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Chapter 8: Problem 41

Write each repeating decimal first as a geometric series and then as a fraction (a ratio of two integers). $$0 . \overline{3}=0.333 \dots$$

Short Answer

Expert verified
Question: Write the repeating decimal 0.3虆 as a fraction. Answer: The repeating decimal 0.3虆 can be represented as the fraction 1/3.

Step by step solution

01

Write the repeating decimal as a geometric series

First, we will represent the repeating decimal \(0.\overline{3}\) as a geometric series. We can notice that: $$0.\overline{3} = 0.3 + 0.03 + 0.003 + \dots$$ This series has a common ratio of \(\frac{1}{10}\), as each term is obtained by multiplying the previous term by this factor. Therefore, the geometric series representation of \(0.\overline{3}\) is: $$0.\overline{3} = \sum_{n=1}^{\infty} 3\left(\frac{1}{10}\right)^n$$
02

Convert the geometric series to a fraction

Next, we will convert the geometric series into a fraction. To do this, we use the formula for the sum of an infinite geometric series: $$S = \frac{a}{1-r}$$ where \(S\) is the sum of the series, \(a\) is the first term, and \(r\) is the common ratio. In our case, \(a = 3 \cdot \left(\frac{1}{10}\right)\) and \(r = \frac{1}{10}\). Plugging these values into the formula, we get: $$0.\overline{3} = \frac{3\left(\frac{1}{10}\right)}{1-\left(\frac{1}{10}\right)}$$ To simplify the fraction, we first calculate the denominator: $$1 - \frac{1}{10} = \frac{9}{10}$$ Now, we can rewrite the fraction as: $$0.\overline{3} = \frac{3\left(\frac{1}{10}\right)}{\frac{9}{10}}$$ Multiplying the numerator and denominator by \(10\) to remove the decimal, we arrive at the final fraction representation: $$0.\overline{3} = \frac{3}{9}$$ We can simplify this fraction by dividing both the numerator and denominator by their greatest common divisor (3): $$0.\overline{3} = \frac{1}{3}$$ So, the repeating decimal \(0.\overline{3}\) can be represented as the fraction \(\frac{1}{3}\).

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Key Concepts

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

Repeating Decimal
A repeating decimal is a decimal number in which a digit or a group of digits repeats indefinitely. Understanding repeating decimals can help in converting them into fractions for more straightforward calculations.
  • Example: In the number \(0.\overline{3}\), the digit "3" is repeating.
  • Understanding repeating patterns is key. They can be represented with an overline indicating the repeated digits.
When working with repeating decimals, be sure to identify the repeating part of the number. Recognizing this pattern allows us to express it as a geometric series. Using geometric series, we can rewrite the repeating decimals into fractions, which can be more convenient to work with. Discovering this pattern is the first step toward a successful transformation into a fraction.
Fraction Conversion
Converting a repeating decimal into a fraction involves writing the decimal as a sum of an infinite series and then using a formula to find its sum. This process simplifies a repeating number into a comprehensible fraction that expresses the value exactly.
  • A repeating decimal, like \(0.\overline{3}\), becomes a series: \(0.3 + 0.03 + 0.003 + \ldots\)
  • This forms a geometric series, where the first term, \(a\), is \(3 \times \frac{1}{10}\), and the common ratio, \(r\), is \(\frac{1}{10}\).
The sum of this infinite geometric series, \(S\), is key. Using the formula \(S = \frac{a}{1-r}\), we can determine the fraction. The series \(0.\overline{3}\) thus converts into the fraction \(\frac{1}{3}\). The greatest common divisor (GCD) of the numerator and denominator can further simplify the fraction to its neatest form.
Infinite Series
Infinite series play a crucial role in expressing repeating decimals. By treating a repeating decimal as an infinite geometric series, we can mathematically sum the parts that appear endless. This is because each term in the series builds on and repeats the previous ones, making them infinite in nature.
  • The series for \(0.\overline{3}\) begins with \(3 \times \frac{1}{10}\) and progresses with each term multiplied by \(\frac{1}{10}\).
  • The geometric series formula, \(S = \frac{a}{1-r}\), is employed to find the sum of such a series.
By understanding the infinite nature of these series, one can better grasp how repeating decimals can transform into neat fractions. This use of infinite series allows for representing complex repeating decimal structures as simple ratios, aiding in both comprehension and practical math problems.

