Absolute convergence P Definition A series an is called absolutely convergent if the series of P absolute values janj is convergent. If the terms of the series an are positive, absolute convergence is the same as convergence. Example Are the following series absolutely convergent? 1 1 X (−1)n+1 X (−1)n ; : n3 n n=1 n=1 P1 (−1)n+1 I To check if the series n=1 n3 is absolutely convergent, we need to P1 1 check if the series of absolute values n=1 n3 is convergent. P1 1 I Since n=1 n3 is a p-series with p = 3 > 1, it converges and therefore P1 (−1)n+1 n=1 n3 is absolutely convergent. P1 (−1)n I To check if the series n=1 n is absolutely convergent, we need to P1 1 check if the series of absolute values n=1 n is convergent. P1 1 I Since n=1 n is a p-series with p = 1, it diverges and therefore P1 (−1)n n=1 n is not absolutely convergent. Annette Pilkington Lecture 28 :Absolute Convergence, Ratio and root test Conditional convergence P Definition A series an is called conditionally convergent if the series is convergent but not absolutely convergent. Which of the series in the above example is conditionally convergent? P1 (−1)n+1 I Since the series n=1 n3 is absolutely convergent, it is not conditionally convergent. P1 (−1)n I Since the series n=1 n is convergent (used the alternating series test P1 1 last day to show this), but the series of absolute values n=1 n is not P1 (−1)n convergent, the series n=1 n is conditionally convergent. Annette Pilkington Lecture 28 :Absolute Convergence, Ratio and root test Absolute conv. implies conv. Theorem If a series is absolutely convergent, then it is convergent, that is P P if janj is convergent, then an is convergent. (A proof is given in your notes) Example Are the following series convergent (test for absolute convergence) 1 1 X (−1)n+1 X sin(n) ; : n3 n4 n=1 n=1 P1 (−1)n+1 I Since n=1 n3 is absolutely convergent, we can conclude that this series is convergent. P1 sin(n) I To check if the series 4 is absolutely convergent, we consider the n=1 n ˛ ˛ series of absolute values P1 ˛ sin(n) ˛. n=1 ˛ n4 ˛ ˛ ˛ Since 0 ≤ j sin(n)j ≤ 1, we have 0 ≤ ˛ sin(n) ˛ ≤ 1 . I ˛ n4 ˛ n4 ˛ ˛ Therefore the series P1 ˛ sin(n) ˛ converges by comparison with the I n=1 ˛ n4 ˛ P1 1 converging p-series n=1 n4 . P1 sin(n) I Therefore the series n=1 n4 is convergent since it is absolutely convergent. Annette Pilkington Lecture 28 :Absolute Convergence, Ratio and root test The Ratio Test This test is useful for determining absolute convergence. P1 Let an be a series (the terms may be positive or negative). n=1 ˛ ˛ an+1 Let L = limn!1 ˛ ˛. ˛ an ˛ P1 I If L < 1, then the series n=1 an converges absolutely (and hence is convergent). P1 I If L > 1 or 1, then the series n=1 an is divergent. I If L = 1, then the Ratio test is inconclusive and we cannot determine if the series converges or diverges using this test. This test is especially useful where factorials and powers of a constant appear in terms of a series. (Note that when the ratio test is inconclusive for an alternating series, the alternating series test may work. ) Example 1 Test the following series for convergence 1 n X n−1 2 (−1) n! n=1 ˛ ˛ ˛ n+1‹ ˛ an+1 2 (n+1)! 2 I limn!1 ˛ ˛ = limn!1 ˛ ‹ ˛ = limn!1 = 0 < 1. ˛ an ˛ ˛ 2n n! ˛ n+1 I Therefore, the series converges. Annette Pilkington Lecture 28 :Absolute Convergence, Ratio and root test Example 2 P1 Ratio Test Let an be a series (the terms may be positive or negative). ˛ n=1˛ an+1 Let L = limn!1 ˛ ˛. ˛ an ˛ P1 If L < 1, then the series n=1 an converges absolutely. P1 If L > 1 or 1, then the series n=1 an is divergent. If L = 1, then the Ratio test is inconclusive. Example 2 Test the following series for convergence 1 X n“ n ” (−1) 5n n=1 ˛ ˛ ˛ ‹ n+1 ˛ an+1 (n+1) 5 n+1 I limn!1 ˛ ˛ = limn!1 ˛ ‹ ˛ = limn!1 = ˛ an ˛ ˛ n 5n ˛ 5n 1 1 5 limn!1(1 + 1=n) = 5 < 1. I Therefore, the series converges. Annette Pilkington Lecture 28 :Absolute Convergence, Ratio and root test Example 3 P1 nn Example 3 Test the following series for convergence n=1 n! ‹ ˛ ˛ ˛ (n+1)n+1 (n+1)! ˛ n ˛ an+1 ˛ ˛ ˛ (n+1)(n+1) n! I limn!1 = limn!1 ‹ = limn!1 · n = ˛ an ˛ ˛ nn n! ˛ (n+1)n! n n n x “ n+1 ” “ 1 ” “ 1 ” limn!1 n = limn!1 1 + n = limx!1 1 + x . “ ”x 1 x ln(1+1=x) limx!1 x ln(1+1=x) I limx!1 1 + x = limx!1 e = e . −1=x2 ln(1+1=x) 0 (1+1=x) limx!1 x ln(1 + 1=x) = limx!1 1=x =( L Hop) limx!1 −1=x2 = 1 limx!1 (1+1=x) = 1: ˛ ˛ x an+1 “ 1 ” 1 I Therefore limn!1 ˛ ˛ = limx!1 1 + = e = e > 1 and the series ˛ an ˛ x P1 nn n=1 n! diverges. Annette Pilkington Lecture 28 :Absolute Convergence, Ratio and root test Example 4 P1 (−1)n Example 4 Test the following series for convergence n=1 n2 I We know already that this series converges absolutely and therefore it converges. (we could also use the alternating series test to deduce this). I Lets see what happens when we apply the ratio test here. ˛ ˛ ˛ ‹ 2 ˛ 2 an+1 1 (n+1) “ n ” I limn!1 ˛ ˛ = limn!1 ˛ ‹ ˛ = limn!1 = ˛ an ˛ ˛ 1 n2 ˛ n+1 2 “ 1 ” limn!1 1+1=n = 1: I Therefore the ratio test is inconclusive here. Annette Pilkington Lecture 28 :Absolute Convergence, Ratio and root test The Root Test P1 Root Test Let n=1 an be a series (the terms may be positive or negative). pn P1 I If limn!1 janj = L < 1, then the series n=1 an converges absolutely (and hence is convergent). pn pn P1 I If limn!1 janj = L > 1 or limn!1 janj = 1, then the series n=1 an is divergent. pn I If limn!1 janj = 1, then the Root test is inconclusive and we cannot determine if the series converges or diverges using this test. n P1 n−1“ 2n ” Example 5 Test the following series for convergence n=1(−1) n+1 r n pn n “ 2n ” 2n 2 I limn!1 janj = limn!1 n+1 = limn!1 n+1 = limn!1 1+1=n = 2 > 1 n P1 n−1“ 2n ” I Therefore by the n th root test, the series n=1(−1) n+1 diverges. Annette Pilkington Lecture 28 :Absolute Convergence, Ratio and root test Example 6 P1 pn Root Test For n=1 an. L = limn!1 janj. P1 If L < 1, then the series n=1 an converges absolutely. P1 If L > 1 or 1, then the series n=1 an is divergent. If L = 1, then the Root test is inconclusive. n P1 “ n ” Example 6 Test the following series for convergence n=1 2n+1 r n pn n “ n ” n 1 I limn!1 janj = limn!1 2n+1 = limn!1 2n+1 = limn!1 2+1=n = 1=2 < 1 n P1 “ n ” I Therefore by the n th root test, the series n=1 2n+1 converges. Annette Pilkington Lecture 28 :Absolute Convergence, Ratio and root test Example 7 P1 pn Root Test For n=1 an. L = limn!1 janj. P1 If L < 1, then the series n=1 an converges absolutely. P1 If L > 1 or 1, then the series n=1 an is divergent. If L = 1, then the Root test is inconclusive. n P1 “ ln n ” Example 7 Test the following series for convergence n=1 n : r n pn n “ ln n ” ln n ln x I limn!1 janj = limn!1 n = limn!1 n = limx!1 x = 0 1=x (L Hop) limx!1 1 = 0 < 1 n P1 “ ln n ” I Therefore by the n th root test, the series n=1 n converges. Annette Pilkington Lecture 28 :Absolute Convergence, Ratio and root test Rearranging sums If we rearrange the terms in a finite sum, the sum remains the same. This is not always the case for infinite sums (infinite series). It can be shown that: P P I If a series an is an absolutely convergent series with an = s, then any P rearrangement of an is convergent with sum s. P I It a series an is a conditionally convergent series, then for any real P number r, there is a rearrangement of an which has sum r. P1 (−1)n I Example The series n=1 2n is absolutely convergent with P1 (−1)n 2 2 n=1 2n = 3 and hence any rearrangement of the terms has sum 3 . Annette Pilkington Lecture 28 :Absolute Convergence, Ratio and root test Rearranging sums P I It a series an is a conditionally convergent series, then for any real P number r, there is a rearrangement of an which has sum r. P1 (−1)n−1 I Example Alternating Harmonic series n=1 n is conditionally convergent, it can be shown that its sum is ln 2, 1 1 1 1 1 1 1 1 1 1 − + − + − + − + − · · · + (−1)n + ··· = ln 2: 2 3 4 5 6 7 8 9 n I Now we rearrange the terms taking the positive terms in blocks of one followed by negative terms in blocks of 2 1 1 1 1 1 1 1 1 1 1 − − + − − + − − + ··· = 2 4 3 6 8 5 10 12 7 “ 1 ” 1 “ 1 1 ” 1 “ 1 1 ” 1 “ 1 1 ” 1 − − + − − + − − + − − · · · = 2 4 3 6 8 5 10 12 7 14 I 1 “ 1 1 1 1 1 1 1 1 1 1 1 − + − + − + − + − · · · + (−1)n + ::: ) = ln 2: 2 2 3 4 5 6 7 8 9 n 2 I Obviously, we could continue in this way to get the series to sum to any number of the form (ln 2)=2n.