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Chapter 9: Problem 49

Write each repeating decimal first as a geometric series and then as a fraction (a ratio of two integers). $$0 . \overline{12}=0.121212 \ldots$$

Short Answer

Expert verified
Question: Convert the repeating decimal \(0.\overline{12}\) to a fraction. Answer: \(0.\overline{12} = \frac{12}{99}\)

Step by step solution

01

Define the repeating decimal as a variable x

Let x be the repeating decimal we want to convert, i.e.: $$x = 0.121212\ldots$$
02

Write the repeating decimal as an infinite geometric series

To write the decimal as a geometric series, we need to recognize the repeating pattern (here, 12) and understand the position of each digit. In this case, every repeating pattern starts at the tenths place and hundredths place. Thus, we can write the decimal as a sum of fractions like this: $$x = \frac{12}{100}+\frac{12}{10000}+\frac{12}{10^6}+\cdots$$ This is an infinite geometric series with first term \(a= \frac{12}{100}\), common ratio \(r= \frac{1}{100}\) and number of terms \(n\to\infty\).
03

Find the sum of the geometric series

To find the sum of an infinite geometric series, we can use the formula: $$S = \frac{a}{1-r}$$ Here, we have \(a=\frac{12}{100}\), and \(r=\frac{1}{100}\). Plugging these into the formula, we have: $$S = \frac{\frac{12}{100}}{1-\frac{1}{100}} = \frac{\frac{12}{100}}{\frac{99}{100}}$$
04

Simplify the fraction and find the sum

To simplify the fraction, we need to multiply both the numerator and the denominator by their lowest common denominator (LCD). In this case, the LCD is 100. Thus, $$S = \frac{\frac{12}{100}}{\frac{99}{100}}\cdot\frac{100}{100} = \frac{12}{99}$$
05

Write the repeating decimal as a fraction

Now we have found the sum of the infinite geometric series, which is the same as x, the repeating decimal. Therefore, $$0.\overline{12} = \frac{12}{99}$$

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Most popular questions from this chapter

The sequence \(\\{n !\\}\) ultimately grows faster than the sequence \(\left\\{b^{n}\right\\},\) for any \(b > 1,\) as \(n \rightarrow \infty .\) However, \(b^{n}\) is generally greater than \(n !\) for small values of \(n\). Use a calculator to determine the smallest value of \(n\) such that \(n ! > b^{n}\) for each of the cases \(b=2, b=e,\) and \(b=10\).

Suppose a function \(f\) is defined by the geometric series \(f(x)=\sum_{k=0}^{\infty} x^{2 k}\) a. Evaluate \(f(0), f(0.2), f(0.5), f(1),\) and \(f(1.5),\) if possible. b. What is the domain of \(f ?\)

Determine whether the following series converge absolutely or conditionally, or diverge. $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1} e^{k}}{(k+1) !}$$

Convergence parameter Find the values of the parameter \(p>0\) for which the following series converge. $$\sum_{k=2}^{\infty} \frac{1}{k \ln k(\ln \ln k)^{p}}$$

Suppose an alternating series \(\sum(-1)^{k} a_{k}\) converges to \(S\) and the sum of the first \(n\) terms of the series is \(S_{n}\) Suppose also that the difference between the magnitudes of consecutive terms decreases with \(k\). It can be shown that for \(n \geq 1,\) $$\left|S-\left[S_{n}+\frac{(-1)^{n+1} a_{n+1}}{2}\right]\right| \leq \frac{1}{2}\left|a_{n+1}-a_{n+2}\right|$$ a. Interpret this inequality and explain why it gives a better approximation to \(S\) than simply using \(S_{n}\) to approximate \(S\). b. For the following series, determine how many terms of the series are needed to approximate its exact value with an error less than \(10^{-6}\) using both \(S_{n}\) and the method explained in part (a). (i) \(\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k}\) (ii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{k \ln k}\) (iii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\sqrt{k}}\)
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