DOCSLIB.ORG

Math 131Infinite Series, Part IV: the Limit Comparisontest

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Math 131Infinite Series, Part IV: the Limit Comparisontest 21 The Limit Comparison Test While the direct comparison test is very useful, there is another comparison test that focuses only on the tails of the series that we want to compare. This makes it more widely applicable and simpler to use. We don’t need to verify that an ≤ bn for all (or most) n. However, it will require our skills in evaluating limits at infinity! THEOREM 14.7 (The Limit Comparison Test). Assume that an > 0 and bn > 0 for all n (or at least all n ≥ k) and that a lim n = L. n!¥ bn ¥ (1) If 0 < L < ¥ (i.e., L is a positive, finite number), then either the series ∑ an and n=1 ¥ ∑ bn both converge or both diverge. n=1 ¥ ¥ (2) If L = 0 and ∑ bn converges, then ∑ an converges. n=1 n=1 ¥ ¥ (3) If L = ¥ and ∑ bn diverges, then ∑ an diverges. n=1 n=1 an The idea of the theorem is since lim = L, then eventually an ≈ Lbn. So if one n!¥ bn of the series converges (diverges) so does the other since the two are ‘essentially’ scalar multiples of each other. ¥ 1 EXAMPLE 14.19. Does converge? ∑ 2 − + n=1 3n n 6 SOLUTION. scrap work: Let’s apply the limit comparison test. Notice that the terms are always positive since the polynomial 3n2 − n + 6 has no roots. In any event, the terms are eventually positive since this an upward-opening parabola. If we focus on ¥ 1 highest powers, then the series looks a like the p-series ∑ 2 which converges. n=1 n 1 1 argument: Since the terms 3n2−n+6 and n2 are positive, we can apply Theorem 14.7. 1 2 a 2 n 1 1 n = 3n −n+6 = = = > lim lim lim 2 lim 0. n!¥ bn n!¥ 1 n!¥ 3n − n + 6 n!¥ − 1 + 6 3 n2 3 n n2 ¥ 1 ¥ 1 Since converges by the p-series test (p = 2 > 1), then converges ∑ 2 ∑ 2 − + n=1 n n=1 3n n 6 by the limit comparison test (Theorem 14.7). note: When the series involve fractions, the first step in the limit process can be done more efficiently. Instead of dividing one fraction by the other, we can multiply one fraction by the reciprocal of the other. For instance, earlier in this example we could have written a 1 n2 n = · lim lim 2 n!¥ bn n!¥ 3n − n + 6 1 and then carried out the rest of the calculation. ¥ 1 EXAMPLE 14.20 (The General Harmonic Series). The series is called the gen- ∑ + n=1 cn d eral harmonic series. If c > 0 does this series converge? SOLUTION. scrap work: Let’s apply the limit comparison test by making the obvi- ¥ 1 ous comparison to the harmonic series ∑ which we know diverges. n=1 n math131 infiniteseries, part iv: the limit comparisontest 22 1 argument: Since the terms cn+d are positive once cn + d > 0, in other words when d 1 n > − c , and since and n is always positive, we can apply Theorem 14.7. a 1 n 1 1 lim n = lim · = lim = > 0, n!¥ b n!¥ cn + d 1 n!¥ d c n c + n ¥ 1 since c > 0. Since ∑ diverges by the p-series test (p = 1), then the general har- n=1 n ¥ 1 monic series diverges by the limit comparison test (Theorem 14.7). ∑ + n=1 cn d p ¥ n EXAMPLE 14.21. Does the series converge? ∑ 3 + n=1 2n 4 p ¥ n SOLUTION. scrap work: This time if we focus on highest powers, is ∑ 3 + n=1 2n 4 ¥ n1/2 ¥ 1 roughly equal to = , which diverges ∑ n3 ∑ n5/2 n=1 n=p1 n 1 argument: Since the terms 2n3+4 and n5/2 are always positive, we can apply Theo- rem 14.7. a n1/2 n5/2 n3 1 1 n = · = = = > lim lim 3 lim 3 lim 0. n!¥ bn n!¥ 2n + 4 1 n!¥ 2n + 4 n!¥ + 4 2 2 n3 p ¥ 1 ¥ n Since converges by the p-series test (p = 5 > 1), then converges ∑ 5/2 2 ∑ 3 + n=1 n n=1 2n 4 by the limit comparison test (Theorem 14.7). ¥ 1 EXAMPLE 14.22. Does the series ∑ p converge? n=1 4n + 5 ¥ 1 SOLUTION. scrap work: The obvious comparison is to the p-series which ∑ 1/2 n=1 n diverges. argument: Since the terms p 1 and 1 are always positive, we can apply Theo- 4n+5 n1/2 rem 14.7. 1/2 1/2 1/2 a 1 n HPwrs n n 1 lim n = lim p · = lim p = lim = > 0. n!