121616949-math.266

# 121616949-math.266 - so it’s generally a good idea to...

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252 Chapter 10 Sequences and Series The obvious first approach, based on what we know, is the integral test. Unfortunately, we can’t compute the required antiderivative. But looking at the series, it would appear that it must converge, because the terms we are adding are smaller than the terms of a p -series, that is, 1 n 2 ln n < 1 n 2 , when n 3. Since adding up the terms 1 /n 2 doesn’t get “too big”, the new series “should” also converge. Let’s make this more precise. The series X n =2 1 n 2 ln n converges if and only if X n =3 1 n 2 ln n converges—all we’ve done is dropped the initial term. We know that X n =3 1 n 2 converges. Looking at two typical partial sums: s n = 1 3 2 ln 3 + 1 4 2 ln 4 + 1 5 2 ln 5 + · · · + 1 n 2 ln n < 1 3 2 + 1 4 2 + 1 5 2 + · · · + 1 n 2 = t n . Since the p -series converges, say to L , and since the terms are positive, t n < L . Since the terms of the new series are positive, the s n form an increasing sequence and s n < t n < L for all n . Hence the sequence { s n } is bounded and so converges. Sometimes, even when the integral test applies, comparison to a known series is easier,
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Unformatted text preview: so it’s generally a good idea to think about doing a comparison before doing the integral test. EXAMPLE 10.32 Does ∞ X n =2 | sin n | n 2 converge? We can’t apply the integral test here, because the terms of this series are not decreasing. Just as in the previous example, however, | sin n | n 2 ≤ 1 n 2 , because | sin n | ≤ 1. Once again the partial sums are non-decreasing and bounded above by ∑ 1 /n 2 = L , so the new series converges. Like the integral test, the comparison test can be used to show both convergence and divergence. In the case of the integral test, a single calculation will conﬁrm whichever is the case. To use the comparison test we must ﬁrst have a good idea as to convergence or divergence and pick the sequence for comparison accordingly....
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