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Chapter 8: Problem 46

Write each of the given repeating decimals as a constant times a geometric series (the geometric series will contain powers of 0.1 ). Use the formula for the sum of a geometric series to express the repeating decimal as a rational number. $$ 0.131313131313 \ldots $$

Short Answer

Expert verified
The repeating decimal \(0.131313\ldots\) is equal to \( \frac{13}{99} \).

Step by step solution

01

Identifying the Pattern

The given decimal is \( 0.13131313\ldots \). We observe that the pattern \( 13 \) repeats indefinitely. This is a repeating decimal where each two-digit block (13) repeats.
02

Expressing as a Geometric Series

Rewrite \( 0.131313\ldots \) by recognizing it as a sum of each of these repeating blocks placed at subsequent positions:\[0.131313\ldots = 0.13 + 0.0013 + 0.000013 + \ldots\]Each term can be written as \( 0.13 \times (0.01)^n \) where \( n = 0, 1, 2, \ldots \). So it takes the form of a geometric series \( 0.13 \times \sum_{n=0}^{ ext{鈭瀩}(0.01)^n \).
03

Applying the Geometric Series Sum Formula

The formula for the sum of an infinite geometric series is \( S = \frac{a}{1 - r} \), where \( a \) is the first term and \( r \) is the common ratio. In our series, \( a = 1 \) and \( r = 0.01 \). Hence:\[S = \frac{1}{1 - 0.01} = \frac{1}{0.99} = \frac{100}{99}\]So, the sum of the series is \( \frac{100}{99} \).
04

Calculating the Decimal as a Rational Number

Now, substitute back with the constant \( 0.13 \):\[0.131313\ldots = 0.13 \times \frac{100}{99} = \frac{13 \times 100}{99 \times 100} = \frac{13}{99}.\]This rational number represents the repeating decimal.

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Key Concepts

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

Geometric Series
A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. In the context of repeating decimals, we convert the decimal into a sum of terms that form such a sequence.
For example, consider the repeating decimal 0.131313... To define its structure, we note that the digits "13" are repeating indefinitely. We can express this decimal as a geometric series:
  • The first term (or initial term) of the series is related to the repeating block, which in this example is 0.13.
  • The series can then be written as a sum of terms like 0.13, 0.0013, 0.000013, and so on.
  • Each subsequent term is a product of multiplying 0.13 by powers of 0.01, showcasing the pattern of a geometric series.
This conversion is crucial for seeing how repeating decimals relate to geometric series, which leads to expressing them as rational numbers.
Rational Numbers
A rational number is any number that can be expressed as the quotient or fraction of two integers, where the numerator is an integer and the denominator is a non-zero integer. Repeating decimals are often represented as rational numbers due to their periodic nature.
In our example with the repeating decimal 0.131313..., by converting it into a geometric series, we can also express it as a rational number. The step-by-step derivation leads us to the fraction:
  • Initially, we represent the repeating decimal as a geometric series with a clear formula for its sum.
  • By applying the geometric series sum formula, we find the exact fraction that matches the repeating pattern.
  • In this case, this results in the fraction \( \frac{13}{99} \), which is a rational expression representing the original decimal.
Understanding this property helps in converting seemingly complex repeating decimals into simpler rational forms.
Geometric Series Sum Formula
The geometric series sum formula is a powerful tool to find the sum of an infinite geometric series. The formula is given by \( S = \frac{a}{1 - r} \), where \( a \) is the first term and \( r \) is the common ratio. This formula applies when the absolute value of \( r \) is less than one.
To apply this to converting repeating decimals, let's look at how it's done for 0.131313...:
  • Here, the first term \( a \) is 1, and the common ratio \( r \) is 0.01.
  • Substituting into the formula gives \( S = \frac{1}{1 - 0.01} = \frac{1}{0.99} \).
  • Compute the fraction to get \( \frac{100}{99} \), showing how the repeating decimal can be summed as a rational number.
This formula helps simplify the process of dealing with repeating decimals and gives a precise method to express them as fractions, showcasing the blend between infinite series and rational numbers.

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Most popular questions from this chapter

In each of Exercises 66-71, evaluate the given series (exactly). $$ \frac{\pi}{4}-\frac{\pi^{3}}{4^{3} \cdot 3 !}+\frac{\pi^{5}}{4^{5} \cdot 5 !}-\frac{\pi^{7}}{4^{7} \cdot 7 !}+\cdots $$

Suppose that \(f^{\prime \prime}\) is continuous on an open interval centered at \(c .\) Use Taylor's Theorem to show that $$ \lim _{h \rightarrow 0} \frac{f(c+h)-2 f(c)+f(c-h)}{h^{2}}=f^{\prime \prime}(c) $$

In each of Exercises 49-54, use Taylor series to calculate the given limit. $$ \lim _{x \rightarrow 0} \frac{1-e^{2 x}}{x} $$

Use partial fractions to calculate the \(N^{\text {th }}\) partial sum \(S_{N}\) of the given series in closed form. Sum the series by finding \(\lim _{N \rightarrow \infty} S_{N}\). $$ \sum_{n=1}^{\infty} \frac{1}{n(n+4)} $$

Find the sum of the given series in closed form. State the radius of convergence \(R\). \(\sum_{n=1}^{\infty} n(n+1) x^{n}\)
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