Integral Test

11 Key excerpts on "Integral Test"

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  eBook - PDF

  A Student's Guide to Infinite Series and Sequences

  • Bernhard W. Bach, Jr.(Authors)
  • 2018(Publication Date)
  • Cambridge University Press

    (Publisher)

  This test works because of the geometric relationship between an in fi nite summation and an improper inte-gral. Both a summation and an integral can be thought of as representing areas. The Integral Test is a comparison test in which we compare the area of an in fi nite summation to the area under a curve, i.e., an integral. This test comes with a caveat: you must be able to solve the integral. 34 In fi nite Series Consider applying the Integral Test to the harmonic series 1 þ 1 2 þ 1 3 þ … þ 1 n þ … ¼ X ∞ n ¼ 1 1 n : Note that this is a positive term series, and the consecutive terms of the series are not increasing. If we allow ourselves to view n as a continuous variable, rather than as a discrete index, then the n th term of the series can be thought of as a function of a continuous variable: a n = f(n ) = 1/ n . Therefore, if the integral ð ∞ f ð n Þ dn ¼ ð ∞ 1 n dn is in fi nite, the harmonic series diverges, and if the integral is fi nite, then the harmonic series converges. Evaluating the integral, we fi nd that ð ∞ 1 n dn ¼ ln n j ∞ ¼ ∞ ; i.e., the integral is in fi nite, and therefore the harmonic series diverges. Note that the integral is only evaluated at the upper limit. The lower limit of the integral can be taken at any point in the series or chosen so as to drop any fi nite number of terms. Neglecting a fi nite number of terms of an in fi nite series will not affect the convergence or divergence of the series. Neglecting the fi rst three terms or the fi rst three million terms may affect the sum of the series but not its convergence or divergence. Recall (from Section 2.3.1 ) that the harmonic series is a special case of the p -series ( p =1). The divergence test established that a p -series diverges for p < 1 but produced an indeterminate result for p = 1. With the use of the Integral Test, we now know that the p -series diverges for p ≤ 1.

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  Calculus

  Concepts and Contexts, Enhanced Edition

  • James Stewart(Author)
  • 2018(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  The Integral Test Suppose is a continuous, positive, decreasing function on and let . Then the series is convergent if and only if the improper integral is convergent. In other words: (a) If is convergent, then is convergent. (b) If is divergent, then is divergent. Note: When we use the Integral Test it is not necessary to start the series or the integral at . For instance, in testing the series Also, it is not necessary that be always decreasing. What is important is that be ulti-mately decreasing, that is, decreasing for larger than some number . Then is convergent, so is convergent by Note 4 of Section 8.2. Using the Integral Test Determine whether the series converges or diverges. SOLUTION The function is positive and continuous for because the logarithm function is continuous. But it is not obvious whether or not is decreasing, so we compute its derivative: Thus when , that is, . It follows that is decreasing when and so we can apply the Integral Test: Since this improper integral is divergent, the series is also divergent by the Integral Test. Convergence of the -series For what values of is the series convergent? SOLUTION If , then . If , then . In either case , so the given series diverges by the Test for Divergence [see (8.2.7)]. lim n l 1 n p 0 lim n l 1 n p 1 p 0 lim n l 1 n p p 0 n 1 1 n p p p EXAMPLE 2 v ln n n lim t l ln t 2 2 y 1 ln x x dx lim t l y t 1 ln x x dx lim t l ln x 2 2 1 t x e f x e ln x 1 f x 0 f x x 1 x ln x x 2 1 ln x x 2 f x 1 f x ln x x n 1 ln n n EXAMPLE 1 v n 1 a n n N a n N x f f y 4 1 x 3 2 dx we use n 4 1 n 3 2 n 1 n 1 a n y 1 f x dx n 1 a n y 1 f x dx x 1 f x dx n 1 a n a n f n 1, f Copyright 2019 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience.

