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Chapter 9: Problem 43

Use the properties of infinite series to evaluate the following series. $$\sum_{k=1}^{\infty} \frac{4}{12^{k}}$$

Short Answer

Expert verified
Answer: The sum of the given infinite geometric series is $\frac{4}{11}$.

Step by step solution

01

Determine the common ratio

The given series is $$\sum_{k=1}^{\infty} \frac{4}{12^k}$$ Let's identify the common ratio \(r\). In a geometric series, the ratio between consecutive terms is constant. Thus, we can find the common ratio by dividing the second term by the first term: $$r = \frac{\frac{4}{12^2}}{\frac{4}{12^1}} = \frac{1}{12}$$ The common ratio is \(\frac{1}{12}\).
02

Check for convergence

For an infinite geometric series to converge, the absolute value of the common ratio must be less than 1: $$|r| < 1$$ In this case, \(|\frac{1}{12}| < 1\), so the series converges.
03

Apply the sum formula

Since we know the series converges, we can apply the infinite geometric series sum formula: $$S = \frac{a}{1-r}$$ where \(S\) is the sum of the series, \(a\) is the first term, and \(r\) is the common ratio. For our series, the first term \(a = \frac{4}{12^1} = \frac{1}{3}\), and the common ratio \(r = \frac{1}{12}\). Plugging these values into the formula, we get: $$S = \frac{\frac{1}{3}}{1-\frac{1}{12}}$$ To simplify the expression, let's find a common denominator for the fractions inside the parenthesis. Since 3 and 12 both divide by 12, let's convert both fractions to have a denominator of 12: $$S = \frac{\frac{4}{12}}{1-\frac{1}{12}} = \frac{\frac{4}{12}}{\frac{11}{12}}$$
04

Simplify the expression

To simplify the expression for the sum, we can divide the fractions by multiplying the numerator by the reciprocal of the denominator: $$S = \frac{4}{12} \cdot \frac{12}{11}$$ The factor of 12 in the numerator and denominator can be canceled: $$S = \frac{4}{11}$$ So, the sum of the given infinite geometric series is \(\frac{4}{11}\).

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Most popular questions from this chapter

Use Exercise 89 to determine how many terms of each series are needed so that the partial sum is within \(10^{-6}\) of the value of the series (that is, to ensure \(R_{n}<10^{-6}\) ). a. \(\sum_{k=0}^{\infty} 0.6^{k}\) b. \(\sum_{k=0}^{\infty} 0.15^{k}\)

A well-known method for approximating \(\sqrt{c}\) for positive real numbers \(c\) consists of the following recurrence relation (based on Newton's method). Let \(a_{0}=c\) and $$a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{c}{a_{n}}\right), \quad \text { for } n=0,1,2,3, \dots$$ a. Use this recurrence relation to approximate \(\sqrt{10} .\) How many terms of the sequence are needed to approximate \(\sqrt{10}\) with an error less than \(0.01 ?\) How many terms of the sequence are needed to approximate \(\sqrt{10}\) with an error less than \(0.0001 ?\) (To compute the error, assume a calculator gives the exact value.) b. Use this recurrence relation to approximate \(\sqrt{c},\) for \(c=2\) \(3, \ldots, 10 .\) Make a table showing how many terms of the sequence are needed to approximate \(\sqrt{c}\) with an error less than \(0.01.\)

Given that \(\sum_{k=1}^{\infty} \frac{1}{k^{4}}=\frac{\pi^{4}}{90},\) show that \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{4}}=\frac{7 \pi^{4}}{720}.\) (Assume the result of Exercise 63.)

Prove that if \(\sum a_{k}\) diverges, then \(\sum c a_{k}\) also diverges, where \(c \neq 0\) is a constant.

Determine whether the following series converge absolutely or conditionally, or diverge. $$\sum_{k=1}^{\infty} \frac{\cos k}{k^{3}}$$
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