Lecture 24 - MATH 118, LECTURE 24: ABSOLUTE CONVERGENCE 1...

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Unformatted text preview: MATH 118, LECTURE 24: ABSOLUTE CONVERGENCE 1 Absolute Convergence All of the series ∑ ∞ n =1 c n we have dealt with so far have had nonnega- tive terms c n . This begs the question of how we determine the conver- gence/divergence of series with an infinite number of both positive and neg- ative terms. To handle this case, we introduce the concept of absolute convergence . A series is said to be absolutely convergent if ∞ summationdisplay n =1 | c n | converges. We notice that the series of absolute values is nonnegative and consequently all of our convergence results apply to this series. This fact is made useful by the following result: Proposition 1.1 (Absolute Convergence Test) . If the series ∑ ∞ n =1 | c n | con- verges, then the series ∑ ∞ n =1 c n converges. Justification: We will divide the series ∑ ∞ n =1 c n into positive and negative components as follows: Let { p n } be the sequence of c n with positive values Let { n n } be the sequence of absolute values of c n with negative values . It is clear from these definitions that ∞ summationdisplay n =1 c n = ∞ summationdisplay n =1 p n- ∞ summationdisplay n =1 n n . What we need to show is that ∑ ∞ n =1 | c n | converging implies that ∑ ∞ n =1 p n and ∑ ∞ n =1 n n converge. Given this decomposition of positive and negative terms in { c n } we have that ∞ summationdisplay n =1 | c n | = ∞ summationdisplay n =1 p n + ∞ summationdisplay n =1 n n = L < ∞ 1 since...
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This note was uploaded on 04/14/2009 for the course ENGINEERIN MATH 118 taught by Professor Soares during the Spring '09 term at Waterloo.

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Lecture 24 - MATH 118, LECTURE 24: ABSOLUTE CONVERGENCE 1...

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