Answer:
The series is convergent and is equal to 1.
General Formulas and Concepts:
Algebra I
- Terms/Coefficients
- Factoring
Pre-Calculus
- Partial Fraction Decomposition
Calculus
Limits
- Limit Rule [Variable Direct Substitution]: [tex]\displaystyle \lim_{x \to c} x = c[/tex]
- Limit Property [Addition/Subtraction]: [tex]\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)[/tex]
Sequences
Series
- Definition of a convergent/divergent series
- Sum of a series: [tex]\displaystyle \lim_{n \to \infty} S_n[/tex]
Telescoping Series: [tex]\displaystyle \sum^\infty_{k = 1} (b_1 - b_{k + 1}) = (b_1 - b_2) + (b_2 - b_3) + (b_3 - b_4) + ... + (b_1 - b_{k + 1}) + ...[/tex]
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle \sum^\infty_{k = 1} \frac{2k + 1}{k^2(k + 1)^2}[/tex]
Step 2: Rewrite Sum
- Factor: [tex]\displaystyle \sum^\infty_{k = 1} \frac{2k + 1}{k^2(k + 1)^2} = \sum^\infty_{k = 1} \frac{2k + 1}{k \cdot k(k + 1)(k + 1)}[/tex]
- Break up [Partial Fraction Decomposition]: [tex]\displaystyle \frac{2k + 1}{k^2(k + 1)^2} = \frac{A}{k} + \frac{B}{k^2} + \frac{C}{k + 1} + \frac{D}{(k + 1)^2}[/tex]
- Simplify [Common Denominator]: [tex]\displaystyle 2k + 1 = Ak(k + 1)^2 + B(k + 1)^2 + Ck^2(k + 1) + Dk^2[/tex]
- [Decomp] Expand: [tex]\displaystyle 2k + 1 = Ak^3 + Ck^3 + A2k^2 + Bl^2 + Ck^2 + Dk^2 + Ak + B2k + B[/tex]
- [Decomp] Factor: [tex]\displaystyle 2k + 1 = k^3(A + C) + k^2(2A + B + C + D) + k(A + 2B) + B[/tex]
- [Decomp] Set up systems: [tex]\displaystyle \begin{cases} A + C = 0 \\ 2A + B + C + D = 0 \\ A + 2B = 2 \\ B = 1 \end{cases}[/tex]
- [Decomp] Solve: [tex]\displaystyle A = 0, \ B = 1, \ C = 0, \ D = -1[/tex]
- [Decomp] Substitute in variables: [tex]\displaystyle \frac{2k + 1}{k^2(k + 1)^2} = \frac{0}{k} + \frac{1}{k^2} + \frac{0}{k + 1} + \frac{-1}{(k + 1)^2}[/tex]
- [Decomp] Simplify: [tex]\displaystyle \frac{2k + 1}{k^2(k + 1)^2} = \frac{1}{k^2} - \frac{1}{(k + 1)^2}[/tex]
- Substitute in decomp [Sum]: [tex]\displaystyle \sum^\infty_{k = 1} \frac{2k + 1}{k^2(k + 1)^2} = \sum^\infty_{k = 1} \bigg( \frac{1}{k^2} - \frac{1}{(k + 1)^2} \bigg)[/tex]
Step 3: Find Sum
- Find Sₙ terms: [tex]\displaystyle \sum^\infty_{k = 1} \bigg( \frac{1}{k^2} - \frac{1}{(k + 1)^2} \bigg) = \bigg( 1 - \frac{1}{4} \bigg) + \bigg( \frac{1}{4} - \frac{1}{9} \bigg) + \bigg( \frac{1}{9} - \frac{1}{16} \bigg) + ... + \bigg( \frac{1}{k^2} - \frac{1}{(k + 1)^2} \bigg) + ...[/tex]
- Find general Sₙ formula: [tex]\displaystyle S_n = 1 - \frac{1}{(k + 1)^2}[/tex]
- Sum of a series: [tex]\displaystyle \sum^\infty_{k = 1} \frac{2k + 1}{k^2(k + 1)^2} = \lim_{n \to \infty} \bigg( 1 - \frac{1}{(k + 1)^2} \bigg)[/tex]
- Evaluate limit [Limit Rule - Variable Direct Substitution]: [tex]\displaystyle \sum^\infty_{k = 1} \frac{2k + 1}{k^2(k + 1)^2} = 1 - 0[/tex]
- Simplify: [tex]\displaystyle \sum^\infty_{k = 1} \frac{2k + 1}{k^2(k + 1)^2} = 1[/tex]
∴ the sum converges to 1 by the Telescoping Series.
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Convergence Tests (BC Only)
Book: College Calculus 10e