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  Calculus

  Single Variable

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2022(Publication Date)
  • Wiley

    (Publisher)

  The following theorem shows that there is a relationship between the convergence of the series and the integral. Theorem 9.4.4: Integral Test Let ∑ u k be a series with positive terms. If f is a function that is decreasing and continuous on an interval [a, +∞) and such that u k = f (k) for all k ≥ a, then ∞  k=1 u k and  +∞ a f (x) dx both converge or both diverge. 1 2 3 4 n + 1 x y 1 2 3 4 n x y . . . . . . y = f (x) y = f (x) u 1 u 2 u 3 u 4 u n u 2 u 3 u 4 u n (b) (a) FIGURE 9.4.1 The proof of the Integral Test is deferred to the end of this section. However, the gist of the proof is captured in Figure 9.4.1: if the integral diverges, then so does the series (Figure 9.4.1a), and if the integral converges, then so does the series (Figure 9.4.1b). Example 4 Show that the Integral Test applies, and use the Integral Test to determine whether the following series converge or diverge: a. ∞  k=1 1 k b. ∞  k=1 1 k 2 Solution (a) We already know that this is the divergent harmonic series, so the Integral Test will simply illustrate another way of establishing the divergence. Note first that the series has positive terms, so the Integral Test is applicable. If we replace k by x in the general term 1 / k, we obtain the function f (x) = 1 / x, which is decreasing and continuous for x ≥ 1 (as required to apply the Integral Test with a = 1). Since  +∞ 1 1 x dx = lim b →+∞  b 1 1 x dx = lim b →+∞ [ln b − ln 1] = +∞ the integral diverges and consequently so does the series. 9.4 Convergence Tests 531 Solution (b) Note first that the series has positive terms, so the Integral Test is applicable. If we replace k by x in the general term 1 / k 2 , we obtain the function f (x) = 1 / x 2 , which is decreasing and continuous for x ≥ 1. Since  +∞ 1 1 x 2 dx = lim b →+∞  b 1 dx x 2 = lim b →+∞  − 1 x  b 1 = lim b →+∞  1 − 1 b  = 1 the integral converges and consequently the series converges by the Integral Test with a = 1.

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  A Radical Approach to Real Analysis

  Second Edition

  • David Bressoud(Author)
  • 2006(Publication Date)
  • American Mathematical Society

    (Publisher)

  Q.E.D. The Integral Test When we first studied the harmonic series in section 2.4, we proved that ∑ ∞ n = 1 1 /n diverges by comparing it to the improper integral ∞ 1 (1 /x ) dx . This is an approach that works whenever a n is the value of a function of n that is positive, decreasing, and asymptotic to 0 as n approaches infinity. The following test for convergence was published by Cauchy in 1827. Theorem 4.13 (The Integral Test). Let f be a positive, decreasing, integrable function for x ≥ 1 . The series ∞ k = 1 f ( k ) converges if and only if we have convergence of the improper integral ∞ 1 f ( x ) dx. Any time we see the symbol ∞ , warning lights should go off. The improper integral actually means the limit lim n →∞ n 1 f ( x ) dx. Proof: Since f is positive for x ≥ 1, it is enough to show that when one of them converges, it provides an upper bound for the other. Since f is decreasing, we know that (see Figure 4.2) f ( k + 1) ≤ k + 1 k f ( x ) dx ≤ f ( k ) . 138 4 The Convergence of Infinite Series Definition: improper integral (unbounded domain) The improper integral ∞ 1 f ( x ) dx is said to converge if there is a number V such that for any error bound , we can always find a response N for which n ≥ N implies that n 1 f ( x ) dx − V < . The number V is called the value of the integral. It follows that N k = 1 f ( k + 1) ≤ N k = 1 k + 1 k f ( x ) dx = N + 1 1 f ( x ) dx ≤ N k = 1 f ( k ) . If the series converges, then the partial integrals are bounded: N + 1 1 f ( x ) dx ≤ N k = 1 f ( k ) ≤ ∞ k = 1 f ( k ) . If the integral converges, then the partial sums are bounded: N + 1 k = 1 f ( k ) ≤ f (1) + N + 1 1 f ( x ) dx ≤ f (1) + ∞ 1 f ( x ) dx. Q.E.D. In section 2.4, we not only proved that the harmonic series diverges, we found an explicit formula for the difference between the partial sum of the first n terms and n 1 dx/x = ln n . The same thing can be done whenever the summand is of the form f ( k ) where f is an analytic function for x > 0.

