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Chapter 9: Problem 52

Write each repeating decimal first as a geometric series and then as a fraction (a ratio of two integers). $$1.00 \overline{39}=1.00393939 \ldots$$

Short Answer

Expert verified
Question: Express the repeating decimal 1.00393939... as a fraction. Answer: The repeating decimal 1.00393939... can be expressed as the fraction \(\frac{3313}{3300}\).

Step by step solution

01

Identify the repeating part

The given repeating decimal is 1.00393939... . So, the repeating part is "39".
02

Express the repeating decimal as a geometric series

We can write the repeating decimal as the sum of its non-repeating part and the repeating part written as a geometric series: $$1.00\overline{39} = 1 + 0.0039 + 0.0039^2 + 0.0039^3 + \cdots$$ This geometric series has a first term "a" and a common ratio "r": $$a = 0.0039 \quad \text{and} \quad r = 10^{-2}$$
03

Find the sum of the geometric series

The formula for the sum S to infinity of a geometric series with first term a and common ratio r, where |r| < 1, is: $$S_\infty = \frac{a}{1 - r}$$ Substitute the values of a and r into the formula: $$S_\infty = \frac{0.0039}{1 - 10^{-2}}$$
04

Evaluate the sum

After evaluating the expression, we get the sum: $$S_\infty = \frac{0.0039}{0.99} \Rightarrow S_\infty = \frac{13}{3300}$$
05

Write the repeating decimal as a fraction

Now combine the non-repeating part and the sum of the geometric series: $$1.00\overline{39} = 1 + \frac{13}{3300} = \frac{3300}{3300} + \frac{13}{3300}$$ $$1.00\overline{39} = \frac{3313}{3300}$$ So, the repeating decimal \(1.00\overline{39}\) can be expressed as the fraction \(\frac{3313}{3300}\).

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Most popular questions from this chapter

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

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

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. $$a_{n+1}=\sqrt{2+a_{n}} ; a_{0}=1, n=0,1,2, \dots$$

For a positive real number \(p,\) how do you interpret \(p^{p^{p \cdot *}},\) where the tower of exponents continues indefinitely? As it stands, the expression is ambiguous. The tower could be built from the top or from the bottom; that is, it could be evaluated by the recurrence relations \(a_{n+1}=p^{a_{n}}\) (building from the bottom) or \(a_{n+1}=a_{n}^{p}(\text { building from the top })\) where \(a_{0}=p\) in either case. The two recurrence relations have very different behaviors that depend on the value of \(p\). a. Use computations with various values of \(p > 0\) to find the values of \(p\) such that the sequence defined by (2) has a limit. Estimate the maximum value of \(p\) for which the sequence has a limit. b. Show that the sequence defined by (1) has a limit for certain values of \(p\). Make a table showing the approximate value of the tower for various values of \(p .\) Estimate the maximum value of \(p\) for which the sequence has a limit.

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