity at \(20^{\circ} \mathrm{C} )\) for various metals. Except at very low temperatures, the resistivity varies almost linearly with temperature and so it is common to approximate the expression for \(\rho(t)\) by its first- or second-degree Taylor polynomial at \(t=20\) . (a) Find expressions for these linear and quadratic approximations. (b) For copper, the tables give \(\alpha=0.0039 /^{\circ} \mathrm{C}\) and \(\rho_{20}=1.7 \times 10^{-8} \Omega-\mathrm{m} .\) Graph the resistivity of copper and the linear and quadratic approximations for \(-250^{\circ} \mathrm{C} \leqslant t \leqslant 1000^{\circ} \mathrm{C}\) (c) For what values of \(t\) does the linear approximation agree with the exponential expression to within one percent?

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