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Chapter 10: Problem 1

Identify a convergence test for each of the following series. If necessary, explain how to simplify or rewrite the series before applying the convergence test. You do not need to carry out the convergence test. $$\sum_{k=1}^{\infty}(-1)^{k}\left(2+\frac{1}{k^{2}}\right)^{k}$$

Short Answer

Expert verified
Question: Determine whether the given series converges or diverges: $$\sum_{k=1}^{\infty}(-1)^{k}\left(2+\frac{1}{k^{2}}\right)^{k}$$ Answer: The convergence or divergence of this series cannot be determined using the Alternating Series Test. Other convergence tests, such as the Ratio Test or the Root Test, need to be considered.

Step by step solution

01

Identify the structure of the series

The given series is: $$\sum_{k=1}^{\infty}(-1)^{k}\left(2+\frac{1}{k^{2}}\right)^{k}$$ The series consists of alternating signs and exponents. It is better to transform the series in order to expose the positive part of it, since this makes the choice of an appropriate convergence test easier.
02

Rewrite the series

Since the given series has alternating signs, we can rewrite the series as: $$\sum_{k=1}^{\infty} (-1)^k a_k$$ where \(a_k = \left(2+\frac{1}{k^{2}}\right)^{k}\). We need to find a convergence test for this modified series.
03

Identify the convergence test

For a series of the form \(\sum_{k=1}^{\infty} (-1)^k a_k\), the Alternating Series Test comes to mind. However, we need to check if the conditions of the Alternating Series Test are satisfied: 1. The terms \(a_k\) are positive and decreasing, i.e., \(a_k > a_{k+1}\) for all \(k\). 2. The limit of \(a_k\) as \(k\) goes to infinity is 0. That is, \(\lim_{k\to\infty} a_k = 0\).
04

Check the conditions for the Alternating Series Test

Let's examine the two conditions for the Alternating Series Test: 1. \(a_k = \left(2+\frac{1}{k^{2}}\right)^{k}\) is positive since \(k^2 \geq 1\) for all \(k \geq 1\), and \((2+\frac{1}{k^{2}})^{k} > 1\). To prove that the \(a_k\) terms are decreasing, we need to show that \(a_k > a_{k+1}\) for all \(k\). Observe that \(\frac{a_{k+1}}{a_k} = \frac{\left(2+\frac{1}{(k+1)^{2}}\right)^{k+1}}{\left(2+\frac{1}{k^{2}}\right)^{k}} = \left(\frac{2+\frac{1}{(k+1)^{2}}}{2+\frac{1}{k^{2}}}\right)^{k} \left( 2+\frac{1}{(k+1)^{2}}\right)\). Since \(k+1>k\), we have \(2+\frac{1}{(k+1)^{2}} < 2+\frac{1}{k^{2}}\) and \(0<\frac{a_{k+1}}{a_k}<1\). This implies that \(a_k>0\) and \(a_{k+1}

