Chap11_Sec6 - 11 INFINITE SEQUENCES AND SERIES INFINITE...

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11 INFINITE SEQUENCES AND SERIES INFINITE SEQUENCES AND SERIES
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11.6 Absolute Convergence and the Ratio and Root tests In this section, we will learn about: Absolute convergence of a series and tests to determine it. INFINITE SEQUENCES AND SERIES
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ABSOLUTE CONVERGENCE Given any series Σ a n , we can consider the corresponding series whose terms are the absolute values of the terms of the original series. 1 2 3 1 ... n n a a a a = = + + +
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ABSOLUTE CONVERGENCE A series Σ a n is called absolutely convergent if the series of absolute values Σ | a n | is convergent. Definition 1
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ABSOLUTE CONVERGENCE Notice that, if Σ a n is a series with positive terms, then | a n | = a n. So, in this case, absolute convergence is the same as convergence.
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ABSOLUTE CONVERGENCE The series is absolutely convergent because is a convergent p -series ( p = 2). Example 1 1 2 2 2 2 1 ( 1) 1 1 1 1 ... 2 3 4 n n n - = - = - + - + 1 2 2 2 2 2 1 1 ( 1) 1 1 1 1 1 ... 2 3 4 n n n n n - = = - = = + + + +
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ABSOLUTE CONVERGENCE We know that the alternating harmonic series is convergent. See Example 1 in Section 11.5. Example 2 1 1 ( 1) 1 1 1 1 ... 2 3 4 n n n - = - = - + - +
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However, it is not absolutely convergent because the corresponding series of absolute values is: This is the harmonic series ( p -series with p = 1) and is, therefore, divergent. ABSOLUTE CONVERGENCE Example 2 1 1 1 ( 1) 1 1 1 1 1 ... 2 3 4 n n n n n - = = - = = + + + +
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CONDITIONAL CONVERGENCE A series Σ a n is called conditionally convergent if it is convergent but not absolutely convergent. Definition 2
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ABSOLUTE CONVERGENCE Example 2 shows that the alternating harmonic series is conditionally convergent. Thus, it is possible for a series to be convergent but not absolutely convergent. However, the next theorem shows that absolute convergence implies convergence.
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ABSOLUTE CONVERGENCE If a series Σ a n is absolutely convergent, then it is convergent. Theorem 3
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Observe that the inequality is true because | a n | is either a n or – a n. 0 2 n n n a a a + ABSOLUTE CONVERGENCE Theorem 3—Proof
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a n is absolutely convergent, then Σ | a n | is convergent. So, Σ 2| a n | is convergent. Thus, by the Comparison Test, Σ ( a n + | a n |) is convergent. ABSOLUTE CONVERGENCE
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This note was uploaded on 01/06/2012 for the course MATH 2414.S01 taught by Professor Alans.grave during the Fall '11 term at Collins.

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Chap11_Sec6 - 11 INFINITE SEQUENCES AND SERIES INFINITE...

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