Homework Help is Here – Start Your Trial Now!
Math
Calculus
00 2 ln k k (2) Σ k=1 O 2 In k k 2 ln k k 2 In k k 2 In k k 1 k > 00 k > 2, Σ = diverges, Σ k k=1 k=1 1 k for k ≥ 2, Σ k=1 1 1 <= for k ≥ 2, Σ converges, k k k=1 1 k for for k ≥ 2, converges, Σ k=1 1 Σε k k=1 2 In k k k=1 diverges, Σ k=1 diverges 2 In k k 2 In k k 2 ln k k converges converges diverges

00 2 ln k k (2) Σ k=1 O 2 In k k 2 ln k k 2 In k k 2 In k k 1 k > 00 k > 2, Σ = diverges, Σ k k=1 k=1 1 k for k ≥ 2, Σ k=1 1 1 <= for k ≥ 2, Σ converges, k k k=1 1 k for for k ≥ 2, converges, Σ k=1 1 Σε k k=1 2 In k k k=1 diverges, Σ k=1 diverges 2 In k k 2 In k k 2 ln k k converges converges diverges

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.2: Arithmetic Sequences
Problem 68E
See similar textbooks
icon
Related questions
Question
In each part, use the comparison test to show that the series converges or diverges. Select the correct comparison test for each series. Σ (a) k=1 O 2 In k k 2 In k k 2 ln k k 2 ln k k 2 In k k <= f V > for k ≥ 2, diverges, 00 1 <for k ≥ 2, k 1 > f k k=1 ∞ for k ≥ 2, converges, k k= 00 k=1 k 00 11 1 k=1 converges, for k ≥ 2, diverges, k=1 2 In k k 00 00 k=1 diverges 2 In k k 2 ln k k 2 In k k converges converges diverges
Transcribed Image Text:In each part, use the comparison test to show that the series converges or diverges. Select the correct comparison test for each series. Σ (a) k=1 O 2 In k k 2 In k k 2 ln k k 2 ln k k 2 In k k <= f V > for k ≥ 2, diverges, 00 1 <for k ≥ 2, k 1 > f k k=1 ∞ for k ≥ 2, converges, k k= 00 k=1 k 00 11 1 k=1 converges, for k ≥ 2, diverges, k=1 2 In k k 00 00 k=1 diverges 2 In k k 2 ln k k 2 In k k converges converges diverges
100 (1) Σ k=1 Ο 0 132 k k³/12 - -/- k k³12 k3/2/ 1 k k3/12 k312 1 - k 1 1 Λ V Σ √k √k √k k=1 √k Σ k=1 - - - √k √k 00 converges, Σ k=1 00 diverges, Σ k=1 00 diverges, Σ k=1 k312 k k312 00 converges, Σ k=1 k k312 3/2 k³ - k και μπορείς χρόνια μετά converges diverges diverges 1 diverges
Transcribed Image Text:100 (1) Σ k=1 Ο 0 132 k k³/12 - -/- k k³12 k3/2/ 1 k k3/12 k312 1 - k 1 1 Λ V Σ √k √k √k k=1 √k Σ k=1 - - - √k √k 00 converges, Σ k=1 00 diverges, Σ k=1 00 diverges, Σ k=1 k312 k k312 00 converges, Σ k=1 k k312 3/2 k³ - k και μπορείς χρόνια μετά converges diverges diverges 1 diverges
Expert Solution
steps

Step by step

Solved in 3 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer