118 - 2 4 6 ... (2 n ) = lim n ( n + 1) 2 | x | n +1 2 4...

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MATH 153 Selected Solutions for § 11.8 Exercise 18 : Find the radius of convergence and interval of convergence of the series. X n =0 n 4 n ( x + 1) n Solution : We start by applying the Ratio Test. lim n →∞ ± ± ± ± n +1 4 n +1 ( x + 1) n +1 n 4 n ( x + 1) n ± ± ± ± = lim n →∞ ( n + 1) | x + 1 | n +1 4 n +1 · 4 n n | x + 1 | n = lim n →∞ ( n + 1) | x + 1 | 4 n = lim n →∞ | x + 1 | 4 . This limit is less than one when x is in ( - 5 , 3). This give a radius of conver- gence of R = 4. We have only to check the end points, x = - 5 and x = 3. If x = - 5, we have the series ( - 1) n n , which clearly diverges as the terms don’t go to 0 (Test for Divergence). Similarly, if x = 3 we have the series n , which also diverges. Thus, the interval of convergence is I = ( - 5 , 3). Exercise 24 : Find the radius of convergence and interval of convergence of the series. X n =2 n 2 x n 2 · 4 · 6 · ... · (2 n ) Solution : We begin with the Ratio Test (because the denominator behaves like a factorial). lim n →∞ ± ± ± ± ± ± ( n +1) 2 x n +1 2 · 4 · 6 · ... · (2( n +1)) n 2 x n
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Unformatted text preview: 2 4 6 ... (2 n ) = lim n ( n + 1) 2 | x | n +1 2 4 ... (2 n ) (2 n + 2) 2 4 ... (2 n ) n 2 | x | n = lim n ( n + 1) 2 | x | n 2 (2 n + 2) = 0 < 1 . Since this limit is less than one for all x , the radius of convergence R is and the interval of convergence is (- , ). 1 2 Exercise 32 : Let p and q be real numbers with p < q . Find a power series whose interval of convergence is (a) ( p,q ) (b) ( p,q ] (c) [ p,q ) (d) [ p,q ] Solution : Let a = p + q 2 be the midpoint of ( p,q ), and let R = q-a be the radius of ( p,q ). (a) : X n =0 x-a R n (b) : X n =0 (-1) n n x-a R n (c) : X n =0 1 n x-a R n (d) : X n =0 1 n 2 x-a R n...
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This note was uploaded on 07/26/2011 for the course MATH 153 taught by Professor Rempe during the Spring '08 term at Ohio State.

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118 - 2 4 6 ... (2 n ) = lim n ( n + 1) 2 | x | n +1 2 4...

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