Question

Find the radius of convergence and the interval of convergence.

Answer 1

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

Calculus

Series Convergence Tests

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

Radius of Convergence (ROC)

• Ratio Test
• Interval bound

Interval of Convergence (IOC)

• Testing endpoints

Step-by-step explanation:

Step 1: Define

$\displaystyle \sum \limit_{n = 1}^\infty \frac{x^n}{2^n(n + 1)^2}$

Step 2: Find ROC

Apply Ratio Test

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

Our ROC is 2.

Step 3: Find IOC

Test endpoints

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

x = -2

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

Test convergence of modified series: Alternating Series Test

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

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]:                                                                                 $\displaystyle \sum \limit_{n = 1}^\infty \frac{2^n}{2^n(n + 1)^2}$
2. [Series] Simplify:                                                                                           $\displaystyle \sum \limit_{n = 1}^\infty \frac{1}{(n + 1)^2}$

Test convergence of modified series: Direct Comparison Test

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

At x = 2, the series is convergent.

∴ IOC for  $\displaystyle \sum \limit_{n = 1}^\infty \frac{x^n}{2^n(n + 1)^2}$  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