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Lecture Comments (1)

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Post by Aiden Song on January 5 at 11:57:45 PM

What happened if L=1 and you have to use another test, how to use? Please specify! Thank you.

Ratio Test

  • Ratio test: testing the limit of 1 future term to 1 current term
  • Apply directly and solve for L

Ratio Test

Does the series ∑3n converges or diverges
  • Apply the Ratio Test, find an + 1
  • an + 1 = 3n + 1
L = limn → ∞ [(3n + 1)/(3n)] = 3L > 1, thus ∑3n diverges
Does the series ∑[1/(2n)] converges or diverges
  • Apply the Ratio Test, find an + 1
  • an + 1 = [1/(2n + 1)]
  • Find and analyze L
= limn → ∞ [(2n)/(2n + 1)] = [1/2]L < 1, thus ∑[1/(2n)] converges
Does the series ∑[n/(3n)] converges or diverges
  • Apply the Ratio Test, find an + 1
  • an + 1 = [(n + 1)/(3n + 1)]
  • Find and analyze L
  • L = limn → ∞ [([(n + 1)/(3n + 1)])/([n/(3n)])]
  • = limn → ∞ [(n + 1)/(3n(3))]( [(3n)/n] )
  • = limn → ∞ [(n + 1)/3n]
  • = [n/3n]
  • = [1/3]
L < 1, thus ∑[1/(2n)] converges
Does the series ∑[(5n)/(n2)] converges or diverges
  • Apply the Ratio Test, find an + 1
  • an + 1 = [(5n + 1)/((n + 1)2)]
  • = [(5n + 1)/(n2 + 2n + 1)]
  • Find and analyze L
  • L = limn → ∞ [([(5n + 1)/(n2 + 2n + 1)])/([(5n)/(n2)])]
  • = limn → ∞ [(5n + 1)/(n2 + 2n + 1)]( [(n2)/(5n)] )
= [(5n2)/(n2)] = 5L > 1, thus the series diverges
Can the Ratio Test be applied to the series ∑[(n + 1)/(n − 1)]?
  • Apply the Ratio Test, find an + 1
  • an + 1 = [(n + 2)/n]
  • Find and analyze L
  • L = limn → ∞ [([(n + 2)/n])/([(n + 1)/(n − 1)])]
  • = limn → ∞ [(n + 2)/n]( [(n − 1)/(n + 1)] )
= [(n2)/(n2)] = 1L = 1, thus another test should be applied.
Does the series ∑[2/((n + 1)!)] converges or diverges
  • Apply the Ratio Test, find an + 1
  • an + 1 = [2/((n + 2)!)]
  • Find and analyze L
  • L = limn → ∞ [([2/((n + 2)!)])/([2/((n + 1)!)])]
  • = limn → ∞ [2/((n + 2)!)]( [((n + 1)!)/2] )
  • = limn → ∞ [1/(n + 2)]
  • = 0
L < 1, thus the series converges
Does the series ∑[(2n)/n!] converges or diverges
  • Apply the Ratio Test, find an + 1
  • an + 1 = [(2n + 1)/((n + 1)!)]
  • Find and analyze L
  • L = limn → ∞ [([(2n + 1)/((n + 1)!)])/([(2n)/n!])]
= limn → ∞ [2/(n + 1)] = 0L < 1, thus the series converges
Does the series ∑[n!/((n − 10)!)] converges or diverges
  • Apply the Ratio Test, find an + 1
  • an + 1 = [((n + 1)!)/((n − 9)!)]
  • Find and analyze L
  • L = limn → ∞ [([((n + 1)!)/((n − 9)!)])/([n!/((n − 10)!)])]
  • = limn → ∞ [((n + 1)!)/((n − 9)!)]( [((n − 10)!)/n!] )
  • = limn → ∞ ( n + 1 )(n − 10)
  • = ∞
L > 1, thus the series diverges
Does the series ∑[(4n)/(6n)] converges or diverges
  • Apply the Ratio Test, find an + 1
  • an + 1 = [(4n + 1)/(6n + 1)]
  • Find and analyze L
  • L = limn → ∞ [([(4n + 1)/(6n + 1)])/([(4n)/(6n)])]
= limn → ∞ [4/6] = [4/6]L < 1, thus the series converges
Does the series ∑[(2n + 1)/(3n − 1)] converges or diverges
  • Apply the Ratio Test, find an + 1
  • an + 1 = [(2n + 2)/(3n)]
  • Find and analyze L
  • L = limn → ∞ [([(2n + 2)/(3n)])/([(2n + 1)/(3n − 1)])]
  • = limn → ∞ [(2n + 2)/(3n)]( [(3n − 1)/(2n + 1)] )
  • = [2/3]
L < 1, thus the series 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

Ratio 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
  • Ratio Test 0:09
  • Example 1 0:57
  • Example 2 2:55
  • Example 3 6:27
  • Example 4 8:36