# 10m45 let a n b n 1 n 1 n for n 1 then a n and b n

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10.M.45: Let a n = b n = ( 1) n +1 n for n 1 . Then a n and b n converge by the alternating series test, but a n b n diverges because it is the harmonic series. 10.M.46: This is a special case of Problem 49 in Section 10.6; the proof used there can be adapted to create a solution of this problem. 1275
10.M.47: Assuming that A exists, we have A = lim n →∞ a n = lim n →∞ 1 + 1 1 + a n = 1 + 1 1 + A because A = 1. Therefore A + A 2 = 2 + A , and it follows that A 2 = 2. Because A 0 (the limit of a sequence of positive numbers cannot be negative), A = 2 . 10.M.48: Part (a): First, F 1 = 1 < 2 1 and F 2 = 1 < 2 2 . If n 2, then F n +1 = F n + F n 1 , so F n > 0 for all n 1. Moreover, if F n 1 < 2 n 1 and F n < 2 n , then F n +1 2 n + 2 n 1 < 2 n + 2 n = 2 n +1 . Therefore, by induction, F n < 2 n for all n 1. Hence n =1 F n x n is dominated by n =1 2 n x n . The latter series converges absolutely for | x | < 1 2 (it is geometric with ratio 2 x ), so the dominated series also converges for such x . Part (b): We use the formula F n +1 F n F n 1 = 0 for n 2. (1 x x 2 )( F 1 x + F 2 x 2 + F 3 x 3 + F 4 x 4 + · · · ) = F 1 x + F 2 x 2 + F 3 x 3 + F 4 x 4 + F 5 x 5 + · · · F 1 x 2 F 2 x 3 F 3 x 4 F 4 x 5 − · · · F 1 x 3 F 2 x 4 F 3 x 5 − · · · · · · = F 1 x + ( F 2 F 1 ) x 2 + ( F 3 F 2 F 1 ) x 3 + ( F 4 F 3 F 2 ) x 4 + ( F 5 F 4 F 3 ) x 5 + · · · = x + (1 1) x 2 + 0 · x 3 + 0 · x 4 + 0 · x 5 + · · · = x. Therefore F ( x ) = x 1 x x 2 . Note: Application of the ratio test to the given power series yields the limit 1 2 ( 1 + 5 ) | x | , so the radius of convergence of the series for F ( x ) is actually 1 τ = 5 1 2 0 . 6180339887498948482045868 . 1276
10.M.49: If a n = 1 n , then the series n =1 ln(1 + a n ) = n =1 ln 1 + 1 n diverges because S k = k n =1 ln 1 + 1 n = k n =1 [ln( n + 1) ln n ] = ln 2 ln 1 + ln 3 ln 2 + ln 4 ln 3 + · · · + ln( k + 1) ln k = ln( k + 1) , and therefore S k + as k + . Alternatively, using the integral test, J = 1 ln 1 + 1 x dx = 1 [ln( x + 1) ln x ] dx = ( x + 1) ln( x + 1) x ln x 1 = + because lim x →∞ [( x + 1) ln( x + 1) x ln x ] lim x →∞ [( x + 1) ln x x ln x ] = lim x →∞ ln x = + and, at the lower limit x = 1 of integration, we have ( x + 1) ln( x + 1) x ln x = ln 4. Therefore, because J = 1 ln 1 + 1 x dx = + , the infinite product n =1 1 + 1 n diverges. 10.M.50: We must test for convergence the infinite series n =1 ln 1 + 1 n 2 . (1) An easy integration by parts yields I = 1 ln 1 + 1 x 2 dx = 1 ln( x 2 + 1) 2 ln x dx = 2 arctan x 2 x ln x + x ln(1 + x 2 ) 1 = π π 2 + ln 2 = π 2 ln 2 . To evaluate the limit of the antiderivative as x + , use l’Hˆ opital’s rule: lim x →∞ x ln(1 + x 2 ) 2 x ln x = lim x →∞ x ln 1 + x 2 x 2 = lim x →∞ ln(1 + x 2 ) ln( x 2 ) 1 x = lim x →∞ 2 x 1 + x 2 2 x 1 x 2 = lim x →∞ 2 x 2 x 3 1 + x 2 = lim x →∞ 2 x + 2 x 3 2 x 3 1 + x 2 = lim x →∞ 2 x 1 + x 2 = 0 .