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Chapter 9: Problem 44

Write each repeating decimal first as a geometric series and then as a fraction (a ratio of two integers). $$0 . \overline{5}=0.555 \ldots$$

Short Answer

Expert verified
Question: Express the repeating decimal 0.555... as a fraction. Answer: The repeating decimal 0.555... can be expressed as the fraction 5/9.

Step by step solution

01

Identify the Repeating Decimal Pattern

In the given repeating decimal, we observe that the digit \(5\) repeats itself indefinitely: $$0 . \overline{5}=0.555 \ldots$$
02

Express the Repeating Decimal as a Geometric Series

To express the repeating decimal as a geometric series, we can start by representing the decimal with repeating digits as an infinite sum. Each term in this sum will be the digit multiplied by the common ratio, which is the place value of the digit (in this case, tenths, hundredths, and so on): $$0 . \overline{5}= 0.5 + 0.05 + 0.005 + \ldots$$ This can be written as a geometric series with the first term \(a\) as \(0.5\) and the common ratio \(r\) as \(0.1\): $$0 . \overline{5}= \sum_{n=0}^{\infty} a \times r^n = 0.5 \times (1 + 0.1 + 0.01 + \ldots)$$
03

Find the Sum of the Geometric Series

To find the sum of the geometric series, we will use the formula: $$S = \frac{a}{1-r}$$ In our case, \(a = 0.5\) and \(r = 0.1\). Plugging in these values: $$S = \frac{0.5}{1 - 0.1}$$
04

Simplify the Fraction

Now, we can simplify the fraction: $$S = \frac{0.5}{0.9}$$ Since we want to express the result as a ratio of two integers, we can multiply both the numerator and the denominator by \(10\), which will eliminate the decimals: $$S = \frac{0.5 \times 10}{0.9 \times 10} = \frac{5}{9}$$ So, the repeating decimal \(0 . \overline{5}\) can be expressed as a fraction, which is \(\frac{5}{9}\).

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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 \(b=10\).

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.)

An early limit Working in the early 1600 s, the mathematicians Wallis, Pascal, and Fermat were attempting to determine the area of the region under the curve \(y=x^{p}\) between \(x=0\) and \(x=1\) where \(p\) is a positive integer. Using arguments that predated the Fundamental Theorem of Calculus, they were able to prove that $$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n-1}\left(\frac{k}{n}\right)^{p}=\frac{1}{p+1}$$ Use what you know about Riemann sums and integrals to verify this limit.

Consider the following sequences defined by a recurrence relation. Use a calculator, analytical methods, and/or graphing to make a conjecture about the value of the limit or determine that the limit does not exist. $$a_{n+1}=2 a_{n}\left(1-a_{n}\right) ; a_{0}=0.3, n=0,1,2, \dots$$

Determine whether the following series converge absolutely or conditionally, or diverge. $$\sum_{k=1}^{\infty} \frac{(-1)^{k} k^{2}}{\sqrt{k^{6}+1}}$$
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