Math Problem Statement

Find the interval of convergence and the radius of convergence for each series. ∞∑n=1 ((3x−1)^n)/(n^2)

Solution

To find the interval of convergence and radius of convergence for the series

n=1(3x1)nn2,\sum_{n=1}^\infty \frac{(3x-1)^n}{n^2},

we proceed as follows:


Step 1: Analyze the given series

The series can be rewritten as:

n=1(3x1)nn2.\sum_{n=1}^\infty \frac{(3x-1)^n}{n^2}.

This is a power series in (3x1)(3x-1). The ratio test or root test can be used to determine the convergence.


Step 2: Apply the root test

The root test examines the limit:

lim supnann,\limsup_{n \to \infty} \sqrt[n]{|a_n|},

where an=(3x1)nn2a_n = \frac{(3x-1)^n}{n^2}.

First, extract the general term: an=3x1nn2.|a_n| = \frac{|3x-1|^n}{n^2}.

Now compute the nth root: [ \sqrt[n]{|a_n|} = \sqrt[n]{\frac{|3x-1|^n}{n^2}} = \frac{|3x-1|}{n^{2/n}}. ]

As nn \to \infty, n2/n1n^{2/n} \to 1, so: lim supnann=3x1.\limsup_{n \to \infty} \sqrt[n]{|a_n|} = |3x-1|.

For convergence of the series, the root test requires: 3x1<1.|3x-1| < 1.


Step 3: Solve for xx in terms of the inequality

The inequality 3x1<1|3x-1| < 1 can be rewritten as: 1<3x1<1.-1 < 3x - 1 < 1.

Add 1 to all parts: 0<3x<2.0 < 3x < 2.

Divide by 3: 0<x<23.0 < x < \frac{2}{3}.

Thus, the radius of convergence is: R=13.R = \frac{1}{3}.


Step 4: Test endpoints

To determine the interval of convergence, we test the endpoints x=0x = 0 and x=23x = \frac{2}{3}:

  1. At x=0x = 0: Substituting x=0x = 0 gives (3x1)n=(1)n(3x-1)^n = (-1)^n: n=1(1)nn2.\sum_{n=1}^\infty \frac{(-1)^n}{n^2}. This series converges by the alternating series test and the p-series test (p=2>1p = 2 > 1).

  2. At x=23x = \frac{2}{3}: Substituting x=23x = \frac{2}{3} gives (3x1)n=1n=1(3x-1)^n = 1^n = 1: n=11n2.\sum_{n=1}^\infty \frac{1}{n^2}. This series converges by the p-series test (p=2>1p = 2 > 1).


Final Answer:

  • Interval of Convergence: [0,23][0, \frac{2}{3}].
  • Radius of Convergence: R=13R = \frac{1}{3}.

Would you like additional details about any step? Here are some related questions to explore:


5 Related Questions:

  1. How does the root test differ from the ratio test in determining convergence?
  2. Why do we need to test endpoints separately when finding the interval of convergence?
  3. What is the significance of the p-series test, and when can it be applied?
  4. How does absolute convergence affect the interval of convergence?
  5. Can a series converge at one endpoint but not the other? Provide examples.

Tip:

Always verify endpoints separately when finding the interval of convergence; convergence behavior at endpoints can differ significantly.

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Math Problem Analysis

Mathematical Concepts

Power Series
Interval of Convergence
Radius of Convergence
Root Test
P-Series Test
Alternating Series Test

Formulas

Root Test: limsup(n → ∞) (|a_n|^(1/n))
Inequality: |3x-1| < 1
P-Series Convergence: Σ(1/n^p) converges for p > 1

Theorems

Root Test
P-Series Test
Alternating Series Test

Suitable Grade Level

Grades 11-12 and College-Level Calculus