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Chapter 8: Problem 7

The associative and distributive laws of addition allow us to add finite sums in any order we want. That is, if \(\sum_{k=0}^{n} a_{k}\) and \(\sum_{k=0}^{n} b_{k}\) are finite sums of real numbers, then $$ \sum_{k=0}^{n} a_{k}+\sum_{k=0}^{n} b_{k}=\sum_{k=0}^{n}\left(a_{k}+b_{k}\right) $$ However, we do need to be careful extending rules like this to infinite series. a. Let \(a_{n}=1+\frac{1}{2^{n}}\) and \(b_{n}=-1\) for each nonnegative integer \(n\). \- Explain why the series \(\sum_{k=0}^{\infty} a_{k}\) and \(\sum_{k=0}^{\infty} b_{k}\) both diverge. \- Explain why the series \(\sum_{k=0}^{\infty}\left(a_{k}+b_{k}\right)\) converges. \- Explain why $$ \sum_{k=0}^{\infty} a_{k}+\sum_{k=0}^{\infty} b_{k} \neq \sum_{k=0}^{\infty}\left(a_{k}+b_{k}\right) $$ This shows that it is possible to have to two divergent series \(\sum_{k=0}^{\infty} a_{k}\) and \(\sum_{k=0}^{\infty} b_{k}\) but yet have the series \(\sum_{k=0}^{\infty}\left(a_{k}+b_{k}\right)\) converge. b. While part (a) shows that we cannot add series term by term in general, we can under reasonable conditions. The problem in part (a) is that we tried to add divergent series. In this exercise we will show that if \(\sum a_{k}\) and \(\sum b_{k}\) are convergent series, then \(\sum\left(a_{k}+b_{k}\right)\) is a convergent series and $$ \sum\left(a_{k}+b_{k}\right)=\sum a_{k}+\sum b_{k} $$ - Let \(A_{n}\) and \(B_{n}\) be the \(n\) th partial sums of the series \(\sum_{k=1}^{\infty} a_{k}\) and \(\sum_{k=1}^{\infty} b_{k}\), respectively. Explain why $$ A_{n}+B_{n}=\sum_{k=1}^{n}\left(a_{k}+b_{k}\right) $$ \- Use the previous result and properties of limits to show that $$ \sum_{k=1}^{\infty}\left(a_{k}+b_{k}\right)=\sum_{k=1}^{\infty} a_{k}+\sum_{k=1}^{\infty} b_{k} . $$ (Note that the starting point of the sum is irrelevant in this problem, so it doesn't matter where we begin the sum.) c. Use the prior result to find the sum of the series \(\sum_{k=0}^{\infty} \frac{2^{k}+3^{k}}{5^{k}}\).

Short Answer

Expert verified
The sum of \( \sum_{k=0}^{\infty} \frac{2^k + 3^k}{5^k} \) is \( \frac{25}{6} \).

Step by step solution

01

Determine Divergence of \ \(\sum_{k=0}^{\infty} a_{k}\ \)

Given that \( a_{k} = 1 + \frac{1}{2^k} \), as \( k \) goes to infinity, \(a_k\) does not approach zero. For a series to converge, its terms must approach zero. Hence, \( \sum_{k=0}^{\infty} a_{k} \) diverges.
02

Determine Divergence of \ \(\sum_{k=0}^{\infty} b_{k}\ \)

Given \( b_{k} = -1 \), each term in the series is constant and not approaching zero. Thus, \( \sum_{k=0}^{\infty} b_{k} \) also diverges.
03

Check Convergence of the Series \ \(\sum_{k=0}^{\infty} (a_{k} + b_{k})\ \)

Combine the terms: \( a_k + b_k = (1 + \frac{1}{2^k}) + (-1) = \frac{1}{2^k} \). Since \( \frac{1}{2^k} \to 0 \) as \( k \to \infty \) and resembles a geometric series with ratio \( \frac{1}{2} \), \( \sum_{k=0}^{\infty} \frac{1}{2^k} \) converges.
04

Show \ \(\sum_{k=0}^{\infty} a_{k} + \sum_{k=0}^{\infty} b_{k} \ eq \ \sum_{k=0}^{\infty} (a_{k} + b_{k})\ \)

