University:
Amherst College
Course:
MATH 121 | Intermediate Calculus
Academic year:

2023
Views:

285

Pages:

2
Author:

Reese Pittman

Absolute Convergence Test (ACT) Given a series ∞ X ∞ X n=1 an , if the Absolute Series ∞ X |an | Converges, then the Original Series n=1 an Converges. n=1 USED: When the Absolute Series is easier to analyze. USED: To avoid analyzing negative signs, or maybe the Alternating Series Test. BENEFITS: Handling series with positive terms is usually easier. Plus, you have more convergence tests available for positive termed series; think IT, CT, LCT. WARNING: If you analyze the Absolute Series, and it is NOT convergent, then this convergence test is Inconclusive and not an option for use. It does not immediately follow up with a conclusion about Conditional Convergence without more work. All you know if the the A.S. diverges is that your original series is NOT Absolutely Convergent (A.C.) NOTE: If the Original Series is already equal to the Absolute series, then Absolute Convergence is the same as Convergence. NOTE: If the Original Series is not the same as the Absolute Series, then this ACT says that Absolute Convergence implies Convergence. APPROACH: • Given the original series, find the Absolute Series by ignoring any negative signs. • Analyze the Absolute Series completely as before. Run all the details of p-Series Test, IT, CT, or LCT. Justify everything. Make a clear conclusion about the convergence of the Absolute Series. • If the A.S. Converges, then you are done, and the O.S. Converges by ACT. • If the A.S. diverges, you have no conclusion about the O.S. You are stuck running a Convergence test (likely AST) on the O.S. EXAMPLES: Determine and state whether each of the following series Converges or Diverges. Name any Convergence test(s) that you use, and justify all of your work. ∞ ∞ X X 1 n 1 1. (−1) → Check the Absolute Series which is a Convergent p-Series with 5 n n5 n=1 n=1 ∞ X 1 p = 5 > 1. Then the Original Series (−1)n 5 is Convergent by ACT . n n=1 NOTE: this technique helps you avoid AST here on the O.S. ∞ X sin2 n sin2 n → Consider the A.S. . 2. (−1) 3 n +5 n3 + 5 n=1 n=1 ∞ X n ∞ X 1 sin2 n sin2 n 1 Bound the terms: 3 ≤ ≤ and is a convergent p-Series with p = 3 > 1. 3 n +5 n3 n3 n n=1 Therefore, the A.S. is Convergent by CT. Finally, the O.S. is Convergent by ACT . ∞ X ∞ ∞ ∞ X n3 + 5n n3 + 5n X n3 X 1 3. (−1) → A.S. ≈ = Conv. p-series p = 5 > 1. 8+7 8+7 8 5 n n n n n=1 n=1 n=1 n=1 n 0 n3 + 5n 5 7 8 6 8 1 + n + 5n n2 Check: lim n + 7 = lim = lim 0= 1 Finite, Non-zero. 1 n→∞ n→∞ n8 + 7 n→∞ 7 1 + n78 n5 Therefore the A.S. is also convergent by LCT (which by definition means that the A.S. is A.C.). Finally, the O.S. is convergent by ACT. ∞ ∞ ∞ π X X arctan n arctan n πX 1 2 (−1) → A.S. ≈ = Constant Mutliple of a 4. 2+9 2+9 2 2 n n n 2 n n=1 n=1 n=1 n=1 Convergent p-series p = 2 > 1 is Convergent. π π arctan n 2 Bound the terms: ≤ 2 ≤ 22 . We analyzed the comparison series above. 2 n +9 n +9 n ∞ X n Finally, the A.S. is convergent by CT which implies that the O.S. is convergent by ACT. LAST NOTE: It is very important to finish the full and complete analysis of the Absolute Series. Make a conclusion about the Convergence of the Absolute Series. THEN you can use that to imply convergence of the O.S. using the ACT.

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