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Chapter 10: Problem 83

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 \(x .\) $$\sum_{n=0}^{\infty}(-1)^{n}(x+1)^{n}$$

Short Answer

Expert verified
The series converges for \(-2 < x < 0\) and the sum is \(\frac{1}{x+2}\).

Step by step solution

01

Identify Series Type

The series \( \sum_{n=0}^{\infty} (-1)^n (x+1)^n \) is a geometric series with first term \( a = 1 \) and common ratio \( r = -(x+1) \).
02

Determine Convergence Criteria

A geometric series converges if the absolute value of the common ratio \( |r| < 1 \). For our series, this means:\[ |-(x+1)| < 1 \]which simplifies to:\[ |x+1| < 1 \].
03

Solve the Inequality

Solve \( |x+1| < 1 \) to find the values of \( x \):\[ -1 < x+1 < 1 \]Subtract 1 from all parts of the inequality to get:\[ -2 < x < 0 \].
04

Find the Sum Formula

For a converging geometric series, the sum \( S \) is given by:\[ S = \frac{a}{1-r} \]Substitute \( a = 1 \) and \( r = -(x+1) \) into the formula:\[ S = \frac{1}{1 + (x+1)} = \frac{1}{x+2} \].

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Key Concepts

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

Convergence Criteria
When working with a geometric series, it is essential to determine whether the series converges or diverges. The criteria for convergence is a fundamental concept in understanding these types of series. Convergence of a geometric series is possible if the absolute value of its common ratio is less than one. In simpler terms, we need:
  • The absolute value of the common ratio, \(|r|\), to be less than 1.
For example, if the common ratio is given as \(- (x + 1)\), then the series converges if \(|-(x+1)|\) is smaller than 1. Once we solve this absolute inequality, we find the range of values for which the series converges. This convergence criterion helps ensure that adding terms of the series together results in a finite sum.
Common Ratio
In a geometric series, each term is a constant multiple of the previous term. This constant multiple is known as the common ratio, denoted by \(r\). Understanding the role of the common ratio is crucial because it controls the behavior of the series.
  • A common ratio \(|r| < 1\) results in a converging series.
  • A common ratio \(|r| \geq 1\) typically leads to divergence.
For the series given in the problem, the common ratio is \(-(x+1)\). This means each term in the series is obtained by multiplying the previous term by \(-(x+1)\). Recognizing the common ratio is the first step in applying convergence criteria and eventually determining the sum of the series.
Absolute Value Inequality
To find out if a geometric series converges, one often deals with absolute value inequalities. Absolute value inequalities appear when we apply convergence criteria.The expression \(|-(x+1)| < 1\) simplifies to \(|x+1| \). The absolute value signs indicate we are concerned with the distance from zero on the number line and not the direction. Solving \(|x+1| < 1\) gives us:
  • Break it into two inequalities: \(-1 < x+1 < 1\).
  • Solve each part to find \(-2 < x < 0\).
This solution tells us the range of \(x\) values that ensure the series converges. Understanding absolute value inequalities is key in analyzing any geometric series.
Sum Formula for Series
Once we establish that a geometric series converges, we can find its sum using a specific formula. This formula simplifies the process of summing an infinite number of terms.The formula to find the sum \(S\) of a convergent geometric series is:\[S = \frac{a}{1-r}\]Here, \(a\) is the first term, and \(r\) is the common ratio. When applying it to our series with \(a = 1\) and \(r = -(x+1)\), the sum becomes:\[S = \frac{1}{1 + (x+1)} = \frac{1}{x+2}\]This sum helps compute how all terms accumulate in a convergent geometric series. Knowing how to use this formula is essential for solving problems involving infinite series.

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Most popular questions from this chapter

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 \(x=0\). \(f(x)=e^{\sin x}\)

Sequences generated by Newton's method Newton's method, applied to a differentiable function \(f(x),\) begins with a starting value \(x_{0}\) and constructs from it a sequence of numbers \(\left\\{x_{n}\right\\}\) that under favorable circumstances converges to a zero of \(f .\) The recursion formula for the sequence is $$ x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} $$ a. Show that the recursion formula for \(f(x)=x^{2}-a, a>0\) can be written as \(x_{n+1}=\left(x_{n}+a / x_{n}\right) / 2\) b. Starting with \(x_{0}=1\) and \(a=3,\) calculate successive terms of the sequence until the display begins to repeat. What number is being approximated? Explain.

Which of the sequences converge, and which diverge? Give reasons for your answers. $$ a_{n}=\frac{2^{n}-1}{2^{n}} $$

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 \(L ?\) $$ a_{n}=\frac{\sin n}{n} $$

Pythagorean triples A triple of positive integers \(a, b,\) and \(c\) is called a Pythagorean triple if \(a^{2}+b^{2}=c^{2} .\) Let \(a\) be an odd positive integer and let $$ b=\left\lfloor\frac{a^{2}}{2}\right\rfloor \text { and } c=\left\lceil\frac{a^{2}}{2}\right\rceil $$ be, respectively, the integer floor and ceiling for \(a^{2} / 2\) a. Show that \(a^{2}+b^{2}=c^{2} .\) (Hint: Let \(a=2 n+1\) and express \(b\) and \(c\) in terms of \(n .\) b. By direct calculation, or by appealing to the accompanying figure, find $$ \lim _{a \rightarrow \infty} \frac{\left\lfloor\frac{a^{2}}{2}\right\rfloor}{ | \frac{a^{2}}{2} \rceil} $$
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