Although both \( \sum_{k=0}^{\infty} a_{k} \) and \( \sum_{k=0}^{\infty} b_{k} \) diverge, \( \sum_{k=0}^{\infty}(a_{k} + b_{k}) \) converges. Therefore, \( \sum_{k=0}^{\infty} a_{k} + \sum_{k=0}^{\infty} b_{k} \) is not equal to \( \sum_{k=0}^{\infty} (a_{k} + b_{k}) \).
05

Demonstrate Sum Formula

For two convergent series \( \sum a_{k} \) and \( \sum b_{k} \), the n-th partial sums are \( A_{n} \) and \( B_{n} \). It follows that \( A_{n} + B_{n} = \sum_{k=1}^{n} (a_{k} + b_{k}) \).
06

Apply Limit Properties

As partial sums \( A_{n} \) and \( B_{n} \) converge to \( \sum_{k=1}^{\infty} a_{k} \) and \( \sum_{k=1}^{\infty} b_{k} \), respectively, the limit of \( A_{n} + B_{n} \) converges to \( \sum_{k=1}^{n} (a_{k}+b_{k}) \). Hence, \[ \sum_{k=1}^{\infty} (a_{k} + b_{k}) = \sum_{k=1}^{\infty} a_{k} + \sum_{k=1}^{\infty} b_{k} \]
07

Find the Sum of \ \( \sum_{k=0}^{\infty} \frac{2^{k} + 3^{k}}{5^{k}} \ \)

Rewrite the series: \( \sum_{k=0}^{\infty} \frac{2^{k} + 3^{k}}{5^{k}} \) as \( \sum_{k=0}^{\infty} \left(\frac{2^{k}}{5^{k}} + \frac{3^{k}}{5^{k}}\right) \). This splits into two geometric series: \( \sum_{k=0}^{\infty} \left(\frac{2}{5}\right)^{k} + \sum_{k=0}^{\infty} \left(\frac{3}{5}\right)^{k} \). Sum of each geometric series is: \[ \sum_{k=0}^{\infty} \left(\frac{2}{5}\right)^{k} = \frac{1}{1-\frac{2}{5}} = \frac{5}{3} \] \[ \sum_{k=0}^{\infty} \left(\frac{3}{5}\right)^{k} = \frac{1}{1-\frac{3}{5}} = \frac{5}{2} \]. Therefore, the total is \( \frac{5}{3} + \frac{5}{2} = \frac{10}{6} + \frac{15}{6} = \frac{25}{6} \).

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Key Concepts

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

Associative Law
The associative law is a fundamental property of addition and multiplication. It states that when adding or multiplying three or more numbers, the result remains the same regardless of how the numbers are grouped. In mathematical terms, for any numbers \(a\), \(b\), and \(c\), we have:
  • \text{Addition}: \text{(a + b) + c = a + (b + c)}
  • \text{Multiplication}: \text{(a \times b) \times c = a \times (b \times c)}

This property tells us that we can rearrange the grouping of the terms while performing the operation without affecting the outcome. It is important to note that the associative law applies specifically to finite sums and products. However, when dealing with infinite series, the associative property does not always hold. This subtle difference is crucial to understand to avoid incorrect conclusions.
Distributive Law
The distributive law connects addition and multiplication and explains how multiplication is distributed over addition. For any numbers \(a\), \(b\), and \(c\), the distributive law states:
  • a \times (b + c) = (a \times b) + (a \times c)

This law is essential because it allows us to break down complex expressions into simpler parts. For example, if you need to calculate \(3 \times (4 + 5)\), you can use the distributive law to first multiply 3 by 4 and 3 by 5 and then add the results: \(3 \times (4 + 5) = (3 \times 4) + (3 \times 5) = 12 + 15 = 27\). The distributive law is always valid for finite sums. However, similar to the associative law, caution is required when dealing with infinite series.
Partial Sums
A partial sum is the sum of the first \(n\) terms of a series. For a given series \(\text{\text{sum}}_{k=1}^{\text{\text{infinity}}}a_k\), the partial sum \(S_n\) is defined as:
  • S_n \text{ = } \text{\text{sum}}_{k=1}^{n}a_k

Partial sums are useful in studying the behavior of infinite series. If the sequence of partial sums converges to a specific number as \(n\) increases, the series is said to converge. In step-by-step solutions, partial sums help simplify the problem by focusing on finite sums that are more manageable. They provide insight into whether an entire infinite series will converge. The concept of partial sums is central to understanding the convergence and divergence of series.
Divergent Series
A series is called divergent if its sequence of partial sums does not converge to a finite limit. This means that the sums keep growing larger without approaching any specific value. A divergent series is characterized by:
  • The terms not approaching zero, or
  • The partial sums approaching infinity or oscillating without settling down to a finite value.

