Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 8: Problem 66

Use the test of your choice to determine whether the following series converge. $$\sum_{k=2}^{\infty} 100 k^{-k}$$

Short Answer

Expert verified
Answer: The Ratio Test is inconclusive in this case, so it's not possible to determine the convergence of the given series. Other tests might be helpful in determining the convergence of the series.

Step by step solution

01

Write down the general term

The general term of the series is given by: $$a_k = 100 k^{-k}$$
02

Apply the Ratio Test

The Formula for the Ratio Test is: $$\lim_{k \rightarrow \infty} \left|\frac{a_{k+1}}{a_k}\right|$$ Now we will substitute the general term into the Ratio Test formula: $$\lim_{k \rightarrow \infty} \left|\frac{100(k+1)^{-(k+1)}}{100 k^{-k}}\right|$$
03

Simplify the expression

We can cancel out the "100" from both the numerator and denominator: $$\lim_{k \rightarrow \infty} \left|\frac{(k+1)^{-(k+1)}}{k^{-k}}\right|$$ Rewrite negative exponents as reciprocals: $$\lim_{k \rightarrow \infty} \left|\frac{\frac{1}{(k+1)^{(k+1)}}}{\frac{1}{k^{k}}}\right|$$ Invert and multiply: $$\lim_{k \rightarrow \infty} \left|\frac{k^k}{(k+1)^{(k+1)}}\right|$$
04

Evaluate the limit

To evaluate the limit, we can take the k-th root of both the numerator and denominator and then let k go to infinity: $$\lim_{k \rightarrow \infty} \left|\frac{\sqrt[k]{k^k}}{\sqrt[k]{(k+1)^{(k+1)}}}\right| = \lim_{k \rightarrow \infty} \left|\frac{k}{k+1}\right|$$ At this point, we can see that as k goes to infinity, the ratio approaches 1, which means the Ratio Test is inconclusive in this case. The Ratio Test doesn't provide enough information to decide if the series converges or diverges. Other tests like the Root Test or the Comparison Test might be helpful in determining the convergence of the series. However, based on the Ratio Test alone, it's not possible to determine the convergence of the given series.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Ratio Test
Understanding the convergence of an infinite series is crucial within the field of calculus. One of the commonly used methods to determine whether a series converges is the Ratio Test. It involves looking at the limit of the absolute ratio of consecutive terms.
To apply the Ratio Test, follow these steps:
  • Identify the general term, usually represented as \(a_k\).
  • Find the ratio \(\frac{a_{k+1}}{a_k}\), which is the next term divided by the current term.
  • Take the limit of this ratio as \(k\) approaches infinity.
  • Observe the limit: if it's less than 1, the series converges; if it's more than 1, the series diverges; if it's equal to 1, the test is inconclusive.
In the given problem, the Ratio Test leads to an inconclusive result because the limit as \(k\) approaches infinity of \(\frac{a_{k+1}}{a_k}\) is equal to 1. This does not determine convergence or divergence, prompting the need to use alternative methods for further analysis.
Keep in mind, the Ratio Test is particularly effective for series with factorial, exponential, or other multiplicative patterns in their terms.
Infinite Series
In calculus, an infinite series is a summation of infinitely many terms. When those terms follow a specific formula for each term based on its position, the series is formally expressed as \(\textstyle\frac{a_1}{a_2}{a_3}...\) or \(\textstyle\frac{\textstyle}{\textstyle}\textstyle\).
For a series to converge, the sum of its terms must approach a finite value as more terms are added. This is a fundamental concept in calculus that helps understand complex mathematical phenomena, such as the behavior of functions over intervals or solving differential equations. The series in the presented problem, \(\textstyle\frac{}{100}k^{-k}\), is an example where understanding convergence is essential. Convergence or divergence can be determined using tests like the Ratio Test, Comparison Test, or others based on the characteristics of the series.
Limits
The concept of limits is foundational in calculus. A limit evaluates the behavior of a function as the input approaches a certain value, but not necessarily reaching that value. This framework allows us to discuss and handle very small changes and trends towards infinity.
Calculating the limit involves several strategies, including simplification, factoring, or special limit laws. In the context of infinite series and the Ratio Test, limits help us decide on the behavior of the series: whether the terms grow indefinitely, shrink to zero, or oscillate without settling to a fixed value.
In the given exercise, the limit was needed to determine if the terms of the series were getting smaller quickly enough for the series to converge. The fact that the limit resulted in 1 left us without a definitive answer, which often happens when approaching the boundaries of a system.
Calculus
The field of calculus encompasses a broad range of concepts including limits, derivatives, integrals, and the study of infinite series. It's a powerful mathematical language that describes change and motion, helping solve problems in physics, engineering, economics, and beyond.
In the context of infinite series, calculus provides tools for analyzing the sum of infinitely many terms. These tools not only determine whether a series converges but also find its sum when it does converge. Different types of series require different approaches, which is why calculus offers a variety of convergence tests.
Understanding the nature of convergence through such tests is more than a mathematical exercise; it plays a vital role in applications ranging from computing complex probabilities to modeling physical phenomena. Through step-by-step analysis using calculus techniques, intricate problems become manageable and discernible.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Here is a fascinating (unsolved) problem known as the hailstone problem (or the Ulam Conjecture or the Collatz Conjecture). It involves sequences in two different ways. First, choose a positive integer \(N\) and call it \(a_{0} .\) This is the seed of a sequence. The rest of the sequence is generated as follows: For \(n=0,1,2, \ldots\) $$a_{n+1}=\left\\{\begin{array}{ll} a_{n} / 2 & \text { if } a_{n} \text { is even } \\ 3 a_{n}+1 & \text { if } a_{n} \text { is odd .} \end{array}\right.$$ However, if \(a_{n}=1\) for any \(n,\) then the sequence terminates. a. Compute the sequence that results from the seeds \(N=2,3\), \(4, \ldots, 10 .\) You should verify that in all these cases, the sequence eventually terminates. The hailstone conjecture (still unproved) states that for all positive integers \(N\), the sequence terminates after a finite number of terms. b. Now define the hailstone sequence \(\left\\{H_{k}\right\\},\) which is the number of terms needed for the sequence \(\left\\{a_{n}\right\\}\) to terminate starting with a seed of \(k\). Verify that \(H_{2}=1, H_{3}=7\), and \(H_{4}=2\). c. Plot as many terms of the hailstone sequence as is feasible. How did the sequence get its name? Does the conjecture appear to be true?