¥ bn n!¥ 4n + 5 1 n!¥ 4n n!¥ 2n1/2 2 ¥ 1 ¥ 1 Since diverges by the p-series test (p = 1 < 1), then the series p ∑ 1/2 2 ∑ n=1 n n=1 4n + 5 diverges by the limit comparison test (Theorem 14.7). ¥ 6n2 · 2n EXAMPLE 14.23. Does the series ∑ 4 converge? n=1 n + 3 ¥ 6n2 · 2n SOLUTION. scrap work: Focusing on highest powers, ∑ 4 is roughly n=1 n + 3 ¥ n2 · 2n ¥ 2n = ∑ 4 ∑ 2 which we saw is divergent in Example 14.5. n=1 n n=1 n 6n2·2n 2n argument: Since the terms n4+3 and n2 are always positive, we can apply Theo- rem 14.7. (Note the use of the reciprocal.) a 6n2 · 2n n2 6n4 6 n = · = = = > lim lim n lim lim 6 0. n!¥ bn n!¥ n4 + 3 2 n!¥ n4 + 3 n!¥ + 4 1 n3 ¥ 2n ¥ 6n2 · 2n Since ∑ 2 diverges from Example 14.5, then the series ∑ 4 diverges by the n=1 n n=1 n + 3 limit comparison test (Theorem 14.7). 23 ¥ (−1)n · (2 ln(en) + 2) EXAMPLE 14.24. Does the series converge? ∑ ( ) 3 n=1 cos np n SOLUTION. scrap work: First simplify the nth term: cos(np) = (−1)n right? And n n ( n) = (−1) ·(2 ln(e )+2) = 2n+2 ¥ 1 2 ln e 2n. So cos(np)n3 n3 . Use a limit comparison test with ∑n=1 n2 . n n (−1) ·(2 ln(e )+2) = 2n+2 1 argument: Since the terms cos(np)n3 n3 and n2 are always positive, we can apply Theorem 14.7. a 2n + 2 n2 2n3 + 2n2 2n3 n = · = HPwrs= = > lim lim 3 lim 3 lim 3 2 0. n!¥ bn n!¥ n 1 n!¥ n n!¥ n ¥ 1 ¥ (−1)n · (2 ln(en) + 2) Since converges by the p-series test, then the series ∑ 2 ∑ ( ) 3 n=1 n n=1 cos np n converges by the limit comparison test (Theorem 14.7). ¥ 1 EXAMPLE 14.25. Does the series ∑ sin 2 converge? n=1 n 1 SOLUTION. scrap work: The terms sin n2 are always positive. Use a limit com- ¥ 1 parison test with ∑n=1 n2 . 1 1 argument: Since the terms cos n2 and n2 are always positive, we can apply Theo- rem 14.7. sin 1 sin 1 cos 1 · − 2 a n2 x2 l’Ho x2 x3 lim n = lim = lim = lim n!¥ n!¥ 1 n!¥ 1 x!¥ 2 bn 2 2 − n x x3 1 = lim cos x!¥ x2 = cos 0 = 1. ¥ 1 ¥ 1 Since ∑ 2 converges by the p-series test, then the series ∑ cos 2 converges by n=1 n n=1 n the first part of the limit comparison test (Theorem 14.7). The Ratio Test One difficulty with using the two comparison tests is that requires you to know something! It requires you to know something about the convergence or di- vergence of a similar or related series. The ratio test does not depend on such knowledge—it is a self-contained or self-referential test and the results depend only on the series under consideration. One of its drawbacks, however, is that the test is often inconclusive in terms of deciding whether there is convergence. How- ever, we will see that this test is extremely useful in dealing with so-called power series, which is our final topic of the term. Ok, here’s the test. ¥ THEOREM 14.8 (The Ratio Test). Assume that ∑ an is a series with positive terms and let n=1 a + r = lim n 1 . n!¥ an ¥ 1. If r < 1, then the series ∑ an converges. n=1 ¥ an+1 2. If r > 1 or lim = ¥, then the series an diverges. n!¥ ∑ an n=1 3. If r = 1, then the test is inconclusive. The series may converge or diverge. math131 infiniteseries, part iv: the limit comparisontest 24 a + Proof. Here’s the idea for the proof of part 1. Suppose that lim n 1 = r < 1. This n!¥ an means that when n is sufficiently large (say n > k), we have an+1 ≈ r ) an+1 ≈ anr an an+2 2 ≈ r ) an+2 ≈ an+1r ≈ anr an+1 an+3 3 ≈ r ) an+3 ≈ an+2r ≈ anr an+2 . ≈ . k an+k ≈ anr So the tail of the series is approximately geometric: ¥ ¥ k ∑ an+k ≈ ∑ anr . k=0 k=0 ¥ But the geometric series converges since r < 1. So ∑ an converges by the (limit) n=1 comparison test. Using the Ratio Test The real utility of this test is that one need not know about another series to deter- mine whether the series under consideration converges. This is very different than with the comparison tests or the integral test where some sort of comparison to another series is required. ¥ n EXAMPLE 14.26. Determine whether ∑ n converges.
