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Limit Comparison Test

  • If is finite and not equal to 0:
    • Series both converge or diverge
  • Save time by choose a series with known convergence properties

Limit Comparison Test

Does the series ∑[1/(7n − 3)] converge?
  • Use bn = [1/n] for Limit Comparison Test
= [n/7n] = [1/7] Thus the series diverges because bn diverges
Does the series ∑[3/(19 − n)] converge?
  • Use bn = [1/n] for Limit Comparison Test
= − [3n/n] = − 3 Thus the series diverges because bn diverges
Does the series ∑[1/(√{n + 1} )] converge?
  • Use bn = [1/(√n )] for Limit Comparison Test
= [(√n )/(√n )] = 1 Thus the series converges because bn converges
Does the series ∑[1/(n2 + 2n + 1)] converge?
  • Use bn = [1/(n2)] for Limit Comparison Test
  • limn → ∞ [([1/(n2 + 2n + 1)] )/([1/(n2)])] = limn → ∞ [(n2)/(n2 + 2n + 1)]
  • = [(n2)/(n2)]
  • = 1
Thus the series converges because bn converges
Is it appropriate to use the Limit Comparison Test on an = n3 + 3n if bn = n2?
Analyze conditions for Limit Comparison Test
No it is not appropriate because limn → ∞ [(an)/(bn)] is infinite
Does the series ∑[(n − 2)/(n3 − 1)] converge?
  • Use bn = [1/(n2)] for Limit Comparison Test
  • limn → ∞ [([(n − 2)/(n3 − 1)])/([1/(n2)])] = limn → ∞ [(n2( n − 2 ))/(n3 − 1)]
= [(n3)/(n3)] = 1 Thus the series converges because bn converges
Does the series ∑[(√n )/((2n − 1)2)] diverge?
  • Use bn = [1/(n3/2)] for Limit Comparison Test
  • limn → ∞ [([(√n )/((2n − 1)2)])/([1/(n3/2)])] = limn → ∞ [(√n (n3/2))/((2n − 1)2)]
= [(n2)/(4n2)] = [1/4] Thus the series converges because bn converges
Does the series ∑[(7n2)/(9n3 + 8n2 + 2n + 3)] diverge?
  • Use bn = [1/n] for Limit Comparison Test
  • limn → ∞ [([(7n2)/(9n3 + 8n2 + 2n + 3)])/([1/n])] = limn → ∞ [(7n3)/(9n3 + 8n2 + 2n + 3)]
  • = [(7n3)/(9n3)]
  • = [7/9]
Thus the series diverges because bn diverges
Does the series ∑[(3n2 + n)/(5n3 − 12n2 − n + 17)] diverge?
  • Use bn = [1/n] for Limit Comparison Test
= [(3n3)/(5n3)] = [3/5] Thus the series diverges because bn diverges
Does the series 3 + [3/4] + [1/3] + [3/16] + ... converge?
  • Analyze the series
  • 3 + [3/4] + [1/3] + [3/16] + ... = ∑[3/(n2)]
  • Use bn = [1/(n2)] for Limit Comparison Test
  • limn → ∞ [([3/(n2)])/([1/(n2)])] = limn → ∞ [(3n2)/(n2)]
  • = 3
Thus the series converges because bn 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

Limit Comparison Test

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
  • Limit Comparison Test 0:09
    • Conditions
  • Example 1 1:01
  • Example 2 2:53
  • Example 3 4:15
  • Example 4 6:19