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Chapter 9: Problem 27

Determine whether or not the series converges, and if so, find its sum. $$ \sum_{n=0}^{\infty}(-1)^{n}(0.3)^{n} $$

Short Answer

Expert verified
The series converges, and its sum is \( \frac{10}{13} \).

Step by step solution

01

Identify the Type of Series

The given series \( \sum_{n=0}^{\infty}(-1)^{n}(0.3)^{n} \) is an alternating series, because each term includes \((-1)^n\), which causes the terms to alternate in sign.
02

Check for Alternating Series Conditions

For an alternating series \( \sum (-1)^n b_n \), the series converges if \( b_n \) is decreasing and \( \lim_{n \to \infty} b_n = 0 \). Here, \( b_n = (0.3)^n \). Since \( 0 < 0.3 < 1 \), \( b_n \) is decreasing and \( \lim_{n \to \infty} (0.3)^n = 0 \). Thus, the series converges by the Alternating Series Test.
03

Recognize Its Geometric Series Form

We notice the series can also be recognized as a geometric series \( \sum_{n=0}^{\infty}ar^n \) with \(a = 1\) and \(r = -0.3\). This is because the alternating factor \((-1)^n\) and the factor \((0.3)^n\) can be combined to produce the common ratio \(-0.3\).
04

Use Geometric Series Sum Formula

A geometric series \( \sum_{n=0}^{\infty} ar^n \) converges to the sum \( \frac{a}{1-r} \) if \(|r| < 1\). Here \(a = 1\) and \(r = -0.3\), so \(|-0.3| < 1\). Therefore, the sum is \( \frac{1}{1-(-0.3)} = \frac{1}{1+0.3} = \frac{1}{1.3} = \frac{10}{13} \).
05

Conclude Convergence and Find Sum

Since both the Alternating Series Test and the Geometric Series Test indicate convergence, the series converges. The sum of the series is \( \frac{10}{13} \).

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Key Concepts

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

Alternating Series Test
The alternating series test is a handy tool when it comes to identifying whether a series converges. To apply this test, a series should have terms that switch signs between positive and negative. This behavior is captured by the expression
  • \((-1)^n b_n\) for each term.
In our example
  • the series \( \sum_{n=0}^{\infty}(-1)^{n}(0.3)^{n} \) alternates due to the factor \((-1)^n\), causing each term to flip from positive to negative as \( n \) increases.
  • The next step is to check that \( b_n = (0.3)^n \) is decreasing and approaches zero as \( n \) becomes very large.
Since 0.3 is between 0 and 1, \( (0.3)^n \) indeed decreases and
  • \( \lim_{n \to \infty} (0.3)^n = 0 \).
Both conditions are fulfilled, so the series converges by the alternating series test.
Geometric Series
A geometric series is characterized by each term being a constant multiple, called the common ratio, of the previous term. Typically, a geometric series looks like
  • \( \sum_{n=0}^{\infty} ar^n \), where \( a \) is the first term and \( r \) is the common ratio.
In the given series
  • \( \sum_{n=0}^{\infty}(-1)^{n}(0.3)^{n} \), we can see it resembles a geometric series by rewriting the terms using the common ratio \(-0.3\).
Here
  • \( a = 1 \) and \( r = -0.3 \).
  • The alternation in sign played by \((-1)^n\) combines with \((0.3)^n\) to form this ratio \(-0.3\).
  • The crucial property for convergence in a geometric series is that the absolute value of the common ratio is less than 1, i.e., \( |r| < 1 \).
  • For our series, \( |-0.3| = 0.3 < 1 \), ensuring its convergence.
Series Sum Formula
Finding the sum of an infinite series might seem tricky at first, but the geometric series sum formula comes to the rescue. If a geometric series
  • \( \sum_{n=0}^{\infty} ar^n \)
converges, its sum is given by
  • \( \frac{a}{1-r} \).
Applying this to our series
  • where \( a = 1 \) and \( r = -0.3 \), we use the formula to find
  • \( \frac{1}{1 - (-0.3)} = \frac{1}{1 + 0.3} = \frac{1}{1.3} = \frac{10}{13} \).
  • Therefore, the sum of the series is \( \frac{10}{13} \).
This method efficiently provides the series' sum, validating our earlier verification of convergence through both the alternating series and geometric series checks.

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Most popular questions from this chapter

a. Find the numerical values of \(\left(\begin{array}{l}6 \\\ 2\end{array}\right)\) and \(\left(\begin{array}{c}1 / 2 \\\ 4\end{array}\right)\). b. Show that $$ \sum_{n=0}^{s}\left(\begin{array}{l} s \\ n \end{array}\right)=2^{s} $$ for any positive integer \(s\). (Hint: Use (2).) c. Show that $$ \sum_{n=0}^{s}\left(\begin{array}{l} s \\ n \end{array}\right)(-1)^{n}=0 $$ for any positive integer \(s\). (Hint: Use (2).)

Determine which series diverge, which converge conditionally, and which converge absolutely. $$ \sum_{n=2}^{\infty}(-1)^{n+1} \frac{1}{n(\ln n)^{2}} $$

Determine whether the series converges or diverges. $$ \sum_{n=1}^{\infty}(-1)^{n} \frac{(\ln n)^{2}}{n} $$

Express the given series \(\sum_{n=1}^{\infty} a_{n}\) in the form \(c+\sum_{n=4}^{\infty} a_{n}\) $$ \sum_{n=1}^{\infty} \frac{1}{n^{2}+1} $$

Use the power series expansion for \(\left(e^{x}-1\right) / x\) to verify that $$ \lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1 $$
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