Start learning today, and be successful in your academic & professional career. Start Today!

• ## Related Books

### Start Learning Now

Our free lessons will get you started (Adobe Flash® required).

### Membership Overview

• *Ask questions and get answers from the community and our teachers!
• Practice questions with step-by-step solutions.
• Track your course viewing progress.
• Learn at your own pace... anytime, anywhere!

### 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.