# Notes when we say lim k a k b k l 0 we mean that the

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Chapter 11 / Exercise 36
Calculus
Stewart
Expert Verified
NOTES When we say lim k →∞ a k b k = L > 0, we mean that the limit exists and is positive. In particular, we mean that lim k →∞ a k b k = ∞ . THEOREM 3.4 (Limit Comparison Test) Suppose that a k , b k > 0 and that for some (finite) value, L , lim k →∞ a k b k = L > 0. Then, either k = 1 a k and k = 1 b k both converge or they both diverge. PROOF If lim k →∞ a k b k = L > 0, this says that we can make a k b k as close to L as desired. So, in particular, we can make a k b k within distance L 2 of L . That is, for some number N > 0, L L 2 < a k b k < L + L 2 , for all k > N or L 2 < a k b k < 3 L 2 . (3.6)
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Chapter 11 / Exercise 36
Calculus
Stewart
Expert Verified
644 CHAPTER 8 . . Infinite Series 8-34
EXAMPLE 3.8 Using the Limit Comparison Test Investigate the convergence or divergence of the series k = 3 1 k 3 5 k . EXAMPLE 3.9 Using the Limit Comparison Test Investigate the convergence or divergence of the series k 2 2 k + 7 k 5 + 5 k 4 3 k 3 + 2 k 1 . 5 10 15 20 1.50
k = 1 FIGURE 8.29
8-35 SECTION 8.3 . . The Integral Test and Comparison Tests 645 Notice that for k large, the general term looks like k 2 k 5 = 1 k 3 (since the terms with the largest exponents tend to dominate the expression, for large values of k ). From the Limit Comparison Test, for b k = 1 k 3 , we have lim k →∞ a k b k = lim k →∞ k 2 2 k + 7 k 5 + 5 k 4 3 k 3 + 2 k 1 1 1 k 3 = lim k →∞ ( k 2 2 k + 7) ( k 5 + 5 k 4 3 k 3 + 2 k 1) k 3 1 = lim k →∞ ( k 5 2 k 4 + 7 k 3 ) ( k 5 + 5 k 4 3 k 3 + 2 k 1) 1 k 5 1 k 5 = lim k →∞ 1