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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 u6 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) 51 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 √ 100001 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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 Spring '08
 REMPE
 Calculus, Geometric Series, lim, b→∞

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