# notes7 - Tests for Convergence II Last time we noticed how...

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Unformatted text preview: Tests for Convergence II Last time we noticed how we could find an upper bound for { s n } provided that all the a k s were not negative. By finding a decreasing continuous function f ( x ) such that f ( k ) = a k and integrating it we saw, s n < Z n f ( x ) dx . And this made sense from the picture Math 9C Summer 2011 (UCR) Pro-Notes October 10, 2011 1 / 9 And this gave us a great way to tell if the infinite series X n =1 a n converges by asking if Z f ( x ) dx converges. Now notice by shifting each rectangle to the right by one in the previous picture, we can use the same function f , but this time the area under the curve of f is smaller than the sum of all the a k s. Math 9C Summer 2011 (UCR) Pro-Notes October 10, 2011 2 / 9 In other words, this time Z n +1 1 f ( x ) dx < s n . This is a really great thing because now, we know that if Z 1 f ( x ) dx diverges to infinity, then so does X k =1 a k ....
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## This note was uploaded on 02/22/2012 for the course MATH 9c taught by Professor C during the Fall '11 term at UC Riverside.

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notes7 - Tests for Convergence II Last time we noticed how...

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