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Chapter 9: Problem 54

Which series converge, and which diverge? Give reasons for your answers. If a series converges, find its sum. $$\sum_{n=0}^{\infty} \frac{\cos n \pi}{5^{n}}$$

Short Answer

Expert verified
The series converges and its sum is \( \frac{5}{6} \).

Step by step solution

01

Understand the Series

The given series is \( \sum_{n=0}^{\infty} \frac{\cos n \pi}{5^{n}} \). Notice that \( \cos n \pi \) alternates between 1 and -1 as \( n \) changes, thus focusing only on the denominator \( 5^{n} \).
02

Recognize the Alternating Series

Since \( \cos n \pi = (-1)^{n} \), the series can be rewritten as \( \sum_{n=0}^{\infty} \frac{(-1)^{n}}{5^{n}} \). This is an alternating series because the term changes sign with each successive term.
03

Check Convergence with Alternating Series Test

An alternating series \( \sum_{n=0}^{\infty} (-1)^n a_n \) converges if \( a_n \) is decreasing and \( \lim_{n \to \infty} a_n = 0 \). Here, \( a_n = \frac{1}{5^{n}} \), which is a decreasing sequence approaching 0 as \( n \) approaches infinity.
04

Determine the Sum Form

Since the series is a form of geometric series \( \sum_{n=0}^{\infty} ar^n \) with \( a = 1 \) and \( r = -\frac{1}{5} \), its sum when \(|r| < 1\) is given by \( \frac{a}{1-r} \).
05

Calculate the Sum of the Series

Apply the sum formula for a geometric series: \( \text{Sum} = \frac{1}{1-(-\frac{1}{5})} = \frac{1}{1+\frac{1}{5}} = \frac{1}{\frac{6}{5}} = \frac{5}{6} \). Thus, the series converges to \( \frac{5}{6} \).

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Key Concepts

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

Alternating Series
An alternating series is a series whose terms alternate between positive and negative. This happens due to the presence of a factor like \((-1)^n\) in each term. Such a factor changes the sign of each term based on whether \(n\) is even or odd. For example, consider the series \sum_{n=0}^{\infty} \frac{(-1)^{n}}{5^{n}}\:\
  • When \(n\) is even, \((-1)^n = 1\), so the term is positive.
  • When \(n\) is odd, \((-1)^n = -1\), so the term flips and becomes negative.
The alternating nature of such series allows them to be analyzed specifically for their convergence properties. In particular, we use the Alternating Series Test to determine if they converge.
Geometric Series
A geometric series is characterized by each term being a constant multiple, or ratio, of the previous term. Its general form is \sum_{n=0}^{\infty} ar^n\. Here, \(a\) is the first term, and \(r\) is the common ratio. An important property of geometric series is its convergence criteria:
  • The series converges if the absolute value of the common ratio is less than 1, i.e., \|r| < 1\.
  • If it converges, the sum of the series is given by \frac{a}{1-r}\.
In the given exercise, the series \sum_{n=0}^{\infty} \frac{(-1)^{n}}{5^{n}}\ can be rewritten in the geometric series form: \sum_{n=0}^{\infty} \left(-\frac{1}{5} \right)^n\. Here, \(a = 1\) and \(r = -\frac{1}{5}\). Since `|r|` is \(\frac{1}{5},\) which is less than 1, the series converges.
Alternating Series Test
The Alternating Series Test is a powerful tool for determining the convergence of alternating series. This test tells us that an alternating series \sum_{n=0}^{\infty} (-1)^n a_n\ converges if two conditions are met:
  • The sequence \((a_n)\) is decreasing, meaning each term is smaller than the previous one.
  • As \(n\) approaches infinity, the term \(a_n\) approaches zero precisely, i.e., \lim_{n \to \infty} a_n = 0\.
In the exercise at hand, we observe that \(a_n = \frac{1}{5^n}\), which satisfies both conditions:
  • \(\frac{1}{5^n}\) is a decreasing sequence because as \(n\) increases, \(5^n\) grows, making each fraction smaller.
  • Furthermore, the limit \lim_{n \to \infty} \frac{1}{5^{n}} = 0\, confirming the convergence of the series.
Thus, using the Alternating Series Test, we affirm that the series converges.

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Most popular questions from this chapter

Is it true that a sequence \(\left\\{a_{n}\right\\}\) of positive numbers must converge if it is bounded from above? Give reasons for your answer.

Use the definition of convergence to prove the given limit. $$\lim _{n \rightarrow \infty}\left(1-\frac{1}{n^{2}}\right)=1$$

Use Taylor's formula with \(a=0\) and \(n=3\) to find the standard cubic approximation of \(f(x)=\) \(1 /(1-x)\) at \(x=0 .\) Give an upper bound for the magnitude of the error in the approximation when \(|x| \leq 0.1\).

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

a. Assuming that \(\lim _{n \rightarrow \infty}\left(1 / n^{c}\right)=0\) if \(c\) is any positive constant, show that $$\lim _{n \rightarrow \infty} \frac{\ln n}{n f}=0$$ if \(c\) is any positive constant. b. Prove that \(\lim _{n \rightarrow \infty}\left(1 / n^{c}\right)=0\) if \(c\) is any positive constant. (Hint: If \(\epsilon=0.001\) and \(c=0.04\), how large should \(N\) be to ensure that \(\left|1 / n^{c}-0\right|<\epsilon\) if \(n>N ?\) )
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