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Chapter 10: Problem 7

In Exercises \(7-14,\) write out the first eight terms of each series to show how the series starts. Then find the sum of the series or show that it diverges. $$ \sum_{n=0}^{\infty} \frac{(-1)^{n}}{4^{n}} $$

Short Answer

Expert verified
The series converges to \( \frac{4}{5} \).

Step by step solution

01

Identify the General Term

The general term of the series is given by \( a_n = \frac{(-1)^n}{4^n} \). This term takes alternating positive and negative values depending on \( n \). We will use this to find the first eight terms of the series.
02

Compute the First Eight Terms

Substitute \( n = 0, 1, 2, \ldots, 7 \) into the general term \( a_n \) to get the first eight terms:\[\begin{align*}a_0 &= \frac{(-1)^0}{4^0} = 1,\a_1 &= \frac{(-1)^1}{4^1} = -\frac{1}{4},\a_2 &= \frac{(-1)^2}{4^2} = \frac{1}{16},\a_3 &= \frac{(-1)^3}{4^3} = -\frac{1}{64},\a_4 &= \frac{(-1)^4}{4^4} = \frac{1}{256},\a_5 &= \frac{(-1)^5}{4^5} = -\frac{1}{1024},\a_6 &= \frac{(-1)^6}{4^6} = \frac{1}{4096},\a_7 &= \frac{(-1)^7}{4^7} = -\frac{1}{16384}.\end{align*}\]
03

Determine the Type of Series

The series appears to alternate in sign, which suggests it might be an alternating geometric series. Recognizing the pattern, this series is a geometric series with the first term \( a = 1 \) and common ratio \( r = -\frac{1}{4} \).
04

Check for Convergence

A geometric series converges if the absolute value of the common ratio \( |r| < 1 \). In this case, \( |r| = \left| -\frac{1}{4} \right| = \frac{1}{4} < 1 \), so the series converges.
05

Calculate the Sum of the Series

The sum \( S \) of an infinite geometric series is given by the formula:\[S = \frac{a}{1 - r}\]where \( a \) is the first term and \( r \) is the common ratio. Substituting into the formula:\[S = \frac{1}{1 - \left(-\frac{1}{4}\right)} = \frac{1}{1 + \frac{1}{4}} = \frac{1}{\frac{5}{4}} = \frac{4}{5}\]The series converges with a sum of \( \frac{4}{5} \).

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Key Concepts

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

Convergence of Series
In mathematics, a series is a sum of the terms of a sequence. For an infinite series, we look at the sum of an unending sequence of terms. To determine whether this sum reaches a finite limit, we consider the concept of convergence.

A series converges if the sequence of partial sums approaches a particular value as the number of terms increases. For a geometric series, this happens when the absolute value of the common ratio, denoted as \( |r| \), is less than 1. If this condition is met, the series converges, meaning it adds up to a finite number.

For example, if the series has a common ratio \( r \) such that \( |r| < 1 \), we conclude that it converges. In the case of our series, \( r = -\frac{1}{4} \), and since \( |-\frac{1}{4}| = \frac{1}{4} < 1 \), the series is confirmed to converge.
Alternating Series
An alternating series is a series where the terms change sign alternately. This means some terms are positive, and others are negative, typically flipping back and forth following a specific pattern. The given series is an excellent example of an alternating series, as the terms switch between positive and negative based on the mathematical expression \((-1)^n\).

Alternating series are important because they provide specific convergence criteria. According to the Alternating Series Test, an alternating series converges if two conditions are met:
  • The absolute values of the terms decrease as you progress through the series.
  • The values of the terms tend toward zero as \(n\) increases.
For the series in question, both these conditions are satisfied. The terms indeed become smaller and their absolute values get ever closer to zero with increasing \(n\).
Sum of Infinite Series
The ability to sum an infinite series to a finite number is fascinating and directly relates to the convergence of that series. The given mathematical example is a geometric series, which allows us to find the sum if it converges.

For a convergent geometric series with a starting term \(a\) and common ratio \(r\), the sum \(S\) of the infinite series can be calculated using the formula:

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

In our series, \(a = 1\) and \(r = -\frac{1}{4}\). Plugging these into the formula gives:

\[ S = \frac{1}{1 - \left(-\frac{1}{4}\right)} = \frac{1}{1 + \frac{1}{4}} = \frac{1}{\frac{5}{4}} = \frac{4}{5} \]

Thus, the series converges to a sum of \(\frac{4}{5}\), demonstrating how infinite components can collectively result in a finite value.

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Most popular questions from this chapter

Show that \(\sum_{n=1}^{\infty} 2^{\left(n^{2}\right)} / n !\) diverges. Recall from the Laws of Exponents that \(2^{\left(n^{2}\right)}=\left(2^{n}\right)^{n}\)

Use series to estimate the integrals' values with an error of magnitude less than \(10^{-5}\) . (The answer section gives the integrals' values rounded to seven decimal places.) \begin{equation} \int_{0}^{0.5} \frac{1}{\sqrt{1+x^{4}}} d x \end{equation}

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 \(x=0\). \(f(x)=1 / \sqrt{1-x^{2}}\)

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \tan \frac{1}{n}\end{equation}

Use series to evaluate the limits. \begin{equation} \lim _{x \rightarrow 2} \frac{x^{2}-4}{\ln (x-1)} \end{equation}
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