/*! 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 27 Use any method to determine whet... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 27

Use any method to determine whether the series converges. $$\sum_{k=1}^{\infty} \frac{k^{2}}{5^{k}}$$

Short Answer

Expert verified
The series converges by the Ratio Test.

Step by step solution

01

Identify the series

The series is given by \( \sum_{k=1}^{\infty} \frac{k^{2}}{5^{k}} \). This series has terms of the form \( \frac{k^{2}}{5^{k}} \) where \( k \) is a positive integer.
02

Apply the Ratio Test

We use the Ratio Test to determine if the series converges. Define \( a_k = \frac{k^2}{5^k} \). The Ratio Test requires us to find \( \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \).
03

Calculate the Ratio

\[ a_{k+1} = \frac{(k+1)^2}{5^{k+1}} \] So the ratio \( \frac{a_{k+1}}{a_k} \) becomes:\[ \frac{a_{k+1}}{a_k} = \frac{(k+1)^2}{5^{k+1}} \times \frac{5^k}{k^2} = \frac{(k+1)^2}{5k^2} \]
04

Simplify and Take the Limit

We take the limit of the ratio as \( k \rightarrow \infty \):\[ \lim_{k \to \infty} \frac{(k+1)^2}{5k^2} \] This simplifies to:\[ \lim_{k \to \infty} \frac{k^2 + 2k + 1}{5k^2} = \lim_{k \to \infty} \frac{1 + \frac{2}{k} + \frac{1}{k^2}}{5} = \frac{1}{5} \]
05

Apply Ratio Test Conclusion

Since \( \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| = \frac{1}{5} < 1 \), by the Ratio Test, the series \( \sum_{k=1}^{\infty} \frac{k^2}{5^k} \) 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!

Key Concepts

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

Convergent series
A convergent series is a sequence of terms whose sum approaches a specific finite value as more terms are added. When a series converges, it means that as you add more and more terms, the total sum of the series stabilizes at a particular number, rather than growing infinitely or oscillating without settling. For example, in the series \( \sum_{k=1}^{\infty} \frac{k^{2}}{5^{k}} \), we found that the series converges using the Ratio Test.

This means:
  • Adding up infinitely many terms of \( \frac{k^{2}}{5^{k}} \) will settle at a specific sum.
  • Even though there are infinitely many terms, each term's contribution becomes negligibly small.
  • The sum stabilizes rather than diverging to infinity or showing erratic behavior.
Understanding convergent series is crucial in mathematical analysis because it helps us evaluate infinite processes and their real-world applications.
Series convergence
Series convergence involves checking if the infinite sum of terms approaches a finite limit. There are several tests to determine this, one of which is the Ratio Test used in our exercise.

Here's a quick overview of the Ratio Test approach:
  • The Ratio Test involves comparing the ratio of consecutive terms \( \frac{a_{k+1}}{a_k} \) as \( k \to \infty \).
  • If the absolute value of this limit is less than 1, the series converges.
  • If it's greater than 1, the series diverges, and if it equals 1, the test is inconclusive.
In our case, for the series \( \sum_{k=1}^{\infty} \frac{k^{2}}{5^{k}} \), the limit of the ratio is \( \frac{1}{5} \), indicating convergence since it's less than 1. The Ratio Test is a powerful tool for analyzing series since it gives a clear-cut criterion for convergence.
Limit of a sequence
The limit of a sequence is the value that terms of the sequence get closer to as the sequence progresses. In mathematical terms, if \( a_k \) is a term in the sequence, as \( k \to \infty \), \( a_k \) approaches the limit. This concept is fundamental when dealing with series and their convergence.

Here's how the limit applies:
  • For \( a_k = \frac{k^2}{5^k} \), the limit of the sequence in the Ratio Test calculation was \( \frac{1}{5} \).
  • This limit helped determine the behavior of the series, showing it converged when assessed against the Ratio Test.
  • A finite limit means we can predict where the sequence is heading, crucial for confirming series convergence.
Understanding the limit of a sequence helps in identifying overall patterns and behaviors of both sequences and series, greatly aiding in mathematical problem-solving and analysis.

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

Find the Taylor polynomials of orders \(n=0,1,2,3\) and 4 about \(x=x_{0},\) and then find the \(n\) th Taylor polynomial for the function in sigma notation. $$\frac{1}{x} ; x_{0}=-1$$

Determine whether the series converges. $$\sum_{k=1}^{\infty} \frac{1}{\sqrt{k^{2}+1}}$$

Prove: (a) If \(f\) is an even function, then all odd powers of \(x\) in its Maclaurin series have coefficient 0 (b) If \(f\) is an odd function, then all even powers of \(x\) in its Maclaurin series have coefficient 0

Find the general term of the series and use the ratio test to show that the series converges. $$1+\frac{1 \cdot 3}{3 !}+\frac{1 \cdot 3 \cdot 5}{5 !}+\frac{1 \cdot 3 \cdot 5 \cdot 7}{7 !}+\cdots$$

(a) Use the Maclaurin series for \(1 /(1-x)\) to find the Maclaurin series for $$f(x)=\frac{x}{1-x^{2}}$$ (b) Use the Maclaurin series obtained in part (a) to find \(f^{(5)}(0)\) and \(f^{(6)}(0)\) (c) What can you say about the value of \(f^{(n)}(0) ?\)
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Probability and Statistics

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.