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# 113 - Solution In order to do this we need to ﬁgure out...

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MATH 153 Selected Solutions for § 11.3 Exercise 18 : Determine whether the geometric series is convergent or diver- gent. X n =0 3 n + 2 n ( n + 1) Solution : As we are in the Integral Test section, that is the test we shall apply. Z 1 3 x + 2 x ( x + 1) dx = lim b →∞ Z b 1 3 x + 2 x ( x + 1) dx = lim b →∞ Z b 1 A x + B x + 1 dx We finish the partial fraction decomposition as follows: 3 x + 2 x ( x + 1) = A x + B x + 1 3 x + 2 = A ( x + 1) + Bx 3 x + 2 = ( A + B ) x + A A + B = 3 , A = 2 A = 2 , B = 1 So to finish the integral, we now have that Z 1 3 x + 2 x ( x + 1) dx = lim b →∞ Z b 1 2 x + 1 x + 1 dx = lim b →∞ 2 ln | x | + ln | x + 1 | b 1 = lim b →∞ 2 ln b - 2 ln 1 + ln( b + 1) - ln 2 = . Thus, since the integral diverges, so does the series by the Integral Test. 1

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2 Exercise 22 : Determine whether the series is convergent or divergent. X k =1 1 n (ln n ) 2 Solution : Again, we apply the Integral Test (with u = ln x ). Z 1 1 x (ln x ) 2 dx = lim b →∞ Z b 1 1 x (ln x ) 2 dx = lim b →∞ Z ln b 0 1 xu 2 x du = lim b →∞ Z ln b 1 u - 2 du = lim b →∞ - 1 u ln b 1 = lim b →∞ - 1 ln b - - 1 1 = 1 < . Thus, since the integral converges, so does the series by the Integral Test.
3 Exercise 35 : Estimate n =1 (2 n + 1) - 6 correct to five decimal places.
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Unformatted text preview: Solution : In order to do this, we need to ﬁgure out how many terms are necessary to add. Recall that ∞ X k = n +1 f ( k ) < Z ∞ n f ( x ) dx. Thus, we want to ﬁnd an integer n such that Z ∞ n (2 x + 1)-6 dx < . 00001 . Z ∞ n (2 x + 1)-6 dx = lim b →∞ Z b n (2 x + 1)-6 dx = lim b →∞ 1 2 Z 2 b +1 2 n +1 u-6 du = 1 2 lim b →∞-1 5 u 5 ± ± ± ± 2 b +1 2 n +1 = 1 2 lim b →∞ ² 1 5(2 b + 1) 5-1 5(2 n + 1) 5 ³ = 1 10(2 n + 1) 5 . Thus, we want to ﬁnd an n for which 1 10(2 n +1) 5 < 1 100000 . Taking the reciprocal gives 10(2 n + 1) 5 > 100000, or (2 n + 1) 5 > 10000. Thus, we want n > 5 √ 10000-1 2 ’ 2 . 65 . Thus, the sum of the ﬁrst n = 3 terms of the series, 1 3 6 + 1 5 6 + 1 7 6 ’ . 001444242 is an approximation to the series that is correct to ﬁve decimal places....
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113 - Solution In order to do this we need to ﬁgure out...

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