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Also Called Cauchy's Condensation Test TMA226 17/18 A NOTE ON THE CONDENSATION TEST The condensation test (also called Cauchy's condensation test) is one of several tests that can be used to determine if a series converges or not. It is closely related to integral test, but in many cases the condensation test requires less work than the integral test. This short note presents a self-contained proof of the condensation test. We also look at the connection to the integral test, and some examples. Preliminaries Let us recall a few important definitions before we start. In particular, we define positive, decreasing and converging series. P P1 Definition 1. Let an = n=1 an be infinite series. P • We say that an is positive if an ≥ 0 for all n. P • We say that an is decreasing if an+1 ≤ an for all n. P P Definition 2. Let an be a positive series. We say that an converges if the partial sums sn = a1 + a2 + ··· + an; n = 1; 2;::: are uniformly bounded. That is, if there is some real number M ≥ 0 so that sn ≤ M 8n: (1) P If an does not converge, then we say that it diverges. P1 P Theorem 1. Let n=1 an be a positive series. Then an converges if and only if there exists a positive real number s such that m X s = lim an: (2) m!1 n=1 P Proof. Assume an converges. Then there is some real number M such that a1 + a2 + ··· + an ≤ M | {z } = sn for all n. Then, because of the least upper bound property of the real numbers, there exists some s 2 R such that s = supfs1; s2; s3;:::g: We will now show that s is the limit of the partial sums. Since s is the least upper bound of all the partial sums, s − is not an upper bound for any > 0. Hence, for any > 0, there is some N such that sN > s − Note that fs1; s2;:::g is an increasing sequence (meaning sn ≤ sn+1 for all n), so we have sn > s − 8n ≥ N: We can rearrange this last inequality to find js − snj = s − sn < 8n ≥ N: Date: May 3, 2018. 1 TMA226 17/18 A NOTE ON THE CONDENSATION TEST 2 Since > 0 was arbitary, this shows that sn converges to s. That is, n X s = lim sn = lim ak: n!1 n!1 k=1 Now renaming the indices gives us the identity (2). Now suppose that the limit (2) exists. Since fs1; s2;:::g is increasing, we have sn ≤ s 8n; which shows that the partial sums are uniformly bounded ((1) holds with M = s) P and therefore the series an converges. Remark 1. Note that we defined a series to be convergent if its partial sums are uniformly bounded. We could instead have defined that the series converges if the limit of the partial sums exists, as is done in the book. Theorem1 shows that the two are equivalent: The limit exists if and only if the partial sums are bounded. For signed series (those that can take both positive and negative values) the situation is more complicated and we need to be more precise in defining convergence. Statement and proof We are now ready to state and prove the condensation theorem. We prove the theorem for a sum starting at index 1. Naturally the theorem holds regardless of the starting index, since it is the infinite tail of the series that determines if it converges or not. P P Theorem 2. Let an be a positive and decreasing series. Then the series an P n converges if and only if the condensed series 2 a2n converges. P P n Proof. Let sn and σn denote the partial sums of an and 2 a2n , respectively. That is, sn = a1 + a2 + a3 + ::: + an n σn = 2a2 + 4a4 + 8a8 + ::: + 2 a2n : Let us first consider sn. Since the series is decreasing, we have ≤ 2a2 z }| { s3 = a1 + a2 + a3 ≤ a1 + 2a2 = a1 + σ1 s7 = a1 + a2 + a3 + a4 + a5 + a6 + a7 ≤ a1 + 2a2 + 4a4 = a1 + σ2: | {z } | {z } ≤ 2a2 ≤ 4a4 4 We want to continue this for n > 7. We have s7 − s3 = 4a , and more generally