Load more

Recommended publications
  • Abel's Identity, 131 Abel's Test, 131–132 Abel's Theorem, 463–464 Absolute Convergence, 113–114 Implication of Condi
    INDEX Abel’s identity, 131 Bolzano-Weierstrass theorem, 66–68 Abel’s test, 131–132 for sequences, 99–100 Abel’s theorem, 463–464 boundary points, 50–52 absolute convergence, 113–114 bounded functions, 142 implication of conditional bounded sets, 42–43 convergence, 114 absolute value, 7 Cantor function, 229–230 reverse triangle inequality, 9 Cantor set, 80–81, 133–134, 383 triangle inequality, 9 Casorati-Weierstrass theorem, 498–499 algebraic properties of Rk, 11–13 Cauchy completeness, 106–107 algebraic properties of continuity, 184 Cauchy condensation test, 117–119 algebraic properties of limits, 91–92 Cauchy integral theorem algebraic properties of series, 110–111 triangle lemma, see triangle lemma algebraic properties of the derivative, Cauchy principal value, 365–366 244 Cauchy sequences, 104–105 alternating series, 115 convergence of, 105–106 alternating series test, 115–116 Cauchy’s inequalities, 437 analytic functions, 481–482 Cauchy’s integral formula, 428–433 complex analytic functions, 483 converse of, 436–437 counterexamples, 482–483 extension to higher derivatives, 437 identity principle, 486 for simple closed contours, 432–433 zeros of, see zeros of complex analytic on circles, 429–430 functions on open connected sets, 430–432 antiderivatives, 361 on star-shaped sets, 429 for f :[a, b] Rp, 370 → Cauchy’s integral theorem, 420–422 of complex functions, 408 consequence, see deformation of on star-shaped sets, 411–413 contours path-independence, 408–409, 415 Cauchy-Riemann equations Archimedean property, 5–6 motivation, 297–300
    [Show full text]
  • Math 142 – Quiz 6 – Solutions 1. (A) by Direct Calculation, Lim 1+3N 2
    Math 142 – Quiz 6 – Solutions 1. (a) By direct calculation, 1 + 3n 1 + 3 0 + 3 3 lim = lim n = = − . n→∞ n→∞ 2 2 − 5n n − 5 0 − 5 5 3 So it is convergent with limit − 5 . (b) By using f(x), where an = f(n), x2 2x 2 lim = lim = lim = 0. x→∞ ex x→∞ ex x→∞ ex (Obtained using L’Hˆopital’s Rule twice). Thus the sequence is convergent with limit 0. (c) By comparison, n − 1 n + cos(n) n − 1 ≤ ≤ 1 and lim = 1 n + 1 n + 1 n→∞ n + 1 n n+cos(n) o so by the Squeeze Theorem, the sequence n+1 converges to 1. 2. (a) By the Ratio Test (or as a geometric series), 32n+2 32n 3232n 7n 9 L = lim (16 )/(16 ) = lim = . n→∞ 7n+1 7n n→∞ 32n 7 7n 7 The ratio is bigger than 1, so the series is divergent. Can also show using the Divergence Test since limn→∞ an = ∞. (b) By the integral test Z ∞ Z t 1 1 t dx = lim dx = lim ln(ln x)|2 = lim ln(ln t) − ln(ln 2) = ∞. 2 x ln x t→∞ 2 x ln x t→∞ t→∞ So the integral diverges and thus by the Integral Test, the series also diverges. (c) By comparison, (n3 − 3n + 1)/(n5 + 2n3) n5 − 3n3 + n2 1 − 3n−2 + n−3 lim = lim = lim = 1. n→∞ 1/n2 n→∞ n5 + 2n3 n→∞ 1 + 2n−2 Since P 1/n2 converges (p-series, p = 2 > 1), by the Limit Comparison Test, the series converges.
