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Chapter 9: Problem 50

Write each repeating decimal first as a geometric series and then as a fraction (a ratio of two integers). $$1 . \overline{25}=1.252525 \ldots$$

Short Answer

Expert verified
Question: Convert the repeating decimal 1.25 (with the 25 repeating) into a fraction. Answer: The repeating decimal 1.25 (with the 25 repeating) can be expressed as the fraction 124/99.

Step by step solution

01

Identify the repeating part and the non-repeating part

In the given decimal, the non-repeating part is 1 and the repeating part is 25. The objective is to convert the repeating decimal $$1.\overline{25}$$ into a fraction.
02

Write the repeating decimal as a sum

We can write the repeating decimal as a sum of the non-repeating part and the repeating part: $$1.\overline{25}=1+\overline{0.25}$$
03

Write the repeating part as a geometric series

We can write the repeating part as a geometric series where the first term (a) is 0.25 and the common ratio (r) is 0.01: $$\overline{0.25}=0.25\cdot\frac{1-0.01^{n-1}}{1-0.01}, \quad n \to \infty$$ Note that, as n approaches infinity, the factor $$0.01^{n-1}$$ becomes negligibly small, so we have: $$\overline{0.25}=0.25\cdot\frac{1}{1-0.01}$$
04

Find the simplified fraction for the repeating part

Now, we'll simplify the fraction $$\frac{0.25}{1-0.01}$$: $$\frac{0.25}{1-0.01}=\frac{0.25}{0.99}=\frac{25}{99}$$
05

Add the non-repeating part to the simplified fraction

Finally, add the non-repeating part (1) to the simplified fraction: $$1.\overline{25}=1+\frac{25}{99}$$
06

Express the sum as a single fraction

To express the sum as a single fraction, we can find a common denominator and perform the addition: $$1+\frac{25}{99}=\frac{99}{99}+\frac{25}{99}=\frac{124}{99}$$
07

Write the final answer

The repeating decimal $$1.\overline{25}$$ can be expressed as the fraction $$\frac{124}{99}$$

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Most popular questions from this chapter

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. \(\sum_{k=0}^{\infty} 0.72^{k}\) b. \(\sum_{k=0}^{\infty}(-0.25)^{k}\)

The prime numbers are those positive integers that are divisible by only 1 and themselves (for example, 2,3,5 7, \(11,13, \ldots .\) ). A celebrated theorem states that the sequence of prime numbers \(\left\\{p_{k}\right\\}\) satisfies \(\lim _{k \rightarrow \infty} p_{k} /(k \ln k)=1 .\) Show that \(\sum_{k=2}^{\infty} \frac{1}{k \ln k}\) diverges, which implies that the series \(\sum_{k=1}^{\infty} \frac{1}{p_{k}}\) diverges.

Determine how many terms of the following convergent series must be summed to be sure that the remainder is less than \(10^{-4} .\) Although you do not need it, the exact value of the series is given in each case. $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{(2 k+1) !}$$

Determine whether the following series converge or diverge. $$\sum_{k=2}^{\infty} \frac{4}{k \ln ^{2} k}$$

Find the limit of the sequence $$\left\\{a_{n}\right\\}_{n=2}^{\infty}=\left\\{\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right) \cdots\left(1-\frac{1}{n}\right)\right\\}.$$
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