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Unformatted text preview: ANSWERS TO SELECTED EXERCISES: HW3 Exercise 7.1.5 Determine all values of p and q for which the following series converges: X k =2 1 k q (ln k ) p Proof : The main idea is that the behavior of this series is determined by the factor 1 k q rather than 1 (ln k ) p ; roughly speaking, polynomial expressions grow a lot faster than logarithmic expressions. So we distinguish three cases: q &lt; 1: Fix an &gt; 0 such that q + &lt; 1. Then compare the given series with X k =2 1 k q + , which diverges. Since lim k (ln k ) p k = 0 , there is a k such that (ln k ) p k &lt; 1, for all k &gt; k or equivalently 1 k &lt; 1 (ln k ) p for all k &gt; k , which yields 1 k q + &lt; 1 k q (ln k ) p for all k &gt; k , hence our series diverges, by the comparison test (Theorem 7.1.2). q &gt; 1: Fix an &gt; 0 such that q- &gt; 1. We compare the given series with X k =2 1 k q- , which converges. Similarly as above, we get a k such that for all k &gt; k the following inequality holds: 1 k q (ln...
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- Winter '08