Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

Write the given decimal as an infinite series, then find the sum of the series, and finally, use the result to write the decimal as a ratio of two integers (see Example 2). $$ 0.21212121 \ldots $$

Step-by-Step Solution

Answer

The decimal 0.21212121... can be written as the ratio \(\frac{7}{33}\).

Step 1: Identify the repeating decimal

The given decimal is \(0.21212121 \ldots\). It has a repeating block of \("21"\). This means we can express the decimal as a repeating series: \(0.21212121 \ldots = 0.21 + 0.0021 + 0.000021 + \ldots\)

Step 2: Express as a geometric series

We notice this series is geometric. The first term \(a\) is \(0.21\), and the common ratio \(r\) is \(0.01\), as each term is \(100\) times smaller than the previous term.

Step 3: Write the series formula

The sum of an infinite geometric series can be expressed with the formula \(S = \frac{a}{1 - r}\), where \(a\) is the first term and \(r\) is the common ratio. In this case, \(a = 0.21\) and \(r = 0.01\).

Step 4: Calculate the sum of the series

Using the formula \(S = \frac{a}{1 - r}\), we have:\[S = \frac{0.21}{1 - 0.01} = \frac{0.21}{0.99}\]

Step 5: Simplify the fraction

To simplify \(\frac{0.21}{0.99}\), multiply both the numerator and the denominator by \(100\) to eliminate the decimals:\[\frac{0.21 \times 100}{0.99 \times 100} = \frac{21}{99}\]Simplify \(\frac{21}{99}\) by dividing both the numerator and the denominator by their greatest common divisor, which is \(3\):\[\frac{21 \div 3}{99 \div 3} = \frac{7}{33}\]

Step 6: Conclusion

The repeating decimal \(0.21212121 \ldots\) can be written as the infinite series \(0.21 + 0.0021 + 0.000021 + \ldots\), with a sum of \(\frac{7}{33}\), thus it can be expressed as the ratio \(\frac{7}{33}\).

Key Concepts

Geometric Series

A geometric series is a series with a constant ratio between successive terms. It is expressed in the form:

• First term, \(a\); and
• Common ratio, \(r\).

Consider the infinite series system, such as the decimal representation in the problem. A geometric series can be smoothly converted into a formula that helps find its sum. This approach is the key when dealing with repeating decimals, as they inherently constitute a geometric sequence.

In our exercise, the decimal \( 0.212121\ldots \) can be rewritten as a geometric series \( 0.21 + 0.0021 + 0.000021 + \ldots \) with:

• First term \(a = 0.21\)
• Common ratio \(r = 0.01\)

Notice how each term is progressively smaller, multiplied by a factor of \(0.01\) compared to its predecessor. The geometric series is powerful because it provides a straightforward way to calculate infinite sums when these two parameters are known.

Repeating Decimal

Repeating decimals, like \( 0.212121\ldots \), are decimals in which a sequence of digits repeats indefinitely. In this case, the repeating block is "21". Repeating decimals can always be expressed as fractions, which means they have a finite representation in terms of the ratio of two integers.

When you encounter a repeating decimal in mathematics, converting it into a fraction typically involves identifying the repeating sequence (in our case, "21") and recognizing it as part of an infinite geometric series. This understanding allows us to sum the series, using the geometric series sum formula, and then express the repeated decimal sequence as a more manageable ratio.

The insight into the repeat pattern is crucial in transforming the infinite decimals into a series, a step that bridges the path to conversion into fractional form.

Sum of Series

To find the sum of an infinite geometric series where the series converges, apply the sum formula:\[S = \frac{a}{1 - r}\]where \(S\) is the sum, \(a\) is the first term, and \(r\) is the common ratio. For our problem's repeating decimal:\(0.21 + 0.0021 + 0.000021 + \ldots\), the parameters are:

• \(a = 0.21\)
• \(r = 0.01\)

Using the formula, substituting the values gives:\[S = \frac{0.21}{1 - 0.01} = \frac{0.21}{0.99}\]To express this in a simplified form, remove decimals by scaling:\[\frac{0.21 \times 100}{0.99 \times 100} = \frac{21}{99}\]Simplifying further by dividing both numerator and denominator by their greatest common divisor (3) results in:\[\frac{7}{33}\]Therefore, the repeating decimal translates to a simple and neat fraction, \(\frac{7}{33}\), making it easy to handle and understand both mathematically and practically. This approach not only provides a sum but also provides a bridge between decimals and fractions.