How to find if a series is convergent or divergent

Asked by admin @ 22/11/2021 in Mathematics viewed by 289 People

INFINITE SERIES

Determine if the series converges or diverges. For convergent series, find the sum of the series.
I know the answer, but I don’t know how to get to it.

Answered by admin @ 22/11/2021

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]: $\displaystyle \lim_{x \to c} x = c$
Limit Property [Addition/Subtraction]: $\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)$

Sequences

Series

Definition of a convergent/divergent series
Sum of a series: $\displaystyle \lim_{n \to \infty} S_n$

Telescoping Series: $\displaystyle \sum^\infty_{k = 1} (b_1 - b_{k + 1}) = (b_1 - b_2) + (b_2 - b_3) + (b_3 - b_4) + ... + (b_1 - b_{k + 1}) + ...$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \sum^\infty_{k = 1} \frac{2k + 1}{k^2(k + 1)^2}$

Step 2: Rewrite Sum

Factor: $\displaystyle \sum^\infty_{k = 1} \frac{2k + 1}{k^2(k + 1)^2} = \sum^\infty_{k = 1} \frac{2k + 1}{k \cdot k(k + 1)(k + 1)}$
Break up [Partial Fraction Decomposition]: $\displaystyle \frac{2k + 1}{k^2(k + 1)^2} = \frac{A}{k} + \frac{B}{k^2} + \frac{C}{k + 1} + \frac{D}{(k + 1)^2}$
Simplify [Common Denominator]: $\displaystyle 2k + 1 = Ak(k + 1)^2 + B(k + 1)^2 + Ck^2(k + 1) + Dk^2$
[Decomp] Expand: $\displaystyle 2k + 1 = Ak^3 + Ck^3 + A2k^2 + Bl^2 + Ck^2 + Dk^2 + Ak + B2k + B$
[Decomp] Factor: $\displaystyle 2k + 1 = k^3(A + C) + k^2(2A + B + C + D) + k(A + 2B) + B$
[Decomp] Set up systems: $\displaystyle \begin{cases} A + C = 0 \\ 2A + B + C + D = 0 \\ A + 2B = 2 \\ B = 1 \end{cases}$
[Decomp] Solve: $\displaystyle A = 0, \ B = 1, \ C = 0, \ D = -1$
[Decomp] Substitute in variables: $\displaystyle \frac{2k + 1}{k^2(k + 1)^2} = \frac{0}{k} + \frac{1}{k^2} + \frac{0}{k + 1} + \frac{-1}{(k + 1)^2}$
[Decomp] Simplify: $\displaystyle \frac{2k + 1}{k^2(k + 1)^2} = \frac{1}{k^2} - \frac{1}{(k + 1)^2}$
Substitute in decomp [Sum]: $\displaystyle \sum^\infty_{k = 1} \frac{2k + 1}{k^2(k + 1)^2} = \sum^\infty_{k = 1} \bigg( \frac{1}{k^2} - \frac{1}{(k + 1)^2} \bigg)$

Step 3: Find Sum

Find Sₙ terms: $\displaystyle \sum^\infty_{k = 1} \bigg( \frac{1}{k^2} - \frac{1}{(k + 1)^2} \bigg) = \bigg( 1 - \frac{1}{4} \bigg) + \bigg( \frac{1}{4} - \frac{1}{9} \bigg) + \bigg( \frac{1}{9} - \frac{1}{16} \bigg) + ... + \bigg( \frac{1}{k^2} - \frac{1}{(k + 1)^2} \bigg) + ...$
Find general Sₙ formula: $\displaystyle S_n = 1 - \frac{1}{(k + 1)^2}$
Sum of a series: $\displaystyle \sum^\infty_{k = 1} \frac{2k + 1}{k^2(k + 1)^2} = \lim_{n \to \infty} \bigg( 1 - \frac{1}{(k + 1)^2} \bigg)$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \sum^\infty_{k = 1} \frac{2k + 1}{k^2(k + 1)^2} = 1 - 0$
Simplify: $\displaystyle \sum^\infty_{k = 1} \frac{2k + 1}{k^2(k + 1)^2} = 1$

∴ 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

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