/*! 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 11 Use the Ratio Test to determine ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 11

Use the Ratio Test to determine whether the following series converge. $$\sum_{k=1}^{\infty} \frac{k^{2}}{4^{k}}$$

Short Answer

Expert verified
Answer: Yes, the series converges.

Step by step solution

01

State the Ratio Test

The Ratio Test states that if we have a series \(\sum_{k=1}^{\infty} a_k\), we should compute the limit of the ratio of consecutive terms as \(k \rightarrow \infty\): $$L = \lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right|$$ If \(L<1\), then the series converges. If \(L>1\), then the series diverges. If \(L=1\), the test is inconclusive.
02

Identify the sequence from the given series

For the given series, we have the sequence \(a_k = \frac{k^{2}}{4^{k}}\).
03

Calculate the ratio of consecutive terms

Next, we need to find the ratio of consecutive terms, that is, \(\frac{a_{k+1}}{a_k}\). This can be written as: $$\frac{a_{k+1}}{a_k} =\frac{\frac{(k+1)^{2}}{4^{k+1}}}{\frac{k^{2}}{4^{k}}}$$
04

Simplify the ratio

Let's simplify the ratio by multiplying numerator and denominator by \(4^k\): $$\frac{a_{k+1}}{a_k}=\frac{(k+1)^{2}}{4k^2}$$
05

Calculate the limit L

Now we need to find the limit \(L\) as \(k \rightarrow \infty\): $$L = \lim_{k\to\infty} \left|\frac{(k+1)^{2}}{4k^2}\right|$$ Divide the numerator and denominator by \(k^2\): $$L = \lim_{k\to\infty} \left|\frac{1+\frac{2}{k}+\frac{1}{k^2}}{4}\right|$$ Taking the limit as \(k \rightarrow \infty\): $$L = \frac{1+0+0}{4} = \frac{1}{4}$$
06

Determine the convergence of the series using the Ratio Test

Since \(L = \frac{1}{4} < 1\), the series converges according to the Ratio Test. Therefore, the given series: $$\sum_{k=1}^{\infty} \frac{k^{2}}{4^{k}}$$ 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 91影视!

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. $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{3 / 2}}$$

Reciprocals of odd squares Assume that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}\)

Evaluate the limit of the following sequences. $$a_{n}=\frac{6^{n}+3^{n}}{6^{n}+n^{100}}$$

For a positive real number \(p,\) how do you interpret \(p^{p^{p \cdot *}},\) where the tower of exponents continues indefinitely? As it stands, the expression is ambiguous. The tower could be built from the top or from the bottom; that is, it could be evaluated by the recurrence relations \(a_{n+1}=p^{a_{n}}\) (building from the bottom) or \(a_{n+1}=a_{n}^{p}(\text { building from the top })\) where \(a_{0}=p\) in either case. The two recurrence relations have very different behaviors that depend on the value of \(p\). a. Use computations with various values of \(p > 0\) to find the values of \(p\) such that the sequence defined by (2) has a limit. Estimate the maximum value of \(p\) for which the sequence has a limit. b. Show that the sequence defined by (1) has a limit for certain values of \(p\). Make a table showing the approximate value of the tower for various values of \(p .\) Estimate the maximum value of \(p\) for which the sequence has a limit.

Given that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6},\) show that \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{2}}=\frac{\pi^{2}}{12}.\) (Assume the result of Exercise 63.)
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Geometry

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.