10sIntegralTest

# 10sIntegralTest - (1 f n = a n for n ∈ N(2 f is...

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INTEGRAL TEST recall If { x n } n N is a non-decreasing sequence (i.e., x n x n +1 ), then either • { x n } n N is bounded above, in which case lim n →∞ x n exists or • { x n } n N is not bounded above, in which case lim n →∞ x n = . positive-term series Deﬁnition : a n is a positive-term series if a n 0 for each n . Explore : Let a n be a positive-term series. (1) Consider its sequence of partial sums { S N } N N where S N = a 1 + a 2 + ... + a N . (2) Recall that n =1 a n = lim N →∞ S N . (3) 0 a n for each n N = S N S N +1 for each N N . (4) So { S N } N N is a non-decreasing sequence. So either: • { S N } N N is bounded above and so lim N →∞ S N exists and so a n converges or • { S N } N N is not bounded above and so lim N →∞ S N = = n =1 a n and so a n diverges. today’s goal Examine the p -series n =1 1 n p . X n =1 1 n p ( converges if p > 1 diverges if p 1 . When p = 1, the p -series n =1 1 n is also called the harmonic series. 1

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integral test Let a n be a positive-term series. Find a function f ( x ) such that
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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 diﬀerent 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 non-decreasing sequences and so each has the choice of either (converging to some ﬁnite number) or (diverging to ∞ ). But a 2 + a 3 + ... + a N ≤ Z N 1 f ( x ) dx ≤ a 1 + a 2 + ... + a N-1 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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10sIntegralTest - (1 f n = a n for n ∈ N(2 f is...

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