/*! 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 31 Determine the convergence or div... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Determine the convergence or divergence of the following series. $$\sum_{k=3}^{\infty} \frac{1}{(k-2)^{4}}$$

Short Answer

Expert verified
Question: Determine the convergence or divergence of the following series: $$\sum_{k=3}^{\infty} \frac{1}{(k-2)^{4}}$$ Answer: The given series is convergent.

Step by step solution

01

Write down the given series and the p-series to compare with

We are given the following series: $$\sum_{k=3}^{\infty} \frac{1}{(k-2)^{4}}$$ We will compare this series with the following convergent p-series: $$\sum_{k=1}^{\infty} \frac{1}{k^4}$$
02

Use the Comparison Test

Let's compare the terms of both series, a_k for the given series and b_k for the p-series: $$a_k = \frac{1}{(k-2)^4} \quad \text{and} \quad b_k = \frac{1}{k^4}$$ We want to show that: $$0 \leq a_k \leq b_k$$ We can easily see that both series have positive terms, as both expressions have even exponents in the denominator. Then: $$0 \leq \frac{1}{(k-2)^4} \leq \frac{1}{k^4}$$ Starting from k = 3, as specified in the given series, we get: $$0 \leq \frac{1}{(3-2)^4} \leq \frac{1}{3^4}$$ The inequality holds, which means: $$\sum_{k=3}^{\infty} \frac{1}{(k-2)^{4}}$$ Is dominated (bounded by) a convergent p-series: $$\sum_{k=1}^{\infty} \frac{1}{k^4}$$
03

Conclude the convergence of the given series

Since we have shown that the given series is dominated by a convergent p-series, we can conclude that the given series: $$\sum_{k=3}^{\infty} \frac{1}{(k-2)^{4}}$$ Is convergent by the Comparison Test.

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!

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

Determine whether the following series converge absolutely or conditionally, or diverge. $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1} e^{k}}{(k+1) !}$$

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 \(\left\\{a_{1}, a_{1} a_{2}, a_{1} a_{2} a_{3}, \dots\right\\}\) a. Show that the infinite product converges (which means its sequence of partial products converges) provided the series \(\sum_{k=1}^{\infty} \ln a_{k}\) converges. b. Consider the infinite product $$P=\prod_{k=2}^{\infty}\left(1-\frac{1}{k^{2}}\right)=\frac{3}{4} \cdot \frac{8}{9} \cdot \frac{15}{16} \cdot \frac{24}{25} \cdots$$ Write out the first few terms of the sequence of partial products, $$P_{n}=\prod_{k=2}^{n}\left(1-\frac{1}{k^{2}}\right)$$ (for example, \(P_{2}=\frac{3}{4}, P_{3}=\frac{2}{3}\) ). Write out enough terms to determine the value of the product, which is \(\lim _{n \rightarrow \infty} P_{n}\). c. Use the results of parts (a) and (b) to evaluate the series $$\sum_{k=2}^{\infty} \ln \left(1-\frac{1}{k^{2}}\right)$$

Consider the following sequences defined by a recurrence relation. Use a calculator, analytical methods, and/or graphing to make a conjecture about the value of the limit or determine that the limit does not exist. $$a_{n+1}=\frac{1}{2}\left(a_{n}+2 / a_{n}\right) ; a_{0}=2, n=0,1,2, \dots$$

The expression where the process continues indefinitely, is called a continued fraction. a. Show that this expression can be built in steps using the recurrence relation \(a_{0}=1, a_{n+1}=1+1 / a_{n},\) for \(n=0,1,2,3, \ldots . .\) Explain why the value of the expression can be interpreted as \(\lim a_{n}\). b. Evaluate the first five terms of the sequence \(\left\\{a_{n}\right\\}\). c. Using computation and/or graphing, estimate the limit of the sequence. d. Assuming the limit exists, use the method of Example 5 to determine the limit exactly. Compare your estimate with \((1+\sqrt{5}) / 2,\) a number known as the golden mean. e. Assuming the limit exists, use the same ideas to determine the value of where \(a\) and \(b\) are positive real numbers.

Given that \(\sum_{k=1}^{\infty} \frac{1}{k^{4}}=\frac{\pi^{4}}{90},\) show that \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{4}}=\frac{7 \pi^{4}}{720}.\) (Assume the result of Exercise 63.)

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.