Math 152 (honors sections) Harold P. Boas boas@tamu.edu Reminder Second examination is Wednesday, November 3. The exam covers through section 10.4. An old exam is posted at http://www.math.tamu.edu/˜boas/courses/ 152-2002c/exam2.pdf Math 152 October 29, 2004 — slide #2 Convergence tests so far ∞ • If an 6→ 0, then ∑ an diverges. n=1 • A geometric series with |ratio| < 1 converges. • Integral test for positive decreasing functions: the improper integral ∞ corresponding series ∑ f (n) either both converge or both diverge. Z ∞ 1 f (x) dx and the n=1 • Inequality comparison test for positive terms: if 0 < a n < bn (at least for n large) and if ∞ ∞ n=1 n=1 ∑ bn converges, then ∑ an converges too. ∞ an exists (finite limit), and if ∑ bn conn→∞ bn n=1 • Limit comparison test for positive terms: if lim ∞ verges, then ∑ an converges too. n=1 Math 152 October 29, 2004 — slide #3 Root test (not in book) ∞ Example: n ∑ 2n converges n=1 ∞ This is not a geometric series, and it is bigger than the convergent geometric series comparison à test does ! not seem to help. n Since n = 2 n1/n 2 n , and since lim n1/n n = 1, we have n < 2 µ 1.1 2 1 , so the 2n n=1 ∑ ¶n when n is large, so we ¶ ∞ µ 1.1 n can use the comparison test after all: compare to the convergent geometric series ∑ . 2 n=1 n→∞ Math 152 October 29, 2004 — slide #4 Root test and ratio test ∞ > 1, then Root test: If 0 < a n , and if lim a1/n < 1, then ∑ an converges. Moreover, if lim a1/n n n n→∞ ∞ ∑ an diverges (because then a n 6→ 0). If lim n→∞ n=1 n=1 a1/n n n→∞ = 1, the test gives no information. Ratio test: Exactly the same as the root test, except look at lim n→∞ an+1 instead of lim a1/n n . n→∞ an 3n (n!)2 Example: ∑ . (2n)! n=1 Á 3 an+1 3n+1 (n + 1)!(n + 1)! 3n (n!)(n!) 3(n + 1)2 lim = lim = lim = < 1, n→∞ an n→∞ n→∞ (2n + 2)! (2n)! (2n + 2)(2n + 1) 4 so the original series converges. ∞ Math 152 October 29, 2004 — slide #5 Series with some negative terms Negative terms can only help with convergence: ∞ ∞ n=1 n=1 if ∑ |an | converges, then so does ∑ an . An absolutely convergent series converges. ∞ Example: ∑ n=1 (−1)n n2 ∞ converges because ∑ n=1 1 n2 converges. ∞ The two series sum to different values, however: ∑ n=1 ∞ Example: ∑ yet 1 n (−1)n n2 = −π 2 12 , ∞ and ∑ n=1 1 n2 = π2 6 . diverges (harmonic series), n=1 ∞ (−1)n ∑ n converges n=1 (in fact to the value ln( 21 )). Math 152 October 29, 2004 — slide #6 Alternating series test ∞ If an ↓ 0, then ∑ (−1)n an converges. n=1 Error estimate: When the alternating series test applies, the sum of the series is trapped between any two consecutive partial sums. (−1)n 1 1 1 1 ∑ n4 = −1 + 24 − 34 + 44 − 54 + · · · n=1 ∞ (−1)n 1 1 1 1 1 1 1 −1 + 4 − 4 + 4 − 4 < ∑ < −1 + − + 4 2 3 4 5 24 34 44 n=1 n ∞ Example: (−1)n < −0.9459 −0.94754 < ∑ 4 n=1 n ∞ ∞ Exact value: ∑ n=1 Math 152 (−1)n n4 = −7π 4 720 ≈ −0.947033. October 29, 2004 — slide #7 Homework • Read section 10.4, pages 605–610. • Do the Suggested Homework problems for section 10.4. Monday we will review for the exam and look at an old exam. Math 152 October 29, 2004 — slide #8