SalasSV_11_02

# SalasSV_11_02 - 11.2 THE INTEGRAL TEST; COMPARISON TESTS...

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² 11.2 THE INTEGRAL TEST; COMPARISON TESTS Here and in the next section we direct our attention to series with nonnegative terms: a k 0 for all k . Note that for such series, the sequence of partial sums is nondecreasing: s n + 1 = n + 1 ³ k = 0 a k = a n + 1 + n ³ k = 0 a k n ³ k = 0 a k = s n . The following theorem is fundamental. THEOREM 11.2.1 A series with nonnegative terms converges iff the sequence of partial sums is bounded. PROOF Assume that the series converges. Then the sequence of partial sums is convergent and therefore bounded (Theorem 10.3.4). Suppose now that the sequence of partial sums is bounded. Since the terms are nonnegative, the sequence is nondecreasing. By being bounded and nondecreasing, the

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644 ± CHAPTER 11 INFINITE SERIES the sequence of partial sums converges (Theorem 10.3.6). This means that the series converges. ± The convergence or divergence of a series can sometimes be deduced from the convergence or divergence of a closely related improper integral. THEOREM 11.2.2 THE INTEGRAL TEST If f is continuous, decreasing, and positive on [1, ), then ± k = 1 f ( k ) converges iff ² 1 f ( x ) dx converges. PROOF In Exercise 69, Section 10.7, you were asked to show that with f continuous, decreasing, and positive on [1, ) ² 1 f ( x ) dx converges iff the sequence ² n 1 f ( x ) dx converges. We assume this result and base our proof on the behavior of the sequence of integrals. To visualize our argument, see Figure 11.2.1. f (2) + • • • + f ( n ) n – 1 n 1 f (1) + • • • + f ( n – 1) 234 n 1 n – 1 n 1234 f ( x ) dx n 1 Figure 11.2.1 Since f decreases on the interval [1, n ] f (2) +···+ f ( n ) is a lower sum for f on [1, n ]. and f (1) f ( n 1) is an upper sum for f on [1, n ]. Consequently (1) f (2) f ( n ) ² n 1 f ( x ) dx and ² n 1 f ( x ) dx f (1) f ( n 1). If the sequence of integrals converges, it is bounded. By the ±rst inequality the sequence of partial sums is bounded and the series is therefore convergent. Suppose now that the sequence of integrals diverges. Since f is positive, the sequence of integrals increases: ² n 1 f ( x ) dx < ² n + 1 1 f ( x ) dx .
11.2 THE INTEGRAL TEST; COMPARISON TESTS ± 645 Since this sequence diverges, it must be unbounded. By the second inequality, the sequence of partial sums must be unbounded and the series is divergent. ± Remark The inequalities established in the proof of Theorem 11.2.2 lead to bounds on the sum of the inFnite series ± k = 1 f ( k ) where f is continuous, decreasing, and positive on [1, ). In particular, it follows from the second inequality in (1) that ² 1 f ( x ) dx ± k = 1 f ( k ), and from the Frst inequality in (1) that ± k = 1 f ( k ) f (1) + ² 1 f ( x ) dx . Combining these two inequalities, we have ² 1 f ( x ) dx ± k = 1 f ( k ) f (1) + ² 1 f ( x ) dx . These inequalities spell out the relation that holds between inFnite series and the corresponding improper integral.

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## This note was uploaded on 10/12/2010 for the course MATH 12345 taught by Professor Smith during the Spring '10 term at University of Houston - Downtown.

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SalasSV_11_02 - 11.2 THE INTEGRAL TEST; COMPARISON TESTS...

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