homework_5 - Homework 5 9.14. Let sn = lim |a an np , so...

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Homework 5 9.14. Let s n = a n n p , so lim | s n +1 s n | is lim | a n p ( n + 1) p | = lim | a || ( n n + 1 ) p | = | a | lim | ( n n + 1 ) p | = | a | lim((1 - 1 n + 1 ) p ) = | a | (lim(1 - 1 n + 1 )) p = | a | (lim(1) - lim( 1 n + 1 )) p = | a | where we used the scalar multiplication theorem for limits once, we used the multiplication theorem for limits p times, and we used the addition theorem for limits once. If | a | < 1 then by 9.12(a) we have that lim s n = 0. If | a | > 1, then a > 1 or a < - 1. If a > 1 then all of the terms of s n are positive, and so we have by 9.12(b) that + = lim | s n | = lim s n . If a < - 1 then | s n | diverges to infinity by 9.12(b), but the even terms of s n are positive while the odd terms of s n are negative. Therefore, the limit does not exist in this case. 9.18. This problem introduces the geometric series . (a) Observe that (1 - a )(1+ a + a 2 + ··· + a n ) = (1+ a + a 2 + ··· + a n ) - ( a + a 2 + ··· + a n +1 ) = 1 - a n +1 so if a 6 = 1 then we can divide the left and right sides of this equation
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homework_5 - Homework 5 9.14. Let sn = lim |a an np , so...

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