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Absolute Convergence

  • If converges then “absolutely converges”
  • Convergence check
    • Test for absolute convergence
    • If not absolutely convergent, test for conditional convergence or divergence

Absolute Convergence

Does the series ∑n = 1 ( − 1)n + 1[1/(n2)] converge absolutely?
  • Determine if ∑n = 1 | an | converge
  • n = 1 | ( − 1)n + 1[1/(n2)] | = ∑n = 1 | [1/(n2)] |
Apply P - series properties
n = 1 | [1/(n2)] | = 0 Thus the series converges completely
Does the series ∑n = 1 ( − 1)n + 1[(n − 1)/(n2 + 2)] converge absolutely?
  • Determine if ∑n = 1 | an | converge
  • n = 1 | ( − 1)n + 1[(n − 1)/(n2 + 2)] | = ∑n = 1 | [(n − 1)/(n2 + 2)] |
  • Apply Harmonic series properties
= [1/n] = diverges
Thus the series does not absolutely converge
Does the series ∑n = 1 [(( − 1)n + 1)/(√n + 3)] absolutely converge?
  • Determine if ∑n = 1 | an | converge
  • n = 1 | [(( − 1)n + 1)/(√n + 3)] | = ∑n = 1 | [1/(√n + 3)] |
  • Apply Comparison Test
  • n = 1 | [1/(√n + 3)] | compared with ∑n = 1 | [1/(√n )] |
  • Apply P - Series properties
  • p = [1/2] < 1 = diverges
Thus the series does not absolutely converge
Does the series ∑n = 1 [(( − 1)n)/(3n − 1)] absolutely converge?
  • Determine if ∑n = 1 | an | converge
  • n = 1 | [(( − 1)n)/(3n − 1)] | = ∑n = 1 | [1/(3n − 1)] |
a = 1r = [1/3]|r| < 1, thus the series absolutely converges
Does the series ∑n = 1 [(( − 1)n + 1)/(4√{n + 2})] absolutely converge?
  • Determine if ∑n = 1 | an | converge
  • n = 1 | [(( − 1)n + 1)/(4√{n + 2})] | = ∑n = 1 | [1/(4√{n + 2})] |
n = 1 | [1/(4√{n + 2})] | = ∑n = 1 | [1/(( n + 2 )1/4)] | p = [1/4] < 1 = diverges
Thus the series does not absolutely converge
Does the series ∑n = 1 [(( − 1)n + 1n2)/(3√{n})] absolutely converge?
  • Determine if ∑n = 1 | an | converge
= ∑n = 1 | n5/3 | = ∞ Thus the series does not absolutely converge
Does the series ∑n = 1 [(2( − 1)n(n + 2)2)/(n2 + 1)] absolutely converge?
  • Determine if ∑n = 1 | an | converge
  • n = 1 | [(2( − 1)n(n + 2)2)/(n2 + 1)] | = ∑n = 1 | [(2(n + 2)2)/(n2 + 1)] |
  • = ∑n = 1 | [(2(n2 + 2n + 4))/(n2 + 1)] |
  • = ∑n = 1 | [(2n2 + 4n + 8)/(n2 + 1)] |
  • = [(2n2)/(n2)]
  • = 2
Thus the series does absolutely converges
Does the series ∑n = 1 [(( − 1)n)/(3n + 1)] absolutely converge?
  • Determine if ∑n = 1 | an | converge
  • n = 1 | [(( − 1)n)/(3n + 1)] | = ∑n = 1 | [1/(3n + 1)] |
Let bn = [1/n]limn∞ [([1/(3n + 1)])/([1/n])] = [1/3] Thus the series absolutely converges
Does the series ∑n = 1 [(( − 1)n + 2(n − 1))/(n3)] absolutely converge?
  • Determine if ∑n = 1 | an | converge
  • n = 1 | [(( − 1)n + 2(n − 1))/(n3)] | = ∑n = 1 | [(n − 1)/(n3)] |
Let bn = [1/(n2)]limn∞ [([(n − 1)/(n3)])/([1/(n2)])] = 1 Thus the series absolutely converges
Does the series ∑n = 1 [(( − 1)n + 12n)/n!] absolutely converge?
  • Determine if ∑n = 1 | an | converge
  • n = 1 | [(( − 1)n + 12n)/n!] | = ∑n = 1 | [(2n)/n!] |
Apply the Ratio Test
n = 1 | [(2n)/n!] | = 0 Thus the series does absolutely converges

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Absolute Convergence

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Absolute Convergence 0:12
  • Example 1 0:52
  • Example 2 3:42
  • Example 3 5:21