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X∞ k=1 k2 k8 k5 7 Answer Keeping just the fastest growing terms on the top and bottom, we have k2 k8 k5 7 ∼ k2 k8 = 1 k6 The series (4) X∞ k=1 1 k6 is a pseries with p = 6 > 1, so is convergent We use the Limit Comparison Test (p 523) to compare the series X∞ k=1 a k = X∞ k=1 k2 k8 k5 7 and X∞ k=1 b k = X∞ k=1 1 k6 5Gc d i "c dm gxm;@m 'c}@ ybc;dm  Á @iyoc;dm a l@bkx`@b c¥ú 9 7 i9 k v ` %9 4The DFT pair X(k) = NX−1 n=0 x(n)e−j2πkn N analysis x(n) = 1 N NX−1 k=0 X(k)ej2πkn N synthesis Alternative formulation X(k) = NX−1 n=0 x(n)Wkn ←−W = e−j2 N π x(n) = 1 N NX−1 k=0 X(k)W−kn EE 524, Fall 04, # 5 3

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