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Chapter 1: Problem 13

Find the interval of convergence, including end-point tests: $$\sum_{n=1}^{\infty} \frac{(x+2)^{n}}{(-3)^{n} \sqrt{n}}$$

Short Answer

Expert verified
The interval of convergence is \(-5 < x \leq 1\).

Step by step solution

01

Identify the general term of the series

The general term for the given series is: \[ a_n = \frac{(x+2)^n}{(-3)^n \sqrt{n}} \]
02

Apply the Ratio Test

To find the interval of convergence, use the Ratio Test. Compute the limit \[ L = \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| \] where \[ a_{n+1} = \frac{(x+2)^{n+1}}{(-3)^{n+1} \sqrt{n+1}}. \]
03

Simplify the Ratio

Simplify the ratio: \[ L = \lim_{{n \to \infty}} \left| \frac{(x+2)^{n+1}}{(-3)^{n+1} \sqrt{n+1}} \cdot \frac{(-3)^n \sqrt{n}}{(x+2)^n} \right| = \lim_{{n \to \infty}} \left| \frac{(x+2)}{-3} \cdot \frac{\sqrt{n}}{\sqrt{n+1}} \right| \]
04

Evaluate the Limit

Evaluate the limit: \[ L = \lim_{{n \to \infty}} \left| \frac{(x+2)}{-3} \cdot \frac{\sqrt{n}}{\sqrt{n+1}} \right| = \left| \frac{x+2}{-3} \right| \cdot \lim_{{n \to \infty}} \left| \frac{\sqrt{n}}{\sqrt{n+1}} \right| = \left| \frac{x+2}{-3} \right| \cdot 1 = \left| \frac{x+2}{3} \right| \]
05

Set Up the Ratio Test Inequality

The Ratio Test requires \[ \left| \frac{x+2}{3} \right| < 1 \] for convergence.
06

Solve the Inequality

Solve the inequality: \[ \left| \frac{x+2}{3} \right| < 1 \] \[ -1 < \frac{x+2}{3} < 1 \] \[ -3 < x+2 < 3 \] \[ -5 < x < 1 \]
07

Test the Endpoints

Check the endpoints \(x = -5\) and \(x = 1\):\ \For \(x = -5\): \The series becomes \[ \sum_{n=1}^{\infty} \frac{(-3)^n}{(-3)^n \sqrt{n}} = \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}, \] which diverges. \ \For \(x = 1\): \The series becomes \[ \sum_{n=1}^{\infty} \frac{(3)^n}{(-3)^{n} \sqrt{n}} = \sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}, \] which converges by the Alternating Series Test.
08

State the Interval of Convergence

The interval of convergence is \[ -5 < x \leq 1 \]

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Key Concepts

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

Ratio Test
The Ratio Test is a useful method to determine the convergence or divergence of an infinite series. It uses the limit of the absolute value of the ratio of successive terms. Specifically, the Ratio Test requires calculating \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). This value will determine the behavior of the series:
If \( L < 1 \), the series converges absolutely.
If \( L > 1 \) or \( L = \infty \), the series diverges.
If \( L = 1 \), the test is inconclusive, and other methods may be needed to determine convergence.
Series Convergence
Series convergence refers to whether the sum of an infinite sequence of terms approaches a finite limit. A series is convergent if the partial sums of the series approach a certain value as the number of terms increases. There are several tests for determining series convergence, such as the Ratio Test, Root Test, and Alternating Series Test. Understanding the behavior and properties of the series' terms is crucial to apply these tests correctly and determine the series' convergence.
Alternating Series Test
The Alternating Series Test (also known as the Leibniz test) is used to determine the convergence of series whose terms alternate in sign. An alternating series has the form \( \sum_{n=1}^{ ty} (-1)^{n} a_n \). For the Alternating Series Test to apply, the following conditions must be met:
  • The absolute value of the terms \( a_n \) must be monotonically decreasing, meaning each successive term is smaller than the preceding one.
  • The limit of the terms \( a_n \) as \( n \) approaches infinity must be zero, i.e., \( \lim_{{n \to \infty}} a_n = 0 \).
If these two conditions are met, the series converges. For example, in the given solution, the series for \( x = 1 \) converges by this test.
Inequality Solving
Inequality solving involves finding the set of all possible values of a variable that satisfy a given inequality condition. This can be applied to the ratio of terms, boundaries of series convergence, and more. In the ratio test for the given problem, the inequality is \( \left|\frac{x+2}{3}\right| < 1 \), which is solved in steps:


  • First, set up the inequality: \( \left|\frac{x+2}{3}\right| < 1 \).
  • Next, solve it step-by-step:
    • Remove the absolute value: \( -1 < \frac{x+2}{3} < 1 \).
    • Multiply through by 3: \( -3 < x+2 < 3 \).
    • Subtract 2 from each part: \( -5 < x < 1 \).
    These steps result in finding the interval \( -5 < x < 1 \) where the series converges. Additionally, testing the endpoints helps to confirm whether the series converges or diverges at specific boundary values.

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    Most popular questions from this chapter

    Use the integral test to find whether the following series converge or diverge. Hint and warning: Do not use lower limits on your integrals (see Problem 16 ). $$\sum_{n=1}^{\infty} \frac{n}{n^{2}+4}$$

    (a) Using computer or tables (or see Chapter 7 , Section 11 ), verify that \(\sum_{n=1}^{\infty} 1 / n^{2}=\) \(\pi^{2} / 6=1.6449+,\) and also verify that the error in approximating the sum of the series by the first five terms is approximately 0.1813 . (b) \(\quad\) By computer or tables verify that \(\sum_{n=1}^{\infty}\left(1 / n^{2}\right)(1 / 2)^{n}=\pi^{2} / 12-(1 / 2)(\ln 2)^{2}=\) \(0.5822+,\) and that the sum of the first five terms is \(0.5815+.\) (c) Prove theorem (14.4). Hint: The error is \(\left|\sum_{N+1}^{\infty} a_{n} x^{n}\right| .\) Use the fact that the absolute value of a sum is less than or equal to the sum of the absolute values. Then use the fact that \(\left|a_{n+1}\right| \leq\left|a_{n}\right|\) to replace all \(a_{n}\) by \(a_{N+1},\) and write the appropriate inequality. Sum the geometric series to get the result.

    Test the following series for convergence or divergence. Decide for yourself which test is easiest to use, but don't forget the preliminary test. Use the facts stated above when they apply. $$\sum_{n=2}^{\infty}(-1)^{n} \frac{n}{n-1}$$

    Test the following series for convergence or divergence. Decide for yourself which test is easiest to use, but don't forget the preliminary test. Use the facts stated above when they apply. $$\sum_{n=1}^{\infty} \frac{n^{5}}{5^{n}}$$

    Test the following series for convergence or divergence. Decide for yourself which test is easiest to use, but don't forget the preliminary test. Use the facts stated above when they apply. $$\frac{1}{2^{2}}-\frac{1}{3^{2}}+\frac{1}{2^{3}}-\frac{1}{3^{3}}+\frac{1}{2^{4}}-\frac{1}{3^{4}}+\cdots$$
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