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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 = qa be the radius of ( p,q ). (a) : X n =0 xa R n (b) : X n =0 (1) n n xa R n (c) : X n =0 1 n xa R n (d) : X n =0 1 n 2 xa 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.
 Spring '08
 REMPE
 Math

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