Tests for Convergence of Series 1) Use the comparison test to confirm the statements in the following exercises. P1 1 P1 1 1. n=4 n diverges, so n=4 n−3 diverges. Answer: Let an = 1=(n − 3), for n ≥ 4. Since n − 3 < n, we have 1=(n − 3) > 1=n, so 1 a > : n n P1 1 P1 1 The harmonic series n=4 n diverges, so the comparison test tells us that the series n=4 n−3 also diverges. P1 1 P1 1 2. n=1 n2 converges, so n=1 n2+2 converges. 2 2 2 2 2 Answer: Let an = 1=(n + 2). Since n + 2 > n , we have 1=(n + 2) < 1=n , so 1 0 < a < : n n2 P1 1 P1 1 The series n=1 n2 converges, so the comparison test tells us that the series n=1 n2+2 also converges. P1 1 P1 e−n 3. n=1 n2 converges, so n=1 n2 converges. −n 2 −n e−n 1 Answer: Let an = e =n . Since e < 1, for n ≥ 1,we have n2 < n2 , so 1 0 < a < : n n2 P1 1 P1 e−n The series n=1 n2 converges, so the comparison test tells us that the series n=1 n2 also converges. 2) Use the comparison test to determine whether the series in the following exercises converge. P1 1 1. n=1 3n+1 n n n n n 1 n Answer: Let an = 1=(3 + 1). Since 3 + 1 > 3 , we have 1=(3 + 1) < 1=3 = 3 , so 1n 0 < a < : n 3 P1 1 P1 1 n Thus we can compare the series n=1 3n+1 with the geometric series n=1 3 : This geometric series P1 1 converges since j1=3j < 1, so the comparison test tells us that n=1 3n+1 also converges. P1 1 2. n=1 n4+en 4 n 4 n 4 Answer: Let an = 1=(n + e ). Since n + e > n , we have 1 1 < ; n4 + en n4 so 1 0 < a < : n n4 P1 1 P1 1 Since the p-series n=1 n4 converges, the comparison test tells us that the series n=1 n4+en also converges. 1 X 1 3. ln n n=2 Answer: Since ln n ≤ n for n ≥ 2, we have 1= ln n ≥ 1=n, so the series diverges by comparison with the harmonic series, P 1=n. P1 n2 4. n=1 n4+1 2 4 4 4 1 1 Answer: Let an = n =(n + 1). Since n + 1 > n , we have n4+1 < n4 , so n2 n2 1 a = < = ; n n4 + 1 n4 n2 therefore 1 0 < a < : n n2 P1 1 P1 n2 Since the p-series n=1 n2 converges, the comparison test tells us that the series n=1 n4+1 converges also. P1 n sin2 n 5. n=1 n3+1 Answer: We know that j sin nj < 1, so n sin2 n n n 1 ≤ < = : n3 + 1 n3 + 1 n3 n2 P1 1 P1 n sin2 n Since the p-series n=1 n2 converges, comparison gives that n=1 n3+1 converges. P1 2n+1 6. n=1 n2n−1 n n n n n Answer: Let an = (2 + 1)=(n2 − 1). Since n2 − 1 < n2 + n = n(2 + 1), we have 2n + 1 2n + 1 1 > = : n2n − 1 n(2n + 1) n P1 2n+1 P1 1 Therefore, we can compare the series n=1 n2n−1 with the divergent harmonic series n=1 n : The comparison P1 2n+1 test tells us that n=1 n2n−1 also diverges. 3) Use the ratio test to decide if the series in the following exercises converge or diverge. P1 1 1. n=1 (2n)! Answer: Since an = 1=(2n)!, replacing n by n + 1 gives an+1 = 1=(2n + 2)!. Thus 1 jan+1j (2n+2)! (2n)! (2n)! 1 = 1 = = = ; janj (2n)! (2n + 2)! (2n + 2)(2n + 1)(2n)! (2n + 2)(2n + 1) so ja j 1 L = lim n+1 = lim = 0: n!1 janj n!1 (2n + 2)(2n + 1) P1 1 Since L = 0, the ratio test tells us that n=1 (2n)! converges. P1 (n!)2 2. n=1 (2n)! 2 2 Answer: Since an = (n!) =(2n)!, replacing n by n + 1 gives an+1 = ((n + 1)!) =(2n + 2)!. Thus, ((n+1)!)2 ja j ((n + 1)!)2 (2n)! n+1 = (2n+2)! = · : (n!)2 2 janj (2n + 2)! (n!) (2n)! However, since (n + 1)! = (n + 1)n! and (2n + 2)! = (2n + 2)(2n + 1)(2n)!, we have 2 2 2 jan+1j (n + 1) (n!) (2n)! (n + 1) n + 1 = 2 = = ; janj (2n + 2)(2n + 1)(2n)!