Series - A Alaca MATH 1005 Winter 2010 1 MATH 1005 WINTER...

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A. Alaca MATH 1005 Winter 2010 1 MATH 1005 WINTER 2010 LECTURE SLIDES Prepared by Ay¸ se Alaca Last modifed: February 25, 2010 (These Slides replace neither the Text Book nor the Lectures) Series The Integral Test The Comparison Tests Alternating Series Absolute Convergence, The Ratio and Root Tests
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A. Alaca MATH 1005 Winter 2010 2 INFINITE SERIES Defnition: Let ( a n ) be an infnite sequence. Then, ± n =1 a n = a 1 + a 2 + ··· + a n + is called an infnite series (or just series). The numbers a 1 ,a 2 , ... are called the terms oF the series. To fnd the sum oF an infnite series we consider the Following sequence oF partial sums: s 1 = a 1 s 2 = a 1 + a 2 s 3 = a 1 + a 2 + a 3 . . . s n - 1 = a 1 + a 2 + a 3 + + a n - 1 s n = a 1 + a 2 + a 3 + + a n - 1 + a n = n ± i =1 a i These partial sums Form a new sequence ( s n ). Note that s n - s n - 1 = a n .
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A. Alaca MATH 1005 Winter 2010 3 Defnition: Let ± n =1 a n be an infnite series and let s n = n ± i =1 a i . IF the sequence ( s n ) is convergent and lim n -→∞ s n = s , then we say that ± n =1 a n is convergent, and we call s the sum oF the series, and we write ± n =1 a n = s. lim n s n = s ⇐⇒ ± n =1 a n = s. IF ( s n ) is divergent, then ± n =1 a n is also divergent. Defnition: The series given by ± n =0 ar n = ± n =1 ar n - 1 = a + ar + ar 2 + ··· + ar n - 1 + ,a ± =0 . is called the geometric series with ratio r . (i) IF r =1: s n = n ± i =1 a = na and lim n →∞ s n = . Thus, ± n =1 ar n - 1 is divergent. (ii) IF r ± Then s n = a + ar + ar 2 + + ar n - 1 , rs n = ar + ar 2 + ar 3 + + ar n , s n - n = a - ar n = s n (1 - r )= a (1 - r n ) , s n = a (1 - r n ) 1 - r . By combining (i) and (ii) we have the Following: ± n =0 ar n = ± n =1 ar n - 1 = lim n s n = lim n a (1 - r n ) 1 - r = a 1 - r , iF - 1 <r< 1 divergent , iF r 1or r ≤- 1 . = a 1 - r , iF | r | < 1 divergent , iF | r |≥ 1 .
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A. Alaca MATH 1005 Winter 2010 4 Example: Is ± n =1 ( - 1) n 5 4 n convergent? Example: Show that ± n =1 1 ( n + 1)( n +2) is convergent and fnd its sum.
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A. Alaca MATH 1005 Winter 2010 5 Example: Show that the harmonic series ± n =1 1 n is divergent. Solution: s 1 =1 s 2 =1+ 1 2 s 4 1 2 + ² 1 3 + 1 4 ³ > 1+ 1 2 + ² 1 4 + 1 4 ³ s 4 > 2 2 s 8 1 2 + ² 1 3 + 1 4 ³ + ² 1 5 + 1 6 + 1 7 + 1 8 ³ > 1 2 + ² 1 4 + 1 4 ³ + ² 1 8 + 1 8 + 1 8 + 1 8 ³ s 8 > 3 2 Similarly, s 16 > 4 2 , s 32 > 5 2 , s 64 > 6 2 . In general, s 2 n > n 2 and so lim n -→∞ s 2 n = .( s n ) is divergent = ± n =1 1 n is divergent. Theorem: If the series ± n =1 a n is convergent, then lim n a n =0. proof: s n - s n - 1 = a n . lim n a n = lim n ( s n - s n - 1 ) = lim n s n - lim n s n - 1 = s - s =0 . Remark: In general, the converse of this theorem is not true. When lim n a n =0, ± n =1 a n may or may not be convergent: lim n 1 n = 0 but ± n =1 1 n is divergent. lim n 1 ( n + 1)( n +2) = 0 and ± n =1 1 ( n + 1)( n = 1 2 .
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A. Alaca MATH 1005 Winter 2010 6 Theorem ( n th term for divergence): If lim n -→∞ a n does not exist or lim n a n ± =0= ± n =1 a n is divergent.
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This note was uploaded on 02/28/2011 for the course MATH 101 taught by Professor Duke during the Spring '11 term at University of Ottawa.

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Series - A Alaca MATH 1005 Winter 2010 1 MATH 1005 WINTER...

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