For example, the series \(\text{\text{sum}}_{k=0}^{\text{\text{infinity o( 1 + \frac{1}{2}}^k)\) diverges because its terms do not get arbitrarily small. Understanding why a series diverges is essential in determining whether certain operations like addition are permissible. In the exercise provided, \(\text{\textorg.sum}}_{k=0}^{\text{\textinf }a_k}\) and \(\text {\text sum }}_{k=0% _{\text{inf}} b_k}\) both diverge, illustrating that we must be careful about extending the associative law to infinite series.
Geometric Series
A geometric series is a series in which each term is obtained by multiplying the previous term by a fixed, non-zero number called the common ratio. A geometric series can be expressed as:
  • \(a + ar + ar^2 + ar^3 + ...\)

where \(a\) is the first term and \(r\) is the common ratio. Geometric series are significant because they have a straightforward formula for the sum when the series converges. If \(|r|<1\), the sum of an infinite geometric series is given by: \text{S} = \frac{a}{1-r}\(
In the example \)\text {\text sum }}_{k=0 \text {[{inf }}{ ( \frac{2^{k}+3^{k}}{5^{k})\( in the original exercise, both embedded geometric series \)\text {\text sum} (frac {2^}{5\() _{k}\) and { (frac {5}}{(frac {3 5)_{k}$ converge, facilitating the determination of the overall sum using the geometric series sum formula. Understanding geometric series and their properties simplified solving the problem steps.

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Most popular questions from this chapter

In the Limit Comparison Test we compared the behavior of a series to one whose behavior we know. In that test we use the limit of the ratio of corresponding terms of the series to determine if the comparison is valid. In this exercise we see how we can compare two series directly, term by term, without using a limit of sequence. First we consider an example. a. Consider the series \(\sum \frac{1}{k^{2}}\) and \(\sum \frac{1}{k^{2}+k}\) We know that the series \(\sum \frac{1}{k^{2}}\) is a \(p\) -series with \(p=2>1\) and so \(\sum \frac{1}{k^{2}}\) converges. In this part of the exercise we will see how to use information about \(\sum \frac{1}{k^{2}}\) to determine information about \(\sum \frac{1}{k^{2}+k} .\) Let \(a_{k}=\frac{1}{k^{2}}\) and \(b_{k}=\frac{1}{k^{2}+k} .\) i) Let \(S_{n}\) be the \(n\) th partial sum of \(\sum \frac{1}{k^{2}}\) and \(T_{n}\) the \(n\) th partial sum of \(\sum \frac{1}{k^{2}+k}\). Which is larger, \(S_{1}\) or \(T_{1}\) ? Why? ii) Recall that $$ S_{2}=S_{1}+a_{2} \text { and } T_{2}=T_{1}+b_{2} $$ Which is larger, \(a_{2}\) or \(b_{2}\) ? Based on that answer, which is larger, \(S_{2}\) or \(T_{2}\) ? iii) Recall that $$ S_{3}=S_{2}+a_{3} \text { and } T_{3}=T_{2}+b_{3} $$ Which is larger, \(a_{3}\) or \(b_{3}\) ? Based on that answer, which is larger, \(S_{3}\) or \(T_{3}\) ? iv) Which is larger, \(a_{n}\) or \(b_{n}\) ? Explain. Based on that answer, which is larger, \(S_{n}\) or \(T_{n}\) ? v) Based on your response to the previous part of this exercise, what relationship do you expect there to be between \(\sum \frac{1}{k^{2}}\) and \(\sum \frac{1}{k^{2}+k} ?\) Do you expect \(\sum \frac{1}{k^{2}+k}\) to converge or diverge? Why? b. The example in the previous part of this exercise illustrates a more general result. Explain why the Direct Comparison Test, stated here, works.

