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Unformatted text preview: 12.3: The Integral Test
The formula
n 2n used to deﬁne the terms in
∞ n=1 n 2n also deﬁnes a function on the real line x f (x) = x 2 so that an = f (n) for all n. The Integral Test compares the area under the graph of f (x) with the area represented by ∞ 2n . n=1 n Theorem (The Integral Test). Suppose f is a continuous, positive, decreasing function on [1, ∞) and f (n) = an. Then the series ∞ an is convergent if and only if n=1 ∞ 1 f (x)dx is convergent. In other words, 1. If 2. If
∞ 1 f (x)dx ∞ 1 f (x)dx is convergent, then so is is divergent, then so is ∞ n=1 an . ∞ n=1 an . 1 12.3: Examples
Determine whether the series is convergent or divergent. 1. 2. 3.
∞ 1 n=1 (2n+1)5 ∞ n2 n=1 n3 +1 ∞ ln(n) n=1 n3 2 12.3: The pSeries
1 For a constant p, the pSeries is ∞ np . Much like n=1 the Geometric Series, we will ﬁnd that some series are just a pSeries in disguise. We can use the Integral Test 1 and the Divergence Test to determine when ∞ np is n=1 convergent or divergent. 1 1. For p < 0, limn→∞ np = ∞, so the Divergence Test. 1 2. For p = 0, limn→∞ np = 1, so Divergence Test. ∞1 n=1 np ∞1 n=1 np diverges by diverges by the For p > 0, we use the integral test with the function 1 f (x) = xp . Section 8.8 Example 4 shows that the im∞1 proper integral 1 xp dx is convergent for p > 1 and divergent when p ≤ 1. Thus, by the Integral Test, we have the following theorem. Theorem (pSeries). The pseries if p > 1 and divergent if p ≤ 1.
∞1 n=1 np is convergent 3 12.3: Examples
Write each sum in the form is convergent or divergent.
1 1. 1 + 1 + 1 + 7 + . . . 3 5 1 1 1 2. 1 + 2√2 + 3√3 + 4√4 + . . . ∞ n=1 an and determine if 4 ...
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This note was uploaded on 04/04/2011 for the course MATH 408 L taught by Professor Zheng during the Fall '10 term at University of Texas at Austin.
 Fall '10
 ZHENG
 Calculus

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