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  Calculus All-in-One For Dummies (+ Chapter Quizzes Online)

  • Mark Ryan(Author)
  • 2023(Publication Date)
  • For Dummies

    (Publisher)

  Look back at Figure 18-1. Now that you know the integral com- parison test, you can appreciate the connection between those integrals and their companion p-series: the divergent harmonic series, n n 1 1 , and the convergent p-series, n n 1 2 1 . Here’s the mumbo jumbo for the integral comparison test. Note the fine print. Integral comparison test: If f x is positive, continuous, and decreasing for all x 1 and if a f n n , then n n a 1 and 1 f x dx either both converge or both diverge. CHAPTER 19 Infinite Series: Welcome to the Outer Limits 597 7 For Problems 7 through 14, determine whether the series converges or diverges. n n n 1 10 0 9 . 8 n n n 1 1 1 10 . 9 1 1001 1 2001 1 3001 1 4001 10 n n n n 1 1 ln 11 n n n 1 3 3 1 ln 12 n n n n 2 1 ln sin 598 UNIT 5 Integration and Infinite Series 13 n n n e 1 2 3 14 n n n 1 3 ! (Given that 1 n! converges.) The two “R” tests: Ratios and roots Unlike the three benchmark tests from the previous section, the ratio and root tests don’t com- pare a new series to a known benchmark. They work by looking only at the nature of the series you’re trying to figure out. They form a cohesive pair because the results of both tests tell you the same thing. If the result is less than 1, the series converges; if it’s more than 1, the series diverges; and if it’s exactly 1, you learn nothing and must try a different test. (As presented here, the ratio and root tests are used for series of positive terms. In other books, you may see a different version of each test that uses the absolute value of the terms. These absolute value versions can be used for series made up of both positive and negative terms. Don’t sweat this; the different versions amount to the same thing.) The ratio test: Given a series u n , consider the limit of the ratio of a term to the previous term, lim n n n u u 1 . If this limit is less than 1, the series converges. If it’s greater than 1 (this includes ), the series diverges.

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  Calculus, Metric Edition

  • James Stewart, Daniel K. Clegg, Saleem Watson, , James Stewart, James Stewart, Daniel K. Clegg, Saleem Watson(Authors)
  • 2020(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. SECTION 11.3 The Integral Test and Estimates of Sums 791 confirmation. Figure 2 shows the curve y - 1 ysx , but this time we use rectangles whose tops lie above the curve. x y 0 2 1 3 4 5 area= 1 oe„ 1 oe„ 1 oe„ 1 oe„ 1 y= 1 oe„ x area= 2 area= 3 area= 4 The base of each rectangle is an interval of length 1. The height is equal to the value of the function y - 1 ysx at the left endpoint of the interval. So the sum of the areas of all the rectangles is 1 s1 1 1 s2 1 1 s3 1 1 s4 1 1 s5 1 ∙ ∙ ∙ - o ` n-1 1 sn This total area is greater than the area under the curve y - 1 ysx for x > 1, which is equal to the integral y ` 1 (1 ysx ) dx. But we know from Example 7.8.4 that this improper integral is divergent. In other words, the area under the curve is infinite. So the sum of the series must be infinite; that is, the series is divergent. The same sort of geometric reasoning that we used for these two series can be used to prove the following test. (The proof is given at the end of this section.) The Integral Test Suppose f is a continuous, positive, decreasing function on f1, `d and let a n - f snd. Then the series  ` n-1 an is convergent if and only if the improper integral y ` 1 f s xd dx is convergent. In other words: (i) If y ` 1 f s xd dx is convergent, then o ` n-1 a n is convergent. (ii) If y ` 1 f s xd dx is divergent, then o ` n-1 a n is divergent. NOTE When we use the Integral Test, it is not necessary to start the series or the integral at n - 1.

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  Calculus

  Resequenced for Students in STEM

  • David Dwyer, Mark Gruenwald(Authors)
  • 2017(Publication Date)
  • Wiley

    (Publisher)

