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In Exercises \(81-86\) , find the values of \(x\) for which the given geometric series converges. Also, find the sum of the series (as a function of \(x )\) for those values of \(x .\) $$\sum_{n=0}^{\infty}\left(-\frac{1}{2}\right)^{n}(x-3)^{n}$$

Short Answer

Expert verified
The series converges for \(1 < x < 5\), and the sum is \(\frac{2}{x-1}\).

Step by step solution

01

Identify the General Formula of a Geometric Series

The general formula for a geometric series is \( \sum_{n=0}^{\infty} ar^n \), where \( a \) is the first term and \( r \) is the common ratio. In this series, \( a = 1 \) and \( r = -\frac{1}{2}(x-3) \).
02

Determine Convergence Criteria

For a geometric series to converge, the absolute value of the common ratio must be less than 1, that is \( |r| < 1 \). Therefore, we have \( \left| -\frac{1}{2}(x-3) \right| < 1 \).
03

Solve the Inequality for Convergence

Solving the inequality \( \left| -\frac{1}{2}(x-3) \right| < 1 \):\( |- \frac{1}{2}(x-3)| < 1 \) implies \( \left| x-3 \right| < 2 \). Hence, the interval of convergence is \( -2 < x-3 < 2 \), which simplifies to \( 1 < x < 5 \).
04

Find the Sum of the Series

If the geometric series converges, its sum is given by \( S = \frac{a}{1-r} \). Here, \( a = 1 \) and \( r = -\frac{1}{2}(x-3) \), so the sum is:\[ S(x) = \frac{1}{1 + \frac{1}{2}(x-3)} = \frac{2}{2+(x-3)} = \frac{2}{x-1}. \]
05

Conclusion

The geometric series converges for \( 1 < x < 5 \). The sum of the series, as a function of \( x \), is \( \frac{2}{x-1} \) within this interval of convergence.

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Key Concepts

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

Series Convergence
In mathematics, the convergence of a series refers to the behavior of the sum of its terms as the number of terms goes to infinity. For a geometric series to converge, the series must meet specific criteria. In our example series, \( \sum_{n=0}^{\infty}\left(-\frac{1}{2}\right)^{n}(x-3)^{n} \), the convergence is contingent upon the common ratio, \( r \), being less than 1 in absolute value.

The common ratio is given by \( r = -\frac{1}{2}(x-3) \). For this series, convergence occurs when:
  • \( |-\frac{1}{2}(x-3)| < 1 \)
  • This simplifies to \( |x-3| < 2 \)
  • Which further simplifies to \( 1 < x < 5 \)
If these conditions are met, the series will converge. When a series converges, it means that as you keep adding more terms, the sum will approach a specific finite value, giving useful insights into the behavior of the series within a particular interval of \( x \).
Sum of Series
The sum of an infinite geometric series that converges can be succinctly expressed using a formula. If the series converges, its sum is determined by dividing the first term by one minus the common ratio. The formula is:

\[S = \frac{a}{1-r}\]

In the context of the given series, we have:
  • The first term, \( a = 1 \)
  • Common ratio, \( r = -\frac{1}{2}(x-3) \)
Substituting these values, the sum \( S(x) \) becomes:

\[S(x) = \frac{1}{1 + \frac{1}{2}(x-3)} = \frac{2}{x-1}\]

This expression is valid for values of \( x \) that satisfy the interval \( 1 < x < 5 \). Within this interval, the sum will provide a reliable representation of the entire series. It is fascinating to observe how an infinite series can converge to such a neat equation!
Interval of Convergence
The interval of convergence for a series is the range of values for which the series will converge. For our geometric series to converge, the calculated conditions create the interval \( 1 < x < 5 \).

Here's how we derive this interval:
  • The inequality \( |x-3| < 2 \) is solved to find \(-2 < x-3 < 2\).
  • This simplifies to the interval \( 1 < x < 5 \).
Understanding the interval of convergence is key because it tells us where the series behaves nicely and approaches a calculable sum.

Outside this interval, the series does not converge, meaning it can grow larger without approaching a finite value. Thus, the interval of convergence defines the boundary within which our derived sum formula \( \frac{2}{x-1} \) is applicable and valid. Knowing this helps in interpreting solutions accurately and applying them to real-world scenarios where convergence is necessary for meaningful results.

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Most popular questions from this chapter

Use a CAS to perform the following steps for the sequences. a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L\) ? b. If the sequence converges, find an integer \(N\) such that \(\quad\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$ a_{n}=\sin n $$

Assume that the series \(\sum a_{n} x^{n}\) converges for \(x=4\) and diverges for \(x=7 .\) Answer true \((T),\) false \((F),\) or not enough information given \((\mathrm{N})\) for the following statements about the series. $$ \begin{array}{l}{\text { a. Converges absolutely for } x=-4} \\ {\text { b. Diverges for } x=5} \\ {\text { c. Converges absolutely for } x=-8.5} \\\ {\text { d. Converges absolutely for } x=-8.5} \\ {\text { e. Diverges for } x=8} \\ {\text { f. Diverges absolutely for } x=0} \\ {\text { g. Converges absolutely for } x=-7.1}\end{array} $$

When \(a\) and \(b\) are real, we define \(e^{(a+i b) x}\) with the equation \begin{equation}e^{(a+i b) x}=e^{a x} \cdot e^{i b x}=e^{a x}(\cos b x+i \sin b x).\end{equation} Differentiate the right-hand side of this equation to show that \begin{equation}\frac{d}{d x} e^{(a+i b) x}=(a+i b) e^{(a+i b) x}.\end{equation} Thus the familiar rule \((d / d x) e^{k x}=k e^{k x}\) holds for \(k\) complex as well as real.

According to the Alternating Series Estimation Theorem, how many terms of the Taylor series for tan \(^{-1} 1\) would you have to add to be sure of finding \(\pi / 4\) with an error of magnitude less than \(10^{-3}\) . Give reasons for your answer.

Quadratic Approximations The Taylor polynomial of order 2 generated by a twice-differentiable function \(f(x)\) at \(x=a\) is called the quadratic approximation of \(f\) at \(x=a .\) Find the (a) linearization (Taylor polynomial of order 1 ) and (b) quadratic approximation of \(f\) at \(x=0\). \(f(x)=\sin x\)

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