# 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 modified: 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 Definition: Let ( a n ) be an infinite sequence. Then, n =1 a n = a 1 + a 2 + · · · + a n + · · · is called an infinite series (or just series). The numbers a 1 , a 2 , ... are called the terms of the series. To find the sum of an infinite 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 .
A. Alaca MATH 1005 Winter 2010 3 Definition: Let n =1 a n be an infinite 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. Definition: 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 = 1: Then s n = a + ar + ar 2 + · · · + ar n - 1 , rs n = ar + ar 2 + ar 3 + · · · + ar n , s n - rs 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 1 or 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 find its sum.
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 + 1 2 + 1 3 + 1 4 > 1 + 1 2 + 1 4 + 1 4 s 4 > 1 + 2 2 s 8 = 1 + 1 2 + 1 3 + 1 4 + 1 5 + 1 6 + 1 7 + 1 8 > 1 + 1 2 + 1 4 + 1 4 + 1 8 + 1 8 + 1 8 + 1 8 s 8 > 1 + 3 2 Similarly, s 16 > 1 + 4 2 , s 32 > 1 + 5 2 , s 64 > 1 + 6 2 . In general, s 2 n > 1 + 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 + 2) = 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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