Suppose you play a game with a friend that involves rolling a standard six- sided die. Before a player can participate in the game, he or she must roll a six with the die. Assume that you roll first and that you and your friend take alternate rolls. In this exercise we will determine the probability that you roll the first six. a. Explain why the probability of rolling a six on any single roll (including your first turn) is \(\frac{1}{6}\) b. If you don't roll a six on your first turn, then in order for you to roll the first six on your second turn, both you and your friend had to fail to roll a six on your first turns, and then you had to succeed in rolling a six on your second turn. Explain why the probability of this event is $$ \left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)=\left(\frac{5}{6}\right)^{2}\left(\frac{1}{6}\right) $$ c. Now suppose you fail to roll the first six on your second turn. Explain why the probability is $$ \left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)=\left(\frac{5}{6}\right)^{4}\left(\frac{1}{6}\right) $$ that you to roll the first six on your third turn. d. The probability of you rolling the first six is the probability that you roll the first six on your first turn plus the probability that you roll the first six on your second turn plus the probability that your roll the first six on your third turn, and so on. Explain why this probability is $$ \frac{1}{6}+\left(\frac{5}{6}\right)^{2}\left(\frac{1}{6}\right)+\left(\frac{5}{6}\right)^{4}\left(\frac{1}{6}\right)+\cdots $$ Find the sum of this series and determine the probability that you roll the first six.

We can use known Taylor series to obtain other Taylor series, and we explore that idea in this exercise, as a preview of work in the following section. a. Calculate the first four derivatives of \(\sin \left(x^{2}\right)\) and hence find the fourth order Taylor polynomial for \(\sin \left(x^{2}\right)\) centered at \(a=0\) b. Part (a) demonstrates the brute force approach to computing Taylor polynomials and series. Now we find an easier method that utilizes a known Taylor series. Recall that the Taylor series centered at 0 for \(f(x)=\sin (x)\) is $$ \sum_{k=0}^{\infty}(-1)^{k} \frac{x^{2 k+1}}{(2 k+1) !} $$ i. Substitute \(x^{2}\) for \(x\) in the Taylor series \((8.5 .7) .\) Write out the first several terms and compare to your work in part (a). Explain why the substitution in this problem should give the Taylor series for \(\sin \left(x^{2}\right)\) centered at \(0 .\) ii. What should we expect the interval of convergence of the series for \(\sin \left(x^{2}\right)\) to be? Explain in detail.

Finding limits of convergent sequences can be a challenge. However, there is a useful tool we can adapt from our study of limits of continuous functions at infinity to use to find limits of sequences. We illustrate in this exercise with the example of the sequence $$ \frac{\ln (n)}{n} $$ a. Calculate the first 10 terms of this sequence. Based on these calculations, do you think the sequence converges or diverges? Why? b. For this sequence, there is a corresponding continuous function \(f\) defined by $$ f(x)=\frac{\ln (x)}{x} $$ Draw the graph of \(f(x)\) on the interval [0,10] and then plot the entries of the sequence on the graph. What conclusion do you think we can draw about the sequence \(\left\\{\frac{\ln (n)}{n}\right\\}\) if \(\lim _{x \rightarrow \infty} f(x)=L ?\) Explain. c. Note that \(f(x)\) has the indeterminate form \(\frac{\infty}{\infty}\) as \(x\) goes to infinity. What idea from differential calculus can we use to calculate \(\lim _{x \rightarrow \infty} f(x) ?\) Use this method to find \(\lim _{x \rightarrow \infty} f(x) .\) What, then, is \(\lim _{n \rightarrow \infty} \frac{\ln (n)}{n} ?\)

Based on the examples we have seen, we might expect that the Taylor series for a function \(f\) always converges to the values \(f(x)\) on its interval of convergence. We explore that idea in more detail in this exercise. Let \(f(x)=\left\\{\begin{array}{ll}e^{-1 / x^{2}} & \text { if } x \neq 0, \\ 0 & \text { if } x=0 .\end{array}\right.\) a. Show, using the definition of the derivative, that \(f^{\prime}(0)=0\). b. It can be shown that \(f^{(n)}(0)=0\) for all \(n \geq 2\). Assuming that this is true, find the Taylor series for \(f\) centered at 0 . c. What is the interval of convergence of the Taylor series centered at 0 for \(f ?\) Explain. For which values of \(x\) the interval of convergence of the Taylor series does the Taylor series converge to \(f(x) ?\)
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