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Chapter 10: Problem 77

In Exercises \(57 - 82 ,\) use any method to determine whether the series converges or diverges. Give reasons for your answer. $$ \sum _ { n = 2 } ^ { \infty } \frac { 1 } { 1 + 2 + 2 ^ { 2 } + \cdots + 2 ^ { n } } $$

Short Answer

Expert verified
The series converges by the Limit Comparison Test.

Step by step solution

01

Recognize the Series Type

Observe that the series is defined as \[\sum _ { n = 2 } ^ { \infty } \frac { 1 } { 1 + 2 + 2 ^ { 2 } + \cdots + 2 ^ { n } }.\] The denominator is a geometric series that sums up to the power of 2. This is crucial in determining convergence or divergence.
02

Simplify the Denominator

The denominator is a geometric series:\[1 + 2 + 2^2 + \cdots + 2^n.\]Using the formula for sum of a geometric series, \(a + ar + ar^2 + \cdots + ar^n = a \frac{r^{n+1} - 1}{r - 1}\), where \(a = 1\) and \(r = 2\), the sum is \(2^{n+1} - 1\). So, the denominator simplifies to \(2^{n+1} - 1\).
03

Re-express the Series

Substituting the simplified denominator, the series becomes:\[\sum _ { n = 2 } ^ { \infty } \frac { 1 }{2^{n+1} - 1}.\]This is crucial to understand how the series behaves as \(n\) approaches infinity.
04

Apply the Limit Comparison Test

Next, compare the series to \(\sum _ { n = 2 } ^ { \infty } \frac { 1 }{2^{n+1}} = \sum _ { n = 2 } ^ { \infty } \frac { 1 }{2 \cdot 2^{n}} = \frac{1}{2} \sum _ { n = 2 } ^ { \infty } \frac { 1 }{2^n}\).We know that \(\sum _ { n = 2 } ^ { \infty } \frac { 1 }{2^n}\) is a convergent geometric series. Therefore, check:\[\lim_{{n \to \infty}} \frac{\frac{1}{2^{n+1} - 1}}{\frac{1}{2^{n+1}}} = \lim_{{n \to \infty}} \frac{2^{n+1}}{2^{n+1} - 1} \approx 1.\]Since the limit is a positive finite number, both series converge or diverge together.
05

Conclude the Series' Behavior

Based on the Limit Comparison Test, since \(\sum _ { n = 2 } ^ { \infty } \frac { 1 }{2^n}\) converges, the given series\[\sum _ { n = 2 } ^ { \infty } \frac { 1 }{2^{n+1} - 1}\]also converges. This definitively determines the series' behavior.

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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 series with a constant ratio between successive terms. For instance, the series \(1 + 2 + 2^2 + \cdots + 2^n\) is a geometric series where the first term \(a\) is 1 and the common ratio \(r\) is 2. Geometric series can be finite or infinite, and there is a simple formula for their sum. If the series is finite, its sum can be calculated with the formula:
  • Sum = \(a \frac{r^{n+1} - 1}{r - 1}\)
where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the last term's exponent. In our original problem, this formula helps simplify the denominator from \(1 + 2 + 2^2 + \cdots + 2^n\) to \(2^{n+1} - 1\). Understanding how to sum a geometric series enables us to re-express the series for further analysis, assisting in evaluating convergence or divergence.
Limit Comparison Test
The limit comparison test is a valuable tool for determining the convergence or divergence of series, especially when dealing with complex expressions. It compares the series in question with another known series:
  • Suppose \(\sum a_n\) is our series.
  • Choose \(\sum b_n\) as a comparison series that is known to converge or diverge.
The test involves evaluating the following limit:\[\lim_{{n \to \infty}} \frac{a_n}{b_n}\]If this limit is a positive finite number, both series \(\sum a_n\) and \(\sum b_n\) converge or diverge together. In this exercise, we compared the re-expressed series with \(\sum \frac{1}{2^n}\), a known convergent geometric series. The limit was found to be approximately 1, confirming that the original series converges by this test.
Infinite Series
Infinite series allow us to explore sums that continue indefinitely. However, unlike finite series, not all infinite series have a finite sum. Understanding whether an infinite series converges (approaches a certain value) or diverges is crucial in mathematics.
  • "Convergence" means the series approaches a specific value as the number of terms increases.
  • "Divergence" means the series does not settle near any single value.
In our problem, by simplifying the series and using the limit comparison test, we conclude that the given series converges. This insight is pivotal for fields such as calculus and mathematical analysis, where infinite series often model real-world situations or complex mathematical functions.

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Most popular questions from this chapter

Which of the sequences converge, and which diverge? Give reasons for your answers. $$ a_{n}=\frac{2^{n}-1}{3^{n}} $$

Estimating Pi The English mathematician Wallis discovered the formula \begin{equation}\frac{\pi}{4}=\frac{2 \cdot 4 \cdot 4 \cdot 6 \cdot 6 \cdot 8 \cdot \cdots}{3 \cdot 3 \cdot 5 \cdot 5 \cdot 7 \cdot 7 \cdot \cdots}\end{equation} Find \(\pi\) to two decimal places with this formula.

Is it true that a sequence \(\left\\{a_{n}\right\\}\) of positive numbers must converge if it is bounded from above? Give reasons for your answer.

Area Consider the sequence \(\\{1 / n\\}_{n=1}^{\infty}\) . On each subinterval \((1 /(n+1), 1 / n)\) within the interval \([0,1],\) erect the rectangle with area \(a_{n}\) having height 1\(/ n\) and width equal to the length of the subinterval. Find the total area \(\sum a_{n}\) of all the rectangles. (Hint: Use the result of Example 5 in Section \(10.2 . )\)

Assume that the series \(\sum a_{n}(x-2)^{n}\) converges for \(x=-1\) and diverges for \(x=6 .\) Answer true \((\mathrm{T})\) , false \((\mathrm{F})\) , or not enough information given \((\mathrm{N})\) for the following statements about the series. $$ \begin{array}{l}{\text { a. Converges absolutely for } x=1} \\ {\text { b. Diverges for } x=-6} \\ {\text { c. Diverges for } x=2} \\ {\text { d. Converges for } x=0} \\ {\text { e. Converges absolutely for } x=5} \\\ {\text { f. Diverges for } x=4.9} \\ {\text { g. Diverges for } x=5.1} \\\ {\text { a. Diverges for } x=5.1} \\ {\text { h. Converges absolutely for } x=4}\end{array} $$
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