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For more information, please see full course syllabus of Calculus BC

• ## Related Books

 0 answersPost by Austin Burns on April 27, 2014For example 2, doesn't the answer 2/e just mean that the series converges, not that the series converges to 2/e?

### Integral Test

• Integral test of convergence:
• Check conditions
• Apply L’Hôpital’s Rule

### Integral Test

Can the Integral Test be applied on the ∑n = 1 [1/(n − 1)] ?
Analyze conditions
If f(n) = [1/(n − 1)], then f(x) is not continuous on the intertval [1,∞] on x = 1 Thus the Integral Test cannot be applied
Can the Integral Test be applied on the ∑n = 1 [(n3)/(n + 1)] ?
Analyze conditions
If f(n) = [(n3)/(n + 1)], then f(1) < f(2) and thus f(x) is increasing.
Thus the Integral Test cannot be applied.
Solve ∫ [(x3)/([(x4)/4] + 1)]dx
• Use substitution with u = [(x4)/4] + 1
• u = [(x4)/4] + 1
• du = x3dx
Integrate with substitution
[(x3)/([(x4)/4] + 1)]dx = ∫ [du/u] = ln|[(x4)/4] + 1| + C
Does the series ∑[(n3)/([(n4)/4] + 1)] converge?
• Integrate f(x) from 1 to ∞
• 1 [(n3)/([(n4)/4] + 1)] = ∫1 [(x3)/([(x4)/4] + 1)]
• Integrate the improper integral
• 1 [(x3)/([(x4)/4] + 1)] = limk∞1k [(x3)/([(x4)/4] + 1)]
• = limk∞ [ ln|[(x4)/4] + 1| ]1k
• = limk∞ [ ln|[(k4)/4] + 1| − ln|[1/4] + 1| ]
• = ∞
Thus the series diverges
Solve ∫ [(x + 61)/(ex)]dx
• Setup integration by parts
• u = x + 61
• du = dx
• dv = [1/(ex)]
• v = − [1/(ex)]
[(x + 61)/(ex)]dx = − [(x + 61)/(ex)] + ∫ [dx/(ex)] = − [(x + 61)/(ex)] − [1/(ex)] + C = [(x + 60)/(ex)] + C
Does the series ∑[(x + 61)/(ex)] converge or diverge?
• Integrate f(x) from 1 to ∞
• 1 [(n3)/([(n4)/4] + 1)] = ∫1 [(x + 61)/(ex)]
• Integrate the improper integral
• 1 [(x + 61)/(ex)] = limk∞1k [(x + 61)/(ex)]
• = limk∞ [ [(x + 60)/(ex)] ]1k
• = limk∞ [ [(k + 60)/(ex)] − [61/(e)] ]
• = limk∞ [ [k/(ex)] − [1/e] ]
• = 0 − [1/e]
• = − [1/e]
Thus the series converges.
Solve ∫ [1/(x3)]dx
Integrate
[1/(x3)]dx = − [(x − 2)/2] + C
Does the series 7 + [7/8] + [7/27] + [7/64] + ... converge or diverge?
• Determine the series formula
• 7 + [7/8] + [7/27] + [7/64] + ... = ∑[7/(n3)]
• Integrate f(x) from 1 to ∞
• 1 [7/(n3)] = ∫1 [7/(x3)]
• = 7∫1 [1/(x3)]
• = 7limk∞ [ − [1/(2x2)] ]1k
• = 7limk∞ [ − [1/(2k2)] + [1/2] ]
• = [7/2]
Thus the series converges
Determine if the series ∑n = 1 [1/(3√{n})] converges with P - series conditions
Analyze pp = [1/3] ≤ 1, thus the series diverges
Determine if the series ∑n = 1 [1/(3√{n})] converges with P - series conditions
• Integrate f(x) from 1 to ∞
• 1 [1/(3√{n})] = ∫1 [1/(3√{x})]
• = limk∞ [ [3/2]x[2/3] ]1
• = [3/2]limk∞ [ k[2/3] − 1 ]
• = ∞
Thus the series diverges.

*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.