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Find the radius of convergence and the interval of convergence.
[tex]\displaystyle \sum \limit_{n=1}^\infty \frac{x^\text{n}}{2^\text{n}(n+1)^2}[/tex]

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Answer:

Radius of Convergence: 2

Interval of Convergence: [-2, 2]

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality

Algebra I

  • Exponential Rule [Multiplying]:                                                                     [tex]\displaystyle b^m \cdot b^n = b^{m + n}[/tex]
  • Exponential Rule [Dividing]:                                                                         [tex]\displaystyle \frac{b^m}{b^n} = b^{m - n}[/tex]

Calculus

Series Convergence Tests

  • P-Series:                                                                                                            [tex]\displaystyle \sum \limit_{n = 1}^\infty \frac{1}{n^p}[/tex]
  • Direct Comparison Test (DCT)
  • Alternating Series Test (AST)
  • Ratio Test:                                                                                                         [tex]\displaystyle \lim_{n \to \infty} \bigg| \frac{a_{n + 1}}{a_n} \bigg|[/tex]

Radius of Convergence (ROC)

  • Ratio Test
  • Interval bound

Interval of Convergence (IOC)

  • Testing endpoints

Step-by-step explanation:

Step 1: Define

[tex]\displaystyle \sum \limit_{n = 1}^\infty \frac{x^n}{2^n(n + 1)^2}[/tex]

Step 2: Find ROC

Apply Ratio Test

  1. [Series] Set up [Ratio Test]:                                                                           [tex]\displaystyle \lim_{n \to \infty} \bigg|\frac{x^{n + 1}}{2^{n + 1}(n + 2)^2} \cdot \frac{2^n(n + 1)^2}{x^n} \bigg|[/tex]
  2. [Ratio Test] Rewrite exponentials [Exponential Rule - Multiplying]:           [tex]\displaystyle \lim_{n \to \infty} \bigg|\frac{x^n \cdot x}{2^n \cdot 2(n + 2)^2} \cdot \frac{2^n(n + 1)^2}{x^n} \bigg|[/tex]
  3. [Ratio Test] Simplify:                                                                                     [tex]\displaystyle \lim_{n \to \infty} \bigg| \frac{x}{2(n + 2)^2} \cdot (n + 1)^2 \bigg|[/tex]
  4. [Ratio Test] Multiply:                                                                                     [tex]\displaystyle \lim_{n \to \infty} \bigg| \frac{x(n + 1)^2}{2(n + 2)^2} \bigg|[/tex]
  5. [Ratio Test] Evaluate limit:                                                                             [tex]\displaystyle \bigg| \frac{x}{2} \bigg| < 1[/tex]
  6. [Ratio Test] Isolate x:                                                                                     [tex]\displaystyle |x| < 2[/tex]

Our ROC is 2.

Step 3: Find IOC

Test endpoints

  1. [ROC] Find interval bound:                                                                           [tex]\displaystyle -2 < x < 2[/tex]

x = -2

  1. Substitute in x [Series]:                                                                                 [tex]\displaystyle \sum \limit_{n = 1}^\infty \frac{(-2)^n}{2^n(n + 1)^2}[/tex]
  2. [Series] Rewrite [Exponential Rules - Multiplying]:                                     [tex]\displaystyle \sum \limit_{n = 1}^\infty \frac{(-1)^n2^n}{2^n(n + 1)^2}[/tex]
  3. [Series] Simplify:                                                                                           [tex]\displaystyle \sum \limit_{n = 1}^\infty \frac{(-1)^n}{(n + 1)^2}[/tex]

Test convergence of modified series: Alternating Series Test

  1. [AST] Condition 1 [Limit Test]:                                                                       [tex]\displaystyle \lim_{n \to \infty} \frac{1}{(n + 1)^2} = 0 \ \checkmark[/tex]
  2. [AST] Condition 2 [aₙ vs bₙ comparison]:                                                     [tex]\displaystyle \frac{1}{(n + 2)^2} \le \frac{1}{(n + 1)^2} \ \checkmark[/tex]

At x = -2, the series is convergent.

∴ Current IOC is -2 ≤ x < 2 or [-2, 2); 2 undetermined

x = 2

  1. Substitute in x [Series]:                                                                                 [tex]\displaystyle \sum \limit_{n = 1}^\infty \frac{2^n}{2^n(n + 1)^2}[/tex]
  2. [Series] Simplify:                                                                                           [tex]\displaystyle \sum \limit_{n = 1}^\infty \frac{1}{(n + 1)^2}[/tex]

Test convergence of modified series: Direct Comparison Test

  1. [DCT] Condition 1 [Define comparing series]:                                             [tex]\displaystyle \sum \limit_{n = 1}^\infty \frac{1}{n^2}[/tex]
  2. [DCT] Condition 1 [Test convergence of comparing series]:                     [tex]\displaystyle p = 2 > 1, \ \sum \limit_{n = 1}^\infty \frac{1}{n^2} \ \text{convergent by p-series}[/tex]
  3. [DCT] Condition 2 [aₙ vs bₙ comparison]:                                                     [tex]\displaystyle \frac{1}{(n + 1)^2} \le \frac{1}{n^2} \ \checkmark[/tex]

At x = 2, the series is convergent.

∴ IOC for  [tex]\displaystyle \sum \limit_{n = 1}^\infty \frac{x^n}{2^n(n + 1)^2}[/tex]  is -2 ≤ x ≤ 2 or [-2, 2]

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Taylor Polynomials and Approximations - Power Series (BC Only)

Book: College Calculus 10e

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