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Chapter 9: Problem 46

Write each repeating decimal first as a geometric series and then as a fraction (a ratio of two integers). $$0 . \overline{27}=0.272727 \ldots$$

Short Answer

Expert verified
Answer: The fraction representation of the repeating decimal \(0.\overline{27}=0.272727\ldots\) is \(\frac{3}{11}\).

Step by step solution

01

Identify the repeating digits and period

In this exercise, the repeating decimal is \(0.\overline{27}=0.272727\ldots\). The repeating digits are "27", and the period (the number of digits in one repetition) is 2.
02

Write the decimal as a geometric series

We can write the repeating decimal as a geometric series with the common ratio \(r=10^{-2}=0.01\): $$0.272727\ldots = 0.27 + 0.27(0.01) + 0.27(0.01)^2 + \ldots$$ This can also be written as: $$0.272727\ldots = \sum_{n=0}^{\infty} 0.27(0.01)^n$$
03

Convert the geometric series into a fraction

To convert the geometric series into a fraction, we can use the formula for the sum of an infinite geometric series: $$\frac{a}{1-r}$$ In our case, the first term \(a=0.27\) and the common ratio \(r=0.01\). Plugging in these values, we get: $$\frac{0.27}{1-0.01}$$
04

Simplify and express the fraction as a ratio of integers

Now, we will simplify the fraction and express it as a ratio of integers: $$\frac{0.27}{0.99} = \frac{27}{99} = \frac{3 \times 9}{11 \times 9} = \frac{3}{11}$$ So, the fraction representation of the repeating decimal \(0.\overline{27}=0.272727\ldots\) is \(\frac{3}{11}\).

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Most popular questions from this chapter

a. Sketch the function \(f(x)=1 / x\) on the interval \([1, n+1]\) where \(n\) is a positive integer. Use this graph to verify that $$ \ln (n+1)<1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}<1+\ln n $$ b. Let \(S_{n}\) be the sum of the first \(n\) terms of the harmonic series, so part (a) says \(\ln (n+1)0,\) for \(n=1,2,3, \ldots\) c. Using a figure similar to that used in part (a), show that $$ \frac{1}{n+1}>\ln (n+2)-\ln (n+1) $$ d. Use parts (a) and (c) to show that \(\left\\{E_{n}\right\\}\) is an increasing sequence \(\left(E_{n+1}>E_{n}\right)\) e. Use part (a) to show that \(\left\\{E_{n}\right\\}\) is bounded above by 1 f. Conclude from parts (d) and (e) that \(\left\\{E_{n}\right\\}\) has a limit less than or equal to \(1 .\) This limit is known as Euler's constant and is denoted \(\gamma\) (the Greek lowercase gamma). g. By computing terms of \(\left\\{E_{n}\right\\},\) estimate the value of \(\gamma\) and compare it to the value \(\gamma \approx 0.5772 .\) (It has been conjectured, but not proved, that \(\gamma\) is irrational.) h. The preceding arguments show that the sum of the first \(n\) terms of the harmonic series satisfy \(S_{n} \approx 0.5772+\ln (n+1)\) How many terms must be summed for the sum to exceed \(10 ?\)

Evaluate the limit of the following sequences. $$a_{n}=\frac{n^{8}+n^{7}}{n^{7}+n^{8} \ln n}$$

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

The Riemann zeta function is the subject of extensive research and is associated with several renowned unsolved problems. It is defined by \(\zeta(x)=\sum_{k=1}^{\infty} \frac{1}{k^{x}}\). When \(x\) is a real number, the zeta function becomes a \(p\) -series. For even positive integers \(p,\) the value of \(\zeta(p)\) is known exactly. For example, $$ \sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}, \quad \sum_{k=1}^{\infty} \frac{1}{k^{4}}=\frac{\pi^{4}}{90}, \quad \text { and } \quad \sum_{k=1}^{\infty} \frac{1}{k^{6}}=\frac{\pi^{6}}{945}, \ldots $$ Use estimation techniques to approximate \(\zeta(3)\) and \(\zeta(5)\) (whose values are not known exactly) with a remainder less than \(10^{-3}\).

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. \(\sum_{k=0}^{\infty} 0.15^{k}\)
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