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Key Concepts

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

Alternating Series Test
The Alternating Series Test is a handy tool for determining the convergence of series that alternate between positive and negative terms. The series is often of the form \( \sum_{k=1}^{\infty} (-1)^k a_k \). This test requires that two conditions are met for the series:
  • The terms \( a_k \) must be positive and show a decreasing trend. This means that \( a_k > a_{k+1} \) for all \( k \).
  • The limit of \( a_k \) as \( k \) goes to infinity must be zero, i.e., \( \lim_{k\to\infty} a_k = 0 \).
Breaking this down, a positive \( a_k \) ensures that we account for the alternating sign in our series, capturing only the magnitude of terms. To visually confirm decreasing terms, you can perform a ratio construction, demonstrating \( \frac{a_{k+1}}{a_k} < 1 \).
For this exercise, the first criterion was met as the terms were shown to be decreasing. However, the second criterion failed as the limit went to infinity rather than zero. Therefore, the Alternating Series Test was not suitable here.
Ratio Test
The Ratio Test is a powerful method used to analyze the convergence of a series. With this test, you take the limit of the absolute value of the ratio of consecutive terms in the series:\[ L = \lim_{k\to\infty} \left| \frac{a_{k+1}}{a_k} \right| \]The rule of thumb for evaluating the limit \( L \) is straightforward:
  • If \( L < 1 \), the series converges absolutely.
  • If \( L > 1 \), the series diverges.
  • If \( L = 1 \), the test is inconclusive. Other methods must then be employed.
Applying the ratio test usually requires simplification of \( a_{k+1} \) and \( a_k \) into a form that allows easy evaluation of the limit. In the context of the original series, after rewriting and simplifying, the Ratio Test can be utilized to understand better whether the series converges or not.
Root Test
The Root Test, also known as the \( n \)-th Root Test, is another method used to determine convergence. This involves examining the \( n \)-th root of the absolute value of the terms in a series:\[ L = \lim_{k\to\infty} \sqrt[k]{|a_k|} \]In this method, if:
  • \( L < 1 \), the series converges.
  • \( L > 1 \), the series diverges.
  • \( L = 1 \), like the Ratio Test, it becomes inconclusive.
The Root Test is especially useful when each term in the series is raised to a power, or when integrating exponential expressions. In this original problem, noticing terms \( \left(2+\frac{1}{k^2}\right)^k \), the Root Test might be more appropriate since it highlights the impact of exponential growth and helps determine whether such growth leads to convergence or divergence.

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Most popular questions from this chapter

Explain why the magnitude of the remainder in an alternating series (with terms that are nonincreasing in magnitude) is less than or equal to the magnitude of the first neglected term.

Infinite products An infinite product \(P=a_{1} a_{2} a_{3} \ldots .,\) which is denoted \(\prod_{k=1}^{\infty} a_{k},\) is the limit of the sequence of partial products \(\left\\{a_{1}, a_{1} a_{2}, a_{1} a_{2} a_{3}, \ldots\right\\} .\) Assume \(a_{k} > 0\) for all \(k\) and \(L\) is a finite constant. a. Evaluate \(\prod_{k=1}^{\infty}\left(\frac{k}{k+1}\right)=\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdots\) b. Show that if \(\sum_{k=1}^{\infty} \ln a_{k}=L,\) then \(\prod_{k=1}^{\infty} a_{k}=e^{L}\) c. Use the result of part (b) to evaluate \(\prod_{k=0}^{\infty} e^{1 / 2^{t}}=e \cdot e^{1 / 2} \cdot e^{1 / 4} \cdot e^{1 / 8} \cdots\)

Binary numbers Humans use the ten digits 0 through 9 to form base- 10 or decimal numbers, whereas computers calculate and store numbers internally as binary numbers-numbers consisting entirely of O's and I's. For this exercise, we consider binary numbers that have the form \(0 . b_{1} b_{2} b_{3} \ldots\) where each of the digits \(b_{1}, b_{2}, b_{3}, \ldots\) is either O or 1 The base- 10 representation of the binary number \(0 . b_{1} b_{2} b_{3} \ldots\) is the infinite series \(\frac{b_{1}}{2^{1}}+\frac{b_{2}}{2^{2}}+\frac{b_{3}}{2^{3}}+\cdots\). Computers can store only a finite number of digits and therefore numbers with nonterminating digits must be rounded or truncated before they can be used and stored by a computer. a. Find the base- 10 representation of the binary number \(0 . \overline{0011}\) b. Suppose a computer rounds the binary number \(0 . \overline{0011}\) to \(b=0.0011001100110011\) before storing it. Find the approximate base- 10 representation of \(b,\) rounding your answer to 8 digits to the right of the decimal place.

Telescoping series For the following telescoping series, find a formula for the nth term of the sequence of partial sums \(\left\\{S_{n}\right\\} .\) Then evaluate lim \(S_{n}\) to obtain the value of the series or state that the series diverges. \(^{n \rightarrow \infty}\). $$\sum_{k=1}^{\infty}\left(\frac{1}{k+2}-\frac{1}{k+3}\right)$$

Suppose an alternating series with terms that are nonincreasing in magnitude converges to a value \(L\). Explain how to estimate the remainder that occurs when the series is terminated after \(n\) terms.
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