  716 CHAPTER 12. INFINITE SERIES Example 4 Applying the Root Test Determine whether the series ∞ X n=1 3 2n n n converges or diverges. Solution Here we have a n = 3 2n n n = (3 2 ) n n n =  9 n  n and so n p |a n | = n s  9 n  n = 9 n Since 9 n → 0 < 1, it follows from the Root Test that the series converges. As the algebraic steps in Example 4 suggest, the Root Test may be especially useful in situations where the terms of a series can be written in the form () n . Summary of Tests for Convergence or Divergence We have considered quite a variety of tests for convergence and divergence, and so one of the challenges is deciding which test to use for a given series. Here are some questions to consider when trying to decide which test(s) to apply. • Is the series geometric or a p-series? If so, apply the criteria for those series. • Do the terms of the series converge to 0? If not, apply the nth-Term Test. • Does the series have positive terms and a form that is similar to a geometric series or a p-series? If so, try a comparison test. The comparison tests are often useful for series whose general nth term is an algebraic function in n. • Does the series have negative terms? If so, consider the Absolute Convergence Test. If the terms alternate in sign, consider the Alternating Series Test. • Does the series involve nth powers or factorials? If so, try the Ratio Test. • Can the nth term be written in the form () n ? If so, try the Root Test. • Does the series satisfy the conditions for the Integral Test, and is the function under consideration easily integrable? If so, try the Integral Test. The following table summarizes the key features of the tests we have considered.

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  Calculus

  One and Several Variables

  • Saturnino L. Salas, Garret J. Etgen, Einar Hille(Authors)
  • 2011(Publication Date)
  • Wiley

    (Publisher)

  Since the terms are nonnegative, the sequence is nondecreasing. By being bounded and nondecreasing, the sequence of partial sums converges. (Theorem 11.3.6.) This means that the series converges. ❏ The convergence or divergence of some series can be deduced from the convergence or divergence of a closely related improper integral. THEOREM 12.3.2 THE Integral Test If f is continuous, positive, and decreasing on [1, ∞), then ∞  k=1 f (k ) converges iff  ∞ 1 f (x ) dx converges. PROOF In Exercise 69, Section 11.7, you were asked to show that if f is continuous, positive, and decreasing on [1, ∞), then  ∞ 1 f (x ) dx converges iff the sequence a n =  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 12.3.1. f (2) + • • • + f (n) • • • n – 1 n 1 f (1) + • • • + f (n – 1) 2 3 4 n 1 • • • n – 1 n 1 2 3 4 f (x) dx n 1 Figure 12.3.1 Let’s suppose that f is continuous, positive, and decreasing on [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, f (2) + · · · + f (n) ≤  n 1 f (x ) dx ≤ f (1) + · · · + f (n − 1). 12.3 THE Integral Test; BASIC COMPARISON, LIMIT COMPARISON ■ 587 If the sequence of integrals converges, it is bounded. Then, by the left inequality, the sequence of partial sums is bounded and the series converges. Suppose now that the sequence of integrals diverges. Since f is positive, the se- quence of integrals increases:  n 1 f (x ) dx <  n+1 1 f (x ) dx . Since this sequence diverges, it must be unbounded. Then, by the right inequality, the sequence of partial sums is unbounded and the series diverges. ❏ Applying the Integral Test Example 1 (The harmonic series) (12.3.3) ∞  k=1 1 k = 1 + 1 2 + 1 3 + 1 4 + · · · diverges.

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  Mathematical Methods for Physicists

  • George B. Arfken(Author)
  • 2013(Publication Date)
  • Academic Press

    (Publisher)

  Taking the limit as i -oo, we have Γ f{x)dx <Σα η < Γ/(χ)άχ + a x . (5.25) Ji «=i Ji 284 INFINITE SERIES Hence the infinite series converges or diverges as the corresponding integral converges or diverges. This Integral Test is particularly useful in setting upper and lower bounds on the remainder of a series after some number of initial terms have been summed. That is, where Σ a n = Σ a n + Σ a n> n = l n = l n = N+l f(x)dx< a n< f(x)dx + a N + l' EXAMPLE 5.2.3 Riemann Zeta Function The Riemann zeta function is defined by M = l We may take/(x) = x~ p and then x~ p dx f ,-p+i -p + 1 = lnx|?, ρφΐ p = l. (5.26) (5.27) The integral and therefore the series are divergent for p < 1, convergent for p > 1. Hence Eq. 5.26 should carry the condition p > 1. This, incidentally, is an independent proof that the harmonic series (p = 1) diverges and diverges logarithmically. The sum of the first million terms ^ 1 » 0 0 0 ' 0 0 0 ^-^ j s on iy 14.392 726 .... This integral comparison may also be used to set an upper limit to the Euler-Mascheroni constant 1 defined by = lim ( Y m ^ ) . Returning to partial sums, s, n C n dx L= Y m x — In n < In n + 1. -=i Ji X (5.28) (5.29) Evaluating the integral on the right, s„ < 1 for all n and therefore y < 1. Exercise 5.2.12 leads to more restrictive bounds. Actually the Euler-Mascheroni constant is 0.577 215 66 .... 'This is the notation of National Bureau of Standards, Handbook of Mathe-matical Functions. Applied Mathematics Series-55 (AMS-55). CONVERGENCE TESTS 285 Kummers Test This is the first of three tests that are somewhat more difficult to apply than the preceding tests. Their importance lies in their power and sensitivity. Fre-quently, at least one of the three will work when the simpler easier tests are indecisive. It must be remembered, however, that these tests, like those pre-viously discussed, are ultimately based on comparisons.

