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Lecture Comments (1)

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Post by Austin Burns on April 27, 2014

For 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.

Answer

Integral 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
  • Integral Test 0:09
  • Example 1 0:54
  • Example 2 4:00
  • Example 3 7:59
  • Example 4 9:47