12.8 Hints Hint 12.1 Use the Cauchy convergence criterion for series In particular, consider |SN +1 − SN | Hint 12.2 CONTINUE Hint 12.3 ∞ n ln(n) n=2 Use the integral test ∞ n=2 ln (nn ) Simplify the summand ∞ ln √ n ln n n=2 Simplify the summand Use the comparison test ∞ n(ln n)(ln(ln n)) n=10 Use the integral test 574 ∞ n=1 ln (2n ) ln (3n ) + Show that the terms in the sum not vanish as n → ∞ ∞ n=0 ln(n + 20) Shift the indices ∞ n=0 4n + 3n − Show that the terms in the sum not vanish as n → ∞ ∞ (Logπ 2)n n=0 This is a geometric series ∞ n=2 n2 − n4 − Simplify the integrand Use the comparison test 10 ∞ n=2 n2 (ln n)n Compare to a geometric series 575 11 ∞ (−1)n ln n=2 n Group pairs of consecutive terms to obtain a series of positive terms 12 ∞ n=2 (n!)2 (2n)! Use the comparison test 13 ∞ n=2 3n + 4n + 5n − 4n − Use the root test 14 ∞ n=2 n! (ln n)n Show that the terms not vanish as n → ∞ 15 ∞ n=2 en ln(n!) Show that the terms not vanish as n → ∞ 16 ∞ n=1 Apply the ratio test 576 (n!)2 (n2 )! 17 ∞ n8 + 4n4 + 3n9 − n5 + 9n n=1 Use the comparison test 18 ∞ 1 − n n+1 n=1 Use the comparison test 19 ∞ n=1 cos(nπ) n Simplify the integrand 20 ∞ n=2 ln n n11/10 Use the integral test Hint 12.4 Group the terms 1 = 2 1 − = 12 1 − = 30 ··· 1− 577 Hint 12.5 Show that |S2n − Sn | > Hint 12.6 The alternating harmonic series is conditionally convergent Let {an } and {bn } be the positive and negative terms in the sum, respectively, ordered in decreasing magnitude Note that both ∞ an and ∞ bn are divergent Devise a n=1 n=1 method for alternately taking terms from {an } and {bn } Hint 12.7 Use the ratio test Hint 12.8 Use the integral test Hint 12.9 Note that sin(nx) = (eınx ) This substitute will yield a finite geometric series Hint 12.10 Let Sn be the sum Consider Sn − zSn Use the finite geometric sum Hint 12.11 The summand is a rational function Find the first few partial sums This a geometric series Hint 12.12 CONTINUE 578 Hint 12.13 CONTINUE ∞ n=0 ∞ n=2 ∞ n=1 ∞ n=1 ∞ n=1 ∞ n=1 ∞ n=0 ∞ n=0 ∞ n=0 zn (z + 3)n Log z ln n z n (z + 2)2 n2 (z − e)n nn z 2n 2nz z n! (n!)2 z ln(n!) n! (z − π)2n+1 nπ n! 579 ∞ 10 n=0 ln n zn Hint 12.14 Hint 12.15 CONTINUE Hint 12.16 Differentiate the geometric series Integrate the geometric series Hint 12.17 The Taylor series is a geometric series Hint 12.18 Hint 12.19 Hint 12.20 1 = z + (z − 1) The right side is the sum of a geometric series Integrate the series for 1/z Differentiate the series for 1/z Integrate the series for Log z 580 Hint 12.21 Evaluate the derivatives of ez at z = Use Taylor’s Theorem Write the cosine and sine in terms of the exponential function Hint 12.22 cos z = − cos(z − π) sin z = − sin(z − π) Hint 12.23 CONTINUE Hint 12.24 CONTINUE Hint 12.25 Hint 12.26 Hint 12.27 Hint 12.28 Hint 12.29 Hint 12.30 CONTINUE 581 12.9 Solutions Solution 12.1 ∞ n=0 an converges only if the partial sums, Sn , are a Cauchy sequence ∀ > ∃N s.t m, n > N ⇒ |Sm − Sn | < , In particular, we can consider m = n + ∀ > ∃N s.t n > N ⇒ |Sn+1 − Sn | < Now we note that Sn+1 − sn = an ∀ > ∃N s.t n > N ⇒ |an | < This is exactly the Cauchy convergence criterion for the sequence {an } Thus we see that limn→∞ an = is a necessary condition for the convergence of the series ∞ an n=0 Solution 12.2 CONTINUE Solution 12.3 ∞ n=2 n ln(n) Since this is a series of positive, monotone decreasing terms, the sum converges or diverges with the integral, ∞ dx = x ln x ∞ ln dξ ξ Since the integral diverges, the series also diverges ∞ n=2 = ln (nn ) 582 ∞ n=2 n ln(n) Thus the series converges absolutely for |z| < 1/4 By the Cauchy-Hadamard formula, the radius of absolute convergence is R= lim sup |n + αn | n = lim sup |α| n |1 + n/αn | = |α| Thus the sum converges absolutely for |z| < 1/|α| Solution 12.15 ∞ kz k k=0 We determine the radius of convergence with the ratio formula k k→∞ k + 1 = lim k→∞ =1 R = lim The series converges absolutely for |z| < ∞ kk z k k=1 598 We determine the radius of convergence with the Cauchy-Hadamard formula R= lim sup k |k k | = lim sup k =0 The series converges only for z = ∞ k=1 k! k z kk We determine the radius of convergence with the ratio formula k!/k k k→∞ (k + 1)!/(k + 1)(k+1) (k + 1)k = lim k→∞ kk k+1 = exp lim k ln k→∞ k ln(k + 1) − ln(k) = exp lim k→∞ 1/k 1/(k + 1) − 1/k = exp lim k→∞ −1/k k = exp lim k→∞ k + = exp(1) =e R = lim 599 The series converges absolutely for |z| < e ∞ (z + ı5)2k (k + 1)2 k=0 We use the ratio formula to determine the domain of convergence (z + ı5)2(k+1) (k + 2)2