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  Anton's Calculus

  Early Transcendentals

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2018(Publication Date)
  • Wiley

    (Publisher)

  Later, we will use some of these tests to study the convergence of Taylor series. THE COMPARISON TEST We will begin with a test that is useful in its own right and is also the building block for other important convergence tests. The underlying idea of this test is to use the known conver- gence or divergence of a series to deduce the convergence or divergence of another series. It is not essential in Theorem 9.5.1 that the condition a k ≤ b k hold for all k, as stated; the conclusions of the theorem remain true if this condition is eventually true. 9.5.1 theorem (The Comparison Test) Let ∑ ∞ k = 1 a k and ∑ ∞ k = 1 b k be series with non- negative terms and suppose that a 1 ≤ b 1 , a 2 ≤ b 2 , a 3 ≤ b 3 , … , a k ≤ b k , … (a) If the “bigger series” Σb k converges, then the “smaller series” Σa k also con- verges. (b) If the “smaller series” Σa k diverges, then the “bigger series” Σb k also diverges. We have left the proof of this theorem for the exercises; however, it is easy to visual- ize why the theorem is true by interpreting the terms in the series as areas of rectangles (Figure 9.5.1). The comparison test states that if the total area ∑ b k is finite, then the total 1 . . . . . . . . . . . . a 1 b 1 2 a 2 b 2 3 a 3 b 3 4 a 4 b 4 5 a 5 b 5 k a k b k For each rectangle, a k denotes the area of the blue portion and b k denotes the combined area of the white and blue portions. Figure 9.5.1 area ∑ a k must also be finite; and if the total area ∑ a k is infinite, then the total area ∑ b k must also be infinite. USING THE COMPARISON TEST There are two steps required for using the comparison test to determine whether a series ∑ u k with positive terms converges: Step 1. Guess at whether the series ∑ u k converges or diverges. Step 2. Find a series that proves the guess to be correct.

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  The Calculus Lifesaver

  All the Tools You Need to Excel at Calculus

  • Adrian Banner(Author)
  • 2009(Publication Date)
  • Princeton University Press

    (Publisher)

  C h a p te r 20 Improper Integrals: Basic Concepts This is a difficult topic, so I’m devoting two chapters to it. This chapter serves as an introduction to improper integrals. The next chapter gets into the details of how to solve problems involving improper integrals. If you are reading this chapter for the first time, you should probably take care to try to understand all the points in it. On the other hand, if you are reviewing for a test, most likely you’ll want to skim over the chapter, noting the boxed formulas and the sections marked as important, and concentrate on the next chapter. Here’s what we’ll actually look at in this chapter: • the definition of improper integrals, convergence, and divergence; • improper integrals over unbounded regions; and • the theoretical basis for the comparison test, the limit comparison test, the p -test, and the absolute convergence test. We’ll revisit all four of these tests in the next chapter and see many examples of how to apply them. 20.1 Convergence and Divergence What is an improper integral, anyway? In Chapter 16, we saw that the integral Z b a f ( x ) dx certainly makes sense if f is a bounded function on [ a, b ] which is continuous except at a finite number of places. If f has infinitely many discontinuities, the integral might still make sense, or it might be totally screwed up (see Section 16.7 of Chapter 16 for an example). What if f isn’t bounded? This means that the values of f ( x ) manage to get really large (positively or neg-atively or both) while x is in the interval [ a, b ]. This sort of thing typically happens when f has a vertical asymptote somewhere in this interval: the function blows up there and can’t be bounded. This causes the above integral to be improper. 432 • Improper Integrals: Basic Concepts There’s a different type of unboundedness that can occur even if f is bounded. The interval [ a, b ] can actually be infinite—something like [0 , ∞ ), [ -7 , ∞ ), ( -∞ , 3] or even ( -∞ , ∞ ).

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