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Lecture 25 : Integral Test Integral Test Integral Test Example Integral Test Example P-Series Integral Test

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Lecture 25 : Integral Test Integral Test Integral Test Example Integral Test Example P-Series Integral Test I Note The result is still true if the condition that f (x) is decreasing on the interval [1; 1) is relaxed to \the function f (x) is decreasing on an interval [M; 1) for some number M ≥ 1." Integral Test Integral Test Example Integral Test Example p-series Integral Test In this section, we see that we can sometimes decide whether a series converges or diverges by comparing it to an improper integral. The analysis in this section P only applies to series an, with positive terms, that is an > 0. Integral Test Suppose f (x) is a positive decreasing continuous function on the interval [1; 1) with f (n) = an: P1 R 1 Then the series n=1 an is convergent if and only if 1 f (x)dx converges, that is: 1 Z 1 X If f (x)dx is convergent, then an is convergent: 1 n=1 1 Z 1 X If f (x)dx is divergent, then an is divergent: 1 n=1 Annette Pilkington Lecture 25 : Integral Test Integral Test Integral Test Example Integral Test Example p-series Integral Test In this section, we see that we can sometimes decide whether a series converges or diverges by comparing it to an improper integral. The analysis in this section P only applies to series an, with positive terms, that is an > 0. Integral Test Suppose f (x) is a positive decreasing continuous function on the interval [1; 1) with f (n) = an: P1 R 1 Then the series n=1 an is convergent if and only if 1 f (x)dx converges, that is: 1 Z 1 X If f (x)dx is convergent, then an is convergent: 1 n=1 1 Z 1 X If f (x)dx is divergent, then an is divergent: 1 n=1 I Note The result is still true if the condition that f (x) is decreasing on the interval [1; 1) is relaxed to \the function f (x) is decreasing on an interval [M; 1) for some number M ≥ 1." Annette Pilkington Lecture 25 : Integral Test I In the picture we compare the P1 1 series n=1 n2 to the improper R 1 1 integral 1 x2 dx. Pn 1 R 1 1 I The n th partial sum is sn = 1 + n=2 n2 < 1 + 1 x2 dx = 1 + 1 = 2: I Since the sequence fsng is increasing (because each an > 0) and bounded, we can conclude that the sequence of partial sums converges and hence the series 1 X 1 converges: n2 i=1 P1 1 R 1 1 I NOTE We are not saying that i=1 n2 = 1 x2 dx here. Integral Test Integral Test Example Integral Test Example p-series Integral Test (Why it works: convergence) We know from a previous lecture that R 1 1 1 xp dx converges if p > 1 and diverges if p ≤ 1: Annette Pilkington Lecture 25 : Integral Test Pn 1 R 1 1 I The n th partial sum is sn = 1 + n=2 n2 < 1 + 1 x2 dx = 1 + 1 = 2: I Since the sequence fsng is increasing (because each an > 0) and bounded, we can conclude that the sequence of partial sums converges and hence the series 1 X 1 converges: n2 i=1 P1 1 R 1 1 I NOTE We are not saying that i=1 n2 = 1 x2 dx here. Integral Test Integral Test Example Integral Test Example p-series Integral Test (Why it works: convergence) We know from a previous lecture that R 1 1 1 xp dx converges if p > 1 and diverges if p ≤ 1: I In the picture we compare the P1 1 series n=1 n2 to the improper R 1 1 integral 1 x2 dx. Annette Pilkington Lecture 25 : Integral Test I Since the sequence fsng is increasing (because each an > 0) and bounded, we can conclude that the sequence of partial sums converges and hence the series 1 X 1 converges: n2 i=1 P1 1 R 1 1 I NOTE We are not saying that i=1 n2 = 1 x2 dx here. Integral Test Integral Test Example Integral Test Example p-series Integral Test (Why it works: convergence) We know from a previous lecture that R 1 1 1 xp dx converges if