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Unformatted text preview: (1) f ( n ) = a n for n N (2) f is decreasing for x 1 (often check this by showing f < 0) (3) f is continuous for x 1 (4) so f ( x ) for x 1 . Then the series n =1 a n and the improper integral R 1 f ( x ) dx either: (1) both converge (to dierent numbers most likely) (2) both diverge . This is because { N n =1 a n } N N and { R N 1 f ( x ) dx } N N are both nondecreasing sequences and so each has the choice of either (converging to some nite number) or (diverging to ). But a 2 + a 3 + ... + a N Z N 1 f ( x ) dx a 1 + a 2 + ... + a N1 Now take the limit as N to see that X n =2 a n Z 1 f ( x ) dx X n =1 a n . 2...
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 Fall '11
 KUSTIN
 Calculus

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