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Most popular questions from this chapter

Consider the alternating series $$ \sum_{k=1}^{\infty}(-1)^{k+1} a_{k}, \text { where } a_{k}=\left\\{\begin{array}{cl} \frac{4}{k+1}, & \text { if } k \text { is odd } \\ \frac{2}{k}, & \text { if } k \text { is even } \end{array}\right. $$ a. Write out the first ten terms of the series, group them in pairs, and show that the even partial sums of the series form the (divergent) harmonic series. b. Show that \(\lim _{k \rightarrow \infty} a_{k}=0\) c. Explain why the series diverges even though the terms of the series approach zero.

Stirling's formula Complete the following steps to find the values of \(p>0\) for which the series \(\sum_{k=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots(2 k-1)}{p^{k} k !}\) converges. a. Use the Ratio Test to show that \(\sum_{k=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots(2 k-1)}{p^{k} k !}\) converges for \(p>2\). b. Use Stirling's formula, \(k !=\sqrt{2 \pi k} k^{k} e^{-k}\) for large \(k,\) to determine whether the series converges when \(p=2\). (Hint: \(1 \cdot 3 \cdot 5 \cdots(2 k-1)=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdots(2 k-1) 2 k}{2 \cdot 4 \cdot 6 \cdots 2 k}\) (See the Guided Project Stirling's formula and \(n\) ? for more on this topic.)

In the following exercises, two sequences are given, one of which initially has smaller values, but eventually "overtakes" the other sequence. Find the sequence with the larger growth rate and the value of \(n\) at which it overtakes the other sequence. $$a_{n}=e^{n / 2} \text { and } b_{n}=n^{5}, n \geq 2$$

Determine whether the following statements are true and give an explanation or counterexample. a. \(\sum_{k=1}^{\infty}\left(\frac{\pi}{e}\right)^{-k}\) is a convergent geometric series. b. If \(a\) is a real number and \(\sum_{k=12}^{\infty} a^{k}\) converges, then \(\sum_{k=1}^{\infty} a^{k}\) converges. If the series \(\sum_{k=1}^{\infty} a^{k}\) converges and \(|a|<|b|,\) then the series \(\sum_{k=1}^{\infty} b^{k}\) converges. d. Viewed as a function of \(r,\) the series \(1+r^{2}+r^{3}+\cdots\) takes on all values in the interval \(\left(\frac{1}{2}, \infty\right)\) e. Viewed as a function of \(r,\) the series \(\sum_{k=1}^{\infty} r^{k}\) takes on all values in the interval \(\left(-\frac{1}{2}, \infty\right)\)

The fractal called the snowflake island (or Koch island ) is constructed as follows: Let \(I_{0}\) be an equilateral triangle with sides of length \(1 .\) The figure \(I_{1}\) is obtained by replacing the middle third of each side of \(I_{0}\) with a new outward equilateral triangle with sides of length \(1 / 3\) (see figure). The process is repeated where \(I_{n+1}\) is obtained by replacing the middle third of each side of \(I_{n}\) with a new outward equilateral triangle with sides of length \(1 / 3^{n+1}\). The limiting figure as \(n \rightarrow \infty\) is called the snowflake island. a. Let \(L_{n}\) be the perimeter of \(I_{n} .\) Show that \(\lim _{n \rightarrow \infty} L_{n}=\infty\) b. Let \(A_{n}\) be the area of \(I_{n} .\) Find \(\lim _{n \rightarrow \infty} A_{n} .\) It exists!
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