p > 1 and diverges if p ≤ 1: I In the picture we compare the P1 1 series n=1 n2 to the improper R 1 1 integral 1 x2 dx. Pn 1 R 1 1 I The n th partial sum is sn = 1 + n=2 n2 < 1 + 1 x2 dx = 1 + 1 = 2: Annette Pilkington Lecture 25 : Integral Test P1 1 R 1 1 I NOTE We are not saying that i=1 n2 = 1 x2 dx here. Integral Test Integral Test Example Integral Test Example p-series Integral Test (Why it works: convergence) We know from a previous lecture that R 1 1 1 xp dx converges if p > 1 and diverges if p ≤ 1: I In the picture we compare the P1 1 series n=1 n2 to the improper R 1 1 integral 1 x2 dx. Pn 1 R 1 1 I The n th partial sum is sn = 1 + n=2 n2 < 1 + 1 x2 dx = 1 + 1 = 2: I Since the sequence fsng is increasing (because each an > 0) and bounded, we can conclude that the sequence of partial sums converges and hence the series 1 X 1 converges: n2 i=1 Annette Pilkington Lecture 25 : Integral Test Integral Test Integral Test Example Integral Test Example p-series Integral Test (Why it works: convergence) We know from a previous lecture that R 1 1 1 xp dx converges if p > 1 and diverges if p ≤ 1: I In the picture we compare the P1 1 series n=1 n2 to the improper R 1 1 integral 1 x2 dx. Pn 1 R 1 1 I The n th partial sum is sn = 1 + n=2 n2 < 1 + 1 x2 dx = 1 + 1 = 2: I Since the sequence fsng is increasing (because each an > 0) and bounded, we can conclude that the sequence of partial sums converges and hence the series 1 X 1 converges: n2 i=1 P1 1 R 1 1 I NOTE We are not saying that i=1 n2 = 1 x2 dx here. Annette Pilkington Lecture 25 : Integral Test I In the picture, we compare the 1 P p1 series n=1 n to the improper 1 R p1 integral 1 x dx. 1 X 1 1 1 1 p = p + p + p + ··· k=1 n 1 2 3 I This time we draw the rectangles so that we get 1 1 1 1 Z n 1 sn > sn−1 = p + p + p + ··· + p > p dx 1 2 3 n − 1 1 x n R p1 I Thus we see that limn!1 sn > limn!1 1 x dx. n R p1 I However, we know that 1 x dx grows without bound and hence since 1 1 R p1 P p1 1 x dx diverges, we can conclude that k=1 n also diverges. Integral Test Integral Test Example Integral Test Example p-series Integral Test (Why it works: divergence) R 1 1 We know that 1 xp dx converges if p > 1 and diverges if p ≤ 1: Annette Pilkington Lecture 25 : Integral Test I This time we draw the rectangles so that we get 1 1 1 1 Z n 1 sn > sn−1 = p + p + p + ··· + p > p dx 1 2 3 n − 1 1 x n R p1 I Thus we see that limn!1 sn > limn!1 1 x dx. n R p1 I However, we know that 1 x dx grows without bound and hence since 1 1 R p1 P p1 1 x dx diverges, we can conclude that k=1 n also diverges. Integral Test Integral Test Example Integral Test Example p-series Integral Test (Why it works: divergence) R 1 1 We know that 1 xp dx converges if p > 1 and diverges if p ≤ 1: I In the picture, we compare the 1 P p1 series n=1 n to the improper 1 R p1 integral 1 x dx. 1 X 1 1 1 1 p = p + p + p + ··· k=1 n 1 2 3 Annette Pilkington Lecture 25 : Integral Test n R p1 I Thus we see that limn!1 sn > limn!1 1 x dx. n R p1 I However, we know that 1 x dx grows without bound and hence since 1 1 R p1 P p1 1 x dx diverges, we can conclude that k=1 n also diverges. Integral Test Integral Test Example Integral Test Example p-series Integral Test (Why it works: divergence) R 1 1 We know that 1 xp dx converges if p > 1 and diverges if p ≤ 1: I In the picture, we compare the 1 P p1 series n=1 n to the improper 1 R p1 integral 1 x dx. 1 X 1 1 1 1 p = p + p + p + ··· k=1 n 1 2 3 I This time we draw the rectangles so that we get 1 1 1 1 Z n 1 sn > sn−1 = p + p + p + ··· + p > p dx 1 2 3 n − 1 1 x Annette Pilkington Lecture 25 : Integral Test n R p1 I However, we know that 1 x dx grows without bound and hence since 1 1 R p1 P p1 1 x dx diverges, we can conclude that k=1 n also diverges.
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