/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 Use the test of your choice to d... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use the test of your choice to determine whether the following series converge. $$\sum_{k=1}^{\infty} \frac{1}{5^{k}-1}$$

Short Answer

Expert verified
Answer: Yes, the series converges.

Step by step solution

01

Choose a geometric series for comparison

Let's compare our series to the geometric series: $$\sum_{k=1}^{\infty} \frac{1}{5^k}$$ This is a geometric series with a common ratio of 1/5, which converges because |1/5| < 1.
02

Show that 0 ≤ a(k) ≤ b(k) for all k

Now we need to show that: $$0 \le \frac{1}{5^k-1} \le \frac{1}{5^k}$$ Since k starts from 1 and increases, we can say that for all k ≥ 1, $$5^k - 1 < {5^k}$$ This is because we are subtracting 1 from the denominator, making the denominator smaller, and the overall fraction larger. Hence, for all k ≥ 1: $$\frac{1}{5^k-1} \ge \frac{1}{5^k}$$ Since all entries in our series are positive, it is also true that $$0 \le \frac{1}{5^k-1}$$ Thus, $$0 \le \frac{1}{5^k-1} \le \frac{1}{5^k}$$ for all k ≥ 1.
03

Apply the Direct Comparison Test

We have shown that: - 0 ≤ a(k) ≤ b(k) for all k, and - The series of b(k) converges (\(\sum_{k=1}^{\infty} \frac{1}{5^k}\) is a convergent geometric series). By the Direct Comparison Test, we can conclude that: $$\sum_{k=1}^{\infty} \frac{1}{5^{k}-1}$$ converges.

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Direct Comparison Test
The Direct Comparison Test is a straightforward method for determining whether an infinite series converges or diverges. It works on the principle that if each term of a series is less than or equal to the corresponding term of a known convergent series, then the original series must also converge.

To use this test, you identify two series, \( a_k \) and \( b_k \), where \( 0 \leq a_k \leq b_k \) for all natural numbers \( k \). If the series \( \sum_{k=1}^{\infty} b_k \) converges, then so does the series \( \sum_{k=1}^{\infty} a_k \). Conversely, if \( \sum_{k=1}^{\infty} a_k \) diverges, and \( a_k \geq b_k \) for all \( k \) (with \( b_k \) being the terms of a divergent series), then \( \sum_{k=1}^{\infty} b_k \) also diverges.

When applying this test, it's important to ensure that you compare the series to a suitable benchmark series, such as a geometric series, whose convergence properties are well-known. Careful selection of your comparison series is key to effectively utilizing the Direct Comparison Test.
Geometric Series
A geometric series is a series of the form \( \sum_{k=0}^{\infty} ar^k \), where \( a \) is the first term, and \( r \) is the common ratio. The series converges if and only if the absolute value of the common ratio \( |r| < 1 \).

The significance of a geometric series lies in its simplicity and the fact that its convergence can be determined by the value of \( r \). When \( r \) meets the convergence criterion, the sum of the series can also be found using the formula \( S = \frac{a}{1-r} \).

Geometric series often serve as a point of reference for the Direct Comparison Test because their behavior is well-understood and can provide a clear benchmark for whether another series might converge or diverge.
Series Convergence
Series convergence refers to the property of an infinite series that its partial sums approach a finite value as the number of terms grows indefinitely. An infinite series \( \sum_{k=1}^{\infty} a_k \) converges if the limit of its partial sums \( S_N = \sum_{k=1}^{N} a_k \) exists and is finite when \( N \) approaches infinity.

If this limit does not exist or is infinite, then the series is said to diverge. Understanding whether a series converges or diverges is crucial in many areas of mathematics, as it often indicates whether a given infinite process will result in a meaningful value.
Infinite Series
An infinite series is the sum of an infinite sequence of terms. It is written in the form \( \sum_{k=1}^{\infty} a_k \) where \( a_k \) represents the terms of the sequence. Infinite series can be either convergent or divergent, and determining which is often a focus of study in calculus and analysis.

Infinite series appear frequently in various mathematical contexts, such as in the representation of functions as power series, in the evaluation of certain integrals, and in the solution of differential equations. The behavior of infinite series can reveal profound insights into the nature of mathematical phenomena and is thus a critical concept in higher mathematics.

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

Evaluate the series \(\sum_{k=1}^{\infty}\left(\frac{1}{2^{k}}-\frac{1}{2^{k+1}}\right)\) two ways. a. Use a telescoping series argument. b. Use a geometric series argument after first simplifying \(\frac{1}{2^{k}}-\frac{1}{2^{k+1}}\)

Use the ideas of Exercise 88 to evaluate the following infinite products. $$\text { a. } \prod_{k=0}^{\infty} e^{1 / 2^{k}}=e \cdot e^{1 / 2} \cdot e^{1 / 4} \cdot e^{1 / 8} \dots$$ $$\text { b. } \prod_{k=2}^{\infty}\left(1-\frac{1}{k}\right)=\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdots$$

The fractal called the snowflake island (or Koch island ) is constructed as follows: Let \(I_{0}\) be an equilateral triangle with sides of length \(1 .\) The figure \(I_{1}\) is obtained by replacing the middle third of each side of \(I_{0}\) with a new outward equilateral triangle with sides of length \(1 / 3\) (see figure). The process is repeated where \(I_{n+1}\) is obtained by replacing the middle third of each side of \(I_{n}\) with a new outward equilateral triangle with sides of length \(1 / 3^{n+1}\). The limiting figure as \(n \rightarrow \infty\) is called the snowflake island. a. Let \(L_{n}\) be the perimeter of \(I_{n} .\) Show that \(\lim _{n \rightarrow \infty} L_{n}=\infty\) b. Let \(A_{n}\) be the area of \(I_{n} .\) Find \(\lim _{n \rightarrow \infty} A_{n} .\) It exists!

Consider the following infinite series. a. Write out the first four terms of the sequence of partial sums. b. Estimate the limit of \(\left\\{S_{n}\right\\}\) or state that it does not exist. $$\sum_{k=1}^{\infty} \frac{3}{10^{k}}$$

The famous Fibonacci sequence was proposed by Leonardo Pisano, also known as Fibonacci, in about \(\mathrm{A.D.} 1200\) as a model for the growth of rabbit populations. It is given by the recurrence relation \(f_{n+1}=f_{n}+f_{n-1},\) for \(n=1,2,3, \ldots,\) where \(f_{0}=1, f_{1}=1 .\) Each term of the sequence is the sum of its two predecessors. a. Write out the first ten terms of the sequence. b. Is the sequence bounded? c. Estimate or determine \(\varphi=\lim _{n \rightarrow \infty} \frac{f_{n+1}}{f_{n}},\) the ratio of the successive terms of the sequence. Provide evidence that \(\varphi=(1+\sqrt{5}) / 2,\) a number known as the golden mean. d. Use induction to verify the remarkable result that $$f_{n}=\frac{1}{\sqrt{5}}\left(\varphi^{n}-(-1)^{n} \varphi^{-n}\right).$$

See all solutions

Recommended explanations on Math Textbooks

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.