Assume that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}\) (Exercises 65 and 66 ) and that the terms of this series may be rearranged without changing the value of the series. Determine the sum of the reciprocals of the squares of the odd positive integers.

A fishery manager knows that her fish population naturally increases at a rate of \(1.5 \%\) per month. At the end of each month, 120 fish are harvested. Let \(F_{n}\) be the fish population after the \(n\) th month, where \(F_{0}=4000\) fish. Assume that this process continues indefinitely. Use infinite series to find the longterm (steady-state) population of the fish.

Evaluate the limit of the following sequences or state that the limit does not exist. $$a_{n}=\cos \left(0.99^{n}\right)+\frac{7^{n}+9^{n}}{63^{n}}$$

For a positive real number \(p,\) the tower of exponents \(p^{p^{p}}\) continues indefinitely and the expression is ambiguous. The tower could be built from the top as the limit of the sequence \(\left\\{p^{p},\left(p^{p}\right)^{p},\left(\left(p^{p}\right)^{p}\right)^{p}, \ldots .\right\\},\) in which case the sequence is defined recursively as \(a_{n+1}=a_{n}^{p}(\text { building from the top })\) where \(a_{1}=p^{p} .\) The tower could also be built from the bottom as the limit of the sequence \(\left\\{p^{p}, p^{\left(p^{p}\right)}, p^{\left(p^{(i)}\right)}, \ldots .\right\\},\) in which case the sequence is defined recursively as \(a_{n+1}=p^{a_{n}}(\text { building from the bottom })\) where again \(a_{1}=p^{p}\). a. Estimate the value of the tower with \(p=0.5\) by building from the top. That is, use tables to estimate the limit of the sequence defined recursively by (1) with \(p=0.5 .\) Estimate the maximum value of \(p > 0\) for which the sequence has a limit. b. Estimate the value of the tower with \(p=1.2\) by building from the bottom. That is, use tables to estimate the limit of the sequence defined recursively by (2) with \(p=1.2 .\) Estimate the maximum value of \(p > 1\) for which the sequence has a limit.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free