(n!) (2n + 2)(2n + 1) 4n + 2 so ja j 1 L = lim n+1 = : n!1 janj 4 P1 (n!)2 Since L < 1, the ratio test tells us that n=1 (2n)! converges. P1 (2n)! 3. n=1 n!(n+1)! Answer: Since an = (2n)!=(n!(n + 1)!), replacing n by n + 1 gives an+1 = (2n + 2)!=((n + 1)!(n + 2)!). Thus, (2n+2)! ja j (2n + 2)! n!(n + 1)! n+1 = (n+1)!(n+2)! = · : (2n)! janj (n + 1)!(n + 2)! (2n)! n!(n+1)! However, since (n + 2)! = (n + 2)(n + 1)n! and (2n + 2)! = (2n + 2)(2n + 1)(2n)!, we have ja j (2n + 2)(2n + 1) 2(2n + 1) n+1 = = ; janj (n + 2)(n + 1) n + 2 so ja j L = lim n+1 = 4: n!1 janj P1 (2n)! Since L > 1, the ratio test tells us that n=1 n!(n+1)! diverges. P1 1 4. n=1 rnn! ; r > 0 n n+1 Answer: Since an = 1=(r n!), replacing n by n + 1 gives an+1 = 1=(r (n + 1)!). Thus 1 n jan+1j rn+1(n+1)! r n! 1 = 1 = n+1 = ; janj rnn! r (n + 1)! r(n + 1) so ja j 1 1 L = lim n+1 = lim = 0: n!1 janj r n!1 n + 1 P1 1 Since L = 0, the ratio test tells us that n=1 rnn! converges for all r > 0. P1 1 5. n=1 nen n n+1 Answer: Since an = 1=(ne ), replacing n by n + 1 gives an+1 = 1=(n + 1)e . Thus 1 n jan+1j (n+1)en+1 ne n 1 = 1 = n+1 = : janj nen (n + 1)e n + 1 e Therefore ja j 1 L = lim n+1 = < 1: n!1 janj e P1 1 Since L < 1, the ratio test tells us that n=1 nen converges. P1 2n 6. n=0 n3+1 n 3 n+1 3 Answer: Since an = 2 =(n + 1), replacing n by n + 1 gives an+1 = 2 =((n + 1) + 1). Thus n+1 2 n+1 3 3 jan+1j (n+1)3+1 2 n + 1 n + 1 = 2n = 3 · n = 2 3 ; janj n3+1 (n + 1) + 1 2 (n + 1) + 1 so ja j L = lim n+1 = 2: n!1 janj P1 2n Since L > 1 the ratio test tells us that the series n=0 n3+1 diverges. 4) Use the integral test to decide whether the following series converge or diverge. 1 X 1 1. n3 n=1 Answer: We use the integral test with f(x) = 1=x3 to determine whether this series converges or diverges. Z 1 1 We determine whether the corresponding improper integral 3 dx converges or diverges: 1 x Z 1 Z b b 1 1 −1 −1 1 1 3 dx = lim 3 dx = lim 2 = lim 2 + = : 1 x b!1 1 x b!1 2x 1 b!1 2b 2 2 1 Z 1 1 X 1 Since the integral dx converges, we conclude from the integral test that the series converges. x3 n3 1 n=1 1 X n 2. n2 + 1 n=1 Answer: We use the integral test with f(x) = x=(x2 +1) to determine whether this series converges or diverges. Z 1 x We determine whether the corresponding improper integral 2 dx converges or diverges: 1 x + 1 Z 1 Z b b x x 1 2 1 2 1 2 dx = lim 2 dx = lim ln(x + 1) = lim ln(b + 1) − ln 2 = 1: 1 x + 1 b!1 1 x + 1 b!1 2 1 b!1 2 2 1 Z 1 x X n Since the integral dx diverges, we conclude from the integral test that the series diverges. x2 + 1 n2 + 1 1 n=1 1 X 1 3. en n=1 Answer : We use the integral test with f(x) = 1=ex to determine whether this series converges or diverges. Z 1 1 To do so we determine whether the corresponding improper integral x dx converges or diverges: 1 e Z 1 Z b b 1 −x −x −b −1 −1 x dx = lim e dx = lim −e = lim −e + e = e : 1 e b!1 1 b!1 1 b!1 1 Z 1 1 X 1 Since the integral dx converges, we conclude from the integral test that the series converges. ex en 1 n=1 We can also observe that this is a geometric series with ratio x = 1=e < 1, and hence it converges. 1 X 1 4. n(ln n)2 n=2 Answer: We use the integral test with f(x) = 1=(x(ln x)2) to determine whether this series converges or Z 1 1 diverges.