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Unformatted text preview: MAT 137Y 200809 Winter Session, Solutions to Problem Set 12 1 (SHE 12.2) 4. Using partial fractions, we have k = 1 ( k + 1 )( k + 3 ) = 1 2 k = 1 k + 1 1 k + 3 . This gives us a telescoping sum; if we look at the partial sums, we have s n = 1 2 1 1 3 + 1 2 1 4 + 1 3 1 5 + + 1 n 1 n + 2 + 1 n + 1 1 n + 3 = 1 2 1 + 1 2 1 n + 2 1 n + 3 Therefore, k = 1 ( k + 1 )( k + 3 ) = lim n s n = lim n 1 2 1 + 1 2 1 n + 2 1 n + 3 = 3 4 . 6. k = ( 1 ) k 5 k = k = 1 5 k = 1 1 + 1 5 = 5 6 . 10. k = 2 3 k 1 4 3 k + 1 = k = 3 k + 1 4 3 k + 7 = 3 4 7 k = 3 4 3 k = 3 4 7 1 1 3 4 3 = 3 15616 . 20. Since lim n 1 4 5 4 n is not zero, it follows by the basic divergence test that the series diverges. 28. The total length removed in the Cantor set is 1 3 + 2 9 + 4 27 + = 1 3 k = ( 2 3 ) k = 1 3 1 1 2 3 = 1. Some points that are in the Cantor Set include 0 , 1 , 1 3 , 2 3 , 1 9 , 2 9 , 7 9 , 8 9 . In fact, for all positive integers n , the point 1 / 3 n is a point on the Cantor set. This leads to the curious paradox that despite the total length removed in the set is 1, there are infinitely many points on the Cantor set!length removed in the set is 1, there are infinitely many points on the Cantor set!...
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This note was uploaded on 04/09/2010 for the course MAT 137 taught by Professor Uppal during the Spring '08 term at University of Toronto Toronto.
 Spring '08
 UPPAL
 Fractions

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