Problem 63
Question
Decimal numbers Any real number in the interval \([0,1]\) can be represented by a decimal (not necessarily unique) as \begin{equation}0 . d_{1} d_{2} d_{3} d_{4} \ldots=\frac{d_{1}}{10}+\frac{d_{2}}{10^{2}}+\frac{d_{3}}{10^{3}}+\frac{d_{4}}{10^{4}}+\cdots,\end{equation} where \(d_{i}\) is one of the integers \(0,1,2,3, \ldots, 9 .\) Prove that the series on the right-hand side always converges.
Step-by-Step Solution
Verified Answer
The decimal series converges by the Comparison Test with a convergent geometric series.
1Step 1: Understand the Series
The series given is a decimal expansion, typically written as \( 0.d_1d_2d_3d_4 \ldots \), which breaks down into the sum \( \frac{d_1}{10} + \frac{d_2}{10^2} + \frac{d_3}{10^3} + \ldots \). Here, each \( d_i \) is a digit from 0 to 9. We need to show that this series converges.
2Step 2: Recognize the Structure of the Series
The series is an infinite series of the form \( \sum_{i=1}^{\infty} \frac{d_i}{10^i} \). We notice that each term is positive (since \( d_i \geq 0 \) and the denominators are positive) and the terms decrease in magnitude due to the increasing power of 10 in the denominator.
3Step 3: Establish Convergence Using the Comparison Test
To prove convergence, we compare this series with a geometric series. We know \( d_i \leq 9 \), so each term \( \frac{d_i}{10^i} \leq \frac{9}{10^i} \). Consider the geometric series \( \sum_{i=1}^{\infty} \frac{9}{10^i} \), which converges because the ratio \( r = \frac{1}{10} < 1 \). By the Comparison Test, since \( \frac{d_i}{10^i} \leq \frac{9}{10^i} \), the original series \( \sum_{i=1}^{\infty} \frac{d_i}{10^i} \) also converges.
4Step 4: Conclusion of Convergence
Since we have shown that each term of the series is bounded by a term in a convergent geometric series, we can conclude that the series \( \sum_{i=1}^{\infty} \frac{d_i}{10^i} \) converges.
Key Concepts
Convergence of SeriesComparison TestGeometric Series
Convergence of Series
When we talk about the convergence of a series, it means determining whether the sum of infinitely many numbers results in a finite number. Think of it like adding numbers forever and checking if the total settles at a certain point. If this happens, the series converges. One way to check this is to look at the limit of the partial sums. If the partial sums approach a limit as the number of terms goes to infinity, the series is convergent. This concept is essential when dealing with decimal expansions, as it ensures that the infinite sum of terms actually represents a real number. In the case of the decimal expansion we are discussing, the series converges because the terms get smaller and smaller, approaching zero, as the powers of 10 in the denominator increase.
Comparison Test
The comparison test is a handy tool used to determine the convergence of a series. It's like checking if one series behaves the same way as another, easier-to-understand series. If each term of the series you're investigating is less than or equal to the corresponding term of a known convergent series, then your series also converges.
For example, in the decimal expansion series, each term is \( \frac{d_i}{10^i} \), where \( d_i \) is a digit between 0 and 9. We compare it to \( \frac{9}{10^i} \), a term from a convergent geometric series. Since \( \frac{d_i}{10^i} \) is less than \( \frac{9}{10^i} \), and we know that the summation of \( \sum_{i=1}^{\infty} \frac{9}{10^i} \) converges, this ensures our original series does too. Simple yet effective!
For example, in the decimal expansion series, each term is \( \frac{d_i}{10^i} \), where \( d_i \) is a digit between 0 and 9. We compare it to \( \frac{9}{10^i} \), a term from a convergent geometric series. Since \( \frac{d_i}{10^i} \) is less than \( \frac{9}{10^i} \), and we know that the summation of \( \sum_{i=1}^{\infty} \frac{9}{10^i} \) converges, this ensures our original series does too. Simple yet effective!
Geometric Series
A geometric series is one where each term is a constant multiple of the previous term. It's like a pattern of numbers that grow or shrink based on the multiplication of a fixed number, called the common ratio. For instance, in the geometric series \( 1 + r + r^2 + r^3 + \ldots \), the common ratio is \( r \). A key characteristic is that it converges if the absolute value of the common ratio is less than 1.
In our case, we examined the geometric series \( \sum_{i=1}^{\infty} \frac{9}{10^i} \). The common ratio here is \( \frac{1}{10} \), which is less than 1, assuring us of convergence. By relating decimal series with this geometric series, we use its properties to demonstrate how the original series converges, confirming that any decimal expansion in the interval [0, 1] represents a real number.
In our case, we examined the geometric series \( \sum_{i=1}^{\infty} \frac{9}{10^i} \). The common ratio here is \( \frac{1}{10} \), which is less than 1, assuring us of convergence. By relating decimal series with this geometric series, we use its properties to demonstrate how the original series converges, confirming that any decimal expansion in the interval [0, 1] represents a real number.
Other exercises in this chapter
Problem 62
Show that the sum of the first 2\(n\) terms of the series $$1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\f
View solution Problem 62
Convergence or Divergence Which of the series in Exercises \(55-62\) converge, and which diverge? Give reasons for your answers. $$\sum_{n=1}^{\infty} \frac{1 \
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Neither the Ratio Test nor the Root Test helps with \(p\) -series. Try them on $$\sum_{n=1}^{\infty} \frac{1}{n^{p}}$$ and show that both tests fail to provide
View solution Problem 63
Which series in Exercises \(49-68\) converge, and which diverge? Give reasons for your answers. If a series converges, find its sum. $$ \sum_{n=1}^{\infty} \fra
View solution