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Final Review Guide

X n 1 ca n c x n 1 a n 2 x n 1 a n b n x n 1 a n x n

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X n =1 ca n = c X n =1 a n 2. X n =1 ( a n ± b n ) = X n =1 a n ± X n =1 b n * Exercises: pp.718-719 1-34 11.4 Series with Positive Terms * The Test for Divergence: If lim n →∞ a n does not exist or lim n →∞ a n 6 = 0, then X n =1 a n diverges. * Warning: It is NOT true in general that lim n →∞ a n = 0 implies that n =1 a n converges. Here is a counterexample, the harmonic series X n =1 1 n does not converge but lim n →∞ 1 n = 0. 11
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* The Integral Test: Suppose f is a continuous, positive, decreasing function on [1 , ). If f ( n ) = a n for all n 1 then X n =1 a n and Z 1 f ( x ) dx either both converge or both diverge. Note: there are several assumptions you need to verify are met before you can use the integral test. * Convergence of p-Series: The p-series X n =1 1 n p converges if p > 1 and diverges if p 1. * The Comparison Test: Suppose a n and b n are series with positive terms. 1. If b n is convergent and a n b n , for all n , then a n is also convergent. 2. If b n is divergent and a n b n , for all n , then a n is also divergent. * When using the comparison test, you need to verify that both series being compared are seris with positive terms. * Exercises: p.729 1-48 11.5 Power Series and Taylor Series * Terms: Taylor series, power series, interval of convergence, radius of convergence * Suppose that we are given the power series X n =0 a n ( x - a ) n Let R = lim n →∞ | a n a n +1 | 1. If R = 0, the series converges only for x = a . 2. If 0 < R < , the series converges for x in the interval ( a - R, a + R ) and diverges for x outside the interval [ a - R, a + R ] 3. If R = , the series converges for all x . * Suppose the function f is defined by f ( x ) = X n =0 a n ( x - a ) n = a 0 + a 1 ( x - 1) + a 2 ( x - 2) 2 + · · · with radius of convergence R > 0. Then f 0 ( x ) = X n =1 na n ( x - a ) n - 1 = a 1 + 2 a 2 ( x - a ) + 3 a 3 ( x - a ) 2 + · · · on the interval ( a - R, a + R ). * Taylor Series Representation of a Function: If a function f has derivatives of all orders in an open interval I = ( a - R, a + R )( R > 0) centered at x = a , then f ( x ) = X n =0 f ( n ) ( a ) n ! ( x - a ) n if and only if lim N →∞ R N ( x ) = 0 for all x in I , where R N ( x ) = f ( x ) - P N ( x ) with P N ( x ) the N th Taylor Polynomial. * Exercises: p.738 1-20 11.6 More on Taylor Series * Know the following Taylor Series: 12
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1. 1 1 - x = 1 + x + x 2 + · · · + x n + · · · = X n =0 x n , ( - 1 < x < 1) 2. e x = 1 + x + 1 2! x 2 + 1 3! x 3 + · · · + 1 n ! x n + · · · = X n =0 1 n ! x n , ( -∞ < x ) 3. ln x = ( x - 1) - 1 2 ( x - 1) 2 + 1 3 ( x - 1) 3 - · · · + ( - 1) n +1 n ( x - 1) n + · · · = X n =1 ( - 1) n +1 n ( x - 1) n , (0 < x 2) * Suppose f ( x ) = X n =0 a n ( x - a ) n = a 0 + a 1 ( x - a ) + a 2 ( x - a ) 2 + · · · + a n ( x - a ) n + · · · , x ( a - R, a + R ) Then Z f ( x ) dx = X n =0 a n n + 1 ( x - a ) n +1 + C = a 0 ( x - a ) + a 1 2 ( x - a ) 2 + a 2 3 ( x - a ) 3 + · · · + a n n + 1 ( x - a ) n +1 + · · · + C, x ( a - R, a + R ) * Know how to manipulate the above expansions using substitution, differentiation, and integration. * Exercises: pp.745 1-29 13
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