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Unformatted text preview: k = 2 ` + 1 is odd, we have ( k !) 1 /k = (1 2 3 ` ( ` + 1) k )  {z } k = 2 ` + 1 factors 1 /k (( ` + 1) k )  {z } ` + 1 factors 1 /k ( ` + 1) ` +1 k ( 1 2 k ) 1 2 (since ` + 1 = k +1 2 > 1 2 k ` + 1 k > 1 2 ) In either case, ( k !) 1 /k ( 1 2 k ) 1 2 as k . Thus r = 0 and we have convergence. example 1.3. See whether or not the the root test applies to the series X k =1 1 2 k +(1) k . solution. We have  a k  1 /k = 1 2 k +(1) k 1 /k = 1 2 ( k +(1) k ) /k = 1 2 1+ 1 k (1) k 1 2 as k Since lim k  a k  1 /k = 1 2 < 1 the series converges. 2...
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This note was uploaded on 03/04/2012 for the course MATH 10250 taught by Professor Himonas during the Fall '08 term at Notre Dame.
 Fall '08
 HIMONAS
 Calculus

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