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Unformatted text preview: 1 Convergence of Series Last time we introduced the first important test: the integral test. The integral test says the following: Theorem: Integral Test Suppose f ( x ) is continuous function which is eventually decreasing and positive, and a n = f ( n ). Then either 1 a n R 1 f ( x ) dx both converge, or they both diverge. This is a convenient test because it ALWAYS gives an definite outcome (assuming that one can test convergence of ONE of the series. Lets look at some examples: Example: The series X n =1 1 n p Converges if p > 1 and diverges if p 1. This is easy to see by applying the integral test. It is easy to see that f ( x ) = 1 x p is decreasing for p > 0, and is continuous on [1 , ). Computing the integral lim R Z R 1 dx x p = lim R x 1 p 1 p  R 1 (1) = lim r 1 1 p (1 R 1 p ) = p > 1 p < 1 (2) In the case p = 1 one gets lim R Z R 1 dx x = lim R ln( x )  R 1 = lim R ln( R ) = Aside: The function f ( p ) = X n =1 1 n p is called the Riemann Zeta function. There is a conjecture (the Riemann hyis called the Riemann Zeta function....
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This note was uploaded on 04/07/2008 for the course MATH 231 taught by Professor Bronski during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 Bronski
 Math, Calculus

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