total of 2n terms z }| { n s2n+1−1 − s2n−1 = a2n + a2n+1 + ··· + a2n+1−1 ≤ 2 a2n for n = 1; 2;:::;. Hence we get n n−1 ≤ 2 a2n ≤ 2 a2n−1 ≤ 2s2 z }| { z }| { z }| { s2n+1−1 = s2n+1−1 − s2n−1 + s2n−1 − s2n−1−1 + ··· + s3 − s1 +s1 (3) n ≤ a1 + 2a2 + 4a4 + ··· + 2 a2n = a1 + σn: P This shows that the partials sums of an can be bounded by the partial sums of P n 2 a2n . We also need a bound in the opposite direction. Note that total of 2n−1 terms total of 2n−1 terms n z }| { z }| { 2 a2n = 2a2n + 2a2n + ··· 2a2n ≤ 2a2n−1+1 + 2a2n−1+2 + ··· 2a2n ; TMA226 17/18 A NOTE ON THE CONDENSATION TEST 3 where we again use the fact that the series is decreasing. This gives us n σn = 2a2 + 4a4 + ··· + 2 a2n ≤ 2a2 + (2a3 + 2a4) + (2a5 + 2a6 + 2a7 + 2a8) + ··· + (2a2n−1+1 + ··· + 2a2n ) ≤ 2a1 + 2a2 + ··· 2a2n = 2s2n : (4) Now combining the inequalities (3) and (4), we have s2n+1−1 − a1 ≤ σn ≤ 2s2n 8n: (5) P This inequality means that the partial sums of an are bounded if and only if the P n P P n partial sums of 2 a2n are bounded, hence an converges if and only if 2 a2n converges. Relation to integral test Recall the integral test, stated below for the sake of completeness. Theorem 3. Let f be a positive and decreasing function on [1; 1). Then 1 X Z 1 f(n) < 1 () f(x) dx < 1: n=1 1 We will show that the condensation test can be inferred from the integral test. Consider a variable substitution x = 2y in the integral. We have Z 1 Z 1 f(x) dx = (log 2) 2yf(2y) dy; 1 0 from which we get Z 1 Z 1 f(x) dx < 1 () 2yf(2y) dy < 1: 1 1 Now apply the integral test to both sides, and we have 1 X Z 1 f(n) < 1 () f(x) dx < 1 n=1 1 1 Z 1 X () 2yf(2y) dy < 1 () 2nf(2n) < 1: 1 n=1 Examples Example 1. Determine if the following series converge or diverge. 1 1 1 X 1 X 1 X 1 (a) (b) (c) n n log n n2 n=1 n=2 n=1 1 1 1 X 1 X 1 X log n (d) (e) (f) (log n)2 n(log n)2 n2 n=2 n=2 n=2 Solution. We use the condensation test on the all the series. (a) With an = 1=n, we have 1 1 1 X n X n 1 X 2 a n = 2 = 1; 2 2n n=1 n=1 n=1 P1 1 which clearly diverges. Hence n=1 n must also diverge. TMA226 17/18 A NOTE ON THE CONDENSATION TEST 4 (b) With an = 1=(n log n), we have 1 1 1 X n X n 1 X 1 2 a n = 2 = ; 2 2n log 2n n log 2 n=2 n=2 n=2 P1 1 which diverges according to (a). Hence n=2 n log n diverges. 2 (c) With an = 1=n , we have 1 1 1 n X n X n 1 X 1 1=2 2 a n = 2 = = = 1: 2 (2n)2 2 1 − 1=2 n=1 n=1 n=1 P1 1 It follows that n=1 n2 converges. 2 (d) With an = 1=(log n) , we have 1 1 1 n X n X n 1 X 2 2 a n = 2 = : 2 (log 2n)2 n2(log 2)2 n=2 n=2 n=2 n 2 This diverges according to the root test. To see this, let bn = 2 =n . Then r n pn n 2 2 bn = = p −! 2 when n −! 1; n2 ( n n)2 p n 2 P1 1 since n ! 1 and x 7! x is continuous. In conclusion, n=2 (log n)2 diverges. 2 (e) With an = 1=(n(log n) ), we have 1 1 1 X n X n 1 X 1 2 a n = 2 = : 2 2n(log 2n)2 n2(log 2)2 n=2 n=2 n=2 P1 1 This converges according to (c), hence n=2 n(log n)2 converges. 2 (f) With an = log n=n , we have 1 1 n 1 X n X n log 2 X n log 2 2 a n = 2 = : 2 (2n)2 2n n=2 n=2 n=2 n This converges according to the root test: With bn = n=2 , we have p r n pn n n 1 b = n = −! when n −! 1: n 2n 2 2 P1 log n Hence n=2 n2 converges. Note that the root test and the ratio test are inconclusive for all the series in example1. We could use the integral test, but to carry out the integrals is more work. Also note that the comparison test could be used for (f) as follows. Sincep (log n)=np ! 0 for all p > 0 when n ! 1, we could find some N so that log n ≤ n when n ≥ N. Then 1 1 p 1 X log n X n X 1 ≤ = n2 n2 n3=2 n=N n=N n=N where the last series on the right is known to converge (using integral test or condensation test).
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