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 0 answersPost by Aiden Song on January 5, 2016What 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.