EXAMPLE 34 Finding the Number of Terms Needed for a Given Accuracy Determine

Example 34 finding the number of terms needed for a

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EXAMPLE 3.4 Finding the Number of Terms Needed for a Given Accuracy Determine the number of terms needed to obtain an approximation to the sum of the series k = 1 1 k 3 correct to within 10 5 . Solution Again, we already used the Integral Test to show that the series in question converges. Then, by Theorem 3.2, we have that the remainder satisfies 0 R n n 1 x 3 dx = lim R →∞ R n 1 x 3 dx = lim R →∞ 1 2 x 2 R n = lim R →∞ 1 2 R 2 + 1 2 n 2 = 1 2 n 2 . So, to ensure that the remainder is less than 10 5 , we require that 0 R n 1 2 n 2 10 5 . Solving this last inequality for n yields n 2 10 5 2 or n 10 5 2 = 100 5 223 . 6 .
8-31 SECTION 8.3 . . The Integral Test and Comparison Tests 641 So, taking n 224 will guarantee the required accuracy and consequently, we have k = 1 1 k 3 224 k = 1 1 k 3 = 1 . 202047, which is correct to within 10 5 , as desired. Comparison Tests We next present two results that allow us to compare a given series with one that is already known to be convergent or divergent, much as we did with improper integrals in section 6.6. THEOREM 3.3 (Comparison Test) Suppose that 0 a k b k , for all k . (i) If k = 1 b k converges, then k = 1 a k converges, too. (ii) If k = 1 a k diverges, then k = 1 b k diverges, too. Intuitively, this theorem should make abundant sense: if the “larger” series converges, then the “smaller” one must also converge. Likewise, if the “smaller” series diverges, then the “larger” one must diverge, too. PROOF Given that 0 a k b k for all k , observe that the n th partial sums of the two series satisfy 0 S n = a 1 + a 2 + ··· + a n b 1 + b 2 + ··· + b n . (i) If k = 1 b k converges (say to B ), this says that 0 S n a 1 + a 2 + ··· + a n b 1 + b 2 + ··· + b n k = 1 b k = B , (3.5) for all n 1. From (3.5), the sequence { S n } n = 1 of partial sums of k = 1 a k is bounded. No- tice that { S n } n = 1 is also increasing. (Why?) Since every bounded, monotonic sequence is convergent (see Theorem 1.4), we get that k = 1 a k is convergent, too. (ii) If k = 1 a k is divergent, we have (since all of the terms of the series are nonnegative) that lim n →∞ ( b 1 + b 2 + ··· + b n ) lim n →∞ ( a 1 + a 2 + ··· + a n ) = ∞ . Thus, k = 1 b k must be divergent, also. You can use the Comparison Test to test the convergence of series that look similar to series that you already know are convergent or divergent (notably, geometric series or p -series).
642 CHAPTER 8 . . Infinite Series 8-32 EXAMPLE 3.5 Using the Comparison Test for a Convergent Series Investigate the convergence or divergence of k = 1 1 k 3 + 5 k . Solution The graph of the first 20 partial sums shown in Figure 8.26 suggests that the series converges to some value near 0.3. To confirm such a conjecture, we must carefully test the series. Note that for large values of k , the general term of the series looks like 1 k 3 , since when k is large, k 3 is much larger than 5 k . This observation is significant, since we already know that k = 1 1 k 3 is a convergent p -series ( p = 3 > 1).

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