1. Checking off homework: 1) 8.8-10.5 Test - Form A - Review Packet 2) 9.1-9.2 Combined Assignment Part 2: 9.1 ( 75, 81, 84, 89-97 eoo ) and 9.2 ( 1-4, 12-15, 25-29 odd, 37, 41, 47, 49 ) 3) 9.1-9.2 Combined Assignment Part 3 : 9.1 ( 51-63 eoo, 79, 119-124 ) and 9.2 ( 61-69 eoo, 83, 85) Bellwork: The 8.8-10.5 Retest has been postponed until Friday. Homework: Read 9.3. 9.3 (7,9,13,21,29,36,39,40,75,79,87-97 odd)

2. Bellwork Solutions – Improper Integration with a convergent integral:

3. 9.3 Notes - The Integral Test and p-series Petrified Forest National Park, Arizona Greg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 2007 Yet more ways for us to determine convergence and divergence of series!

4. 9.3 Notes - The Integral Test and p-series Petrified Forest National Park, Arizona Greg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 2007 Yet more ways for us to determine convergence and divergence of series! Aren’t you excited???

5. Remember that when we first studied integrals, we used a summation of rectangles to approximate the area under a curve: This leads to: Theorem 9.10 The Integral Test If is a positive sequence and where is a continuous, positive decreasing function, then: and both converge or both diverge.

6. Example 1: Does converge? Since the integral converges, the series must converge. (but not necessarily to 2.) If you showed this amount of work on the AP test, you would have a point deducted for not showing that a n is positive and decreasing.

7. Theorem 9.11 p-series Test converges if, diverges if. We could show this with the integral test. If this test seems backward after the ratio and nth root tests (which we will cover soon), remember that larger values of p would make the denominators increase faster and the terms decrease faster.

8. the harmonic series: diverges. (It is a p-series with p=1.) It diverges very slowly, but it diverges. Because the p-series is so easy to evaluate, we use it to compare to other series.

9. Integral Test and p-series Very often comparing the n th term of a series a n with the corresponding function f(x) will yield useful information about a n. If f(x) is continuous, positive and decreasing for x  1 and a n = f(n), then The integral test uses such a comparison.

10. Test the convergence or divergence of the series The corresponding function is: For x  1, this function is positive and decreasing. Therefore, we can use the integral test. x 2 + 1 = u 2xdx = du xdx = du/2 When x = 1, u = 2 When x = , u =  =  The integral diverges and hence the series diverges.

11. Test the convergence or divergence of the series The corresponding function is: For x  1, this function is positive and decreasing. Therefore, we can use the integral test. Use the table method for integration. udv xe -x 1-e -x 0e -x (+) (-) Since the integral converges, the series also converges.

12. Test the convergence or divergence of the series The corresponding function is: For x  1, this function is positive and decreasing. Therefore, we can use the integral test. Since the integral converges, the series also converges.

13. Test the convergence or divergence of the series The corresponding function is: For x  1, this function is positive and decreasing. Therefore, we can use the integral test. x 2 + 3 = u 2xdx = du xdx = du/2 When x = 1, u = 4 When x = , u =  =  The integral diverges and hence the series diverges.

14. p-series and the Harmonic series A series of the form:is called a p-series. Using integral theorem, it can be shown that a p-series converges for p > 1 A p-series diverges for 0 < p  1 When p = 1, the series is called a harmonic series. A harmonic series is a diverging series. This can be shown using integral theorem.

15. Determine the convergence or divergence of: This is a p-series with p = 4/3 Since p > 1, this is a converging p-series. Determine the convergence or divergence of: This is a p-series with 0 < p < 1.Therefore, the series diverges.

16. Partial sum S N as an approximation of S Very often the sum of an infinite series can be approximated taking the sum of the first n terms. Of course there will be a small error in this approximation. The error is a measure of the remainder which is the sum of all the remaining terms. We can use the following theorem to obtain the remainder R N in the approximation. If f is a positive and decreasing function such that a n = f(n), then the error involved in approximating S, the sum of the series, as S N, the sum of the first n terms of the series, will be less than or equal to: In other words, If R N = S – S N, then:

17. Approximate the sum of the series using the first 10 terms. 1/((n + 1)(ln(n + 1))^3), n, 1, 10, 1,), ENTER  1.9821 2 nd, STAT, right arrow twice, 5, 2 nd, STAT, right arrow,5 and type in The error in this approximation is given by: ln(x + 1) = u When x = 10, u = ln 11 When x = b, u = ln(b + 1)

18. Approximate the sum of the series using the first 10 terms.

19. Approximate the sum of the series using the first 4 terms. = 0.5713 S 4 = e -1 + e -2 + e -3 + e -4 The error in this approximation is given by: = 0.0183

20. Find N such that the error in approximating S using S N is less than 0.001 (R N  0.001) for the series: If we use N terms to approximate the sum of the series, the remainder R N is given by We need to find the value of N such that this remainder is less than 0.001

21. Determine the convergence or divergence of the series: Here we use the integral test. The corresponding function is: Since the integral converges, the series also converges. For x > 1, this function is positive and decreasing.

22. Determine the convergence or divergence of the series: Do this on your own. Answer: This is a geometric series with r = 1.075 Since r > 1, this geometric series diverges.

23. Determine the convergence or divergence of the series: Each of these is a p-series with p > 1 Therefore, the series converges. Now try using the integral test to show that the series converges.

24. Determine the convergence or divergence of the series: Here we use the integral test. The corresponding function is: Use integration by parts For x  2, this function is positive and decreasing.

25. Determine the convergence or divergence of the series: Since the integral converges, the series also converges.