    [Show full text]
  • 11.3-11.4 Integral and Comparison Tests
    11.3-11.4 Integral and Comparison Tests The Integral Test: Suppose a function f(x) is continuous, positive, and decreasing on [1; 1). Let an 1 P R 1 be defined by an = f(n). Then, the series an and the improper integral 1 f(x) dx either BOTH n=1 CONVERGE OR BOTH DIVERGE. Notes: • For the integral test, when we say that f must be decreasing, it is actually enough that f is EVENTUALLY ALWAYS DECREASING. In other words, as long as f is always decreasing after a certain point, the \decreasing" requirement is satisfied. • If the improper integral converges to a value A, this does NOT mean the sum of the series is A. Why? The integral of a function will give us all the area under a continuous curve, while the series is a sum of distinct, separate terms. • The index and interval do not always need to start with 1. Examples: Determine whether the following series converge or diverge. 1 n2 • X n2 + 9 n=1 1 2 • X n2 + 9 n=3 1 1 n • X n2 + 1 n=1 1 ln n • X n n=2 Z 1 1 p-series: We saw in Section 8.9 that the integral p dx converges if p > 1 and diverges if p ≤ 1. So, by 1 x 1 1 the Integral Test, the p-series X converges if p > 1 and diverges if p ≤ 1. np n=1 Notes: 1 1 • When p = 1, the series X is called the harmonic series. n n=1 • Any constant multiple of a convergent p-series is also convergent.
    [Show full text]
  • 3.3 Convergence Tests for Infinite Series
    3.3 Convergence Tests for Infinite Series 3.3.1 The integral test We may plot the sequence an in the Cartesian plane, with independent variable n and dependent variable a: n X The sum an can then be represented geometrically as the area of a collection of rectangles with n=1 height an and width 1. This geometric viewpoint suggests that we compare this sum to an integral. If an can be represented as a continuous function of n, for real numbers n, not just integers, and if the m X sequence an is decreasing, then an looks a bit like area under the curve a = a(n). n=1 In particular, m m+2 X Z m+1 X an > an dn > an n=1 n=1 n=2 For example, let us examine the first 10 terms of the harmonic series 10 X 1 1 1 1 1 1 1 1 1 1 = 1 + + + + + + + + + : n 2 3 4 5 6 7 8 9 10 1 1 1 If we draw the curve y = x (or a = n ) we see that 10 11 10 X 1 Z 11 dx X 1 X 1 1 > > = − 1 + : n x n n 11 1 1 2 1 (See Figure 1, copied from Wikipedia) Z 11 dx Now = ln(11) − ln(1) = ln(11) so 1 x 10 X 1 1 1 1 1 1 1 1 1 1 = 1 + + + + + + + + + > ln(11) n 2 3 4 5 6 7 8 9 10 1 and 1 1 1 1 1 1 1 1 1 1 1 + + + + + + + + + < ln(11) + (1 − ): 2 3 4 5 6 7 8 9 10 11 Z dx So we may bound our series, above and below, with some version of the integral : x If we allow the sum to turn into an infinite series, we turn the integral into an improper integral.
    [Show full text]
  • The Binomial Series 16.3   
    The Binomial Series 16.3 Introduction In this Section we examine an important example of an infinite series, the binomial series: p(p − 1) p(p − 1)(p − 2) 1 + px + x2 + x3 + ··· 2! 3! We show that this series is only convergent if |x| < 1 and that in this case the series sums to the value (1 + x)p. As a special case of the binomial series we consider the situation when p is a positive integer n. In this case the infinite series reduces to a finite series and we obtain, by replacing x with b , the binomial theorem: a n(n − 1) (b + a)n = bn + nbn−1a + bn−2a2 + ··· + an. 2! Finally, we use the binomial series to obtain various polynomial expressions for (1 + x)p when x is ‘small’. ' • understand the factorial notation $ Prerequisites • have knowledge of the ratio test for convergence of infinite series. Before starting this Section you should ... • understand the use of inequalities '& • recognise and use the binomial series $% Learning Outcomes • state and use the binomial theorem On completion you should be able to ... • use the binomial series to obtain numerical approximations & % 26 HELM (2008): Workbook 16: Sequences and Series ® 1. The binomial series A very important infinite series which occurs often in applications and in algebra has the form: p(p − 1) p(p − 1)(p − 2) 1 + px + x2 + x3 + ··· 2! 3! in which p is a given number and x is a variable. By using the ratio test it can be shown that this series converges, irrespective of the value of p, as long as |x| < 1.
    [Show full text]
  • 1 Improper Integrals
    July 14, 2019 MAT136 { Week 6 Justin Ko 1 Improper Integrals In this section, we will introduce the notion of integrals over intervals of infinite length or integrals of functions with an infinite discontinuity. These definite integrals are called improper integrals, and are understood as the limits of the integrals we introduced in Week 1. Definition 1. We define two types of improper integrals: 1. Infinite Region: If f is continuous on [a; 1) or (−∞; b], the integral over an infinite domain is defined as the respective limit of integrals over finite intervals, Z 1 Z t Z b Z b f(x) dx = lim f(x) dx; f(x) dx = lim f(x) dx: a t!1 a −∞ t→−∞ t R 1 R a If both a f(x) dx < 1 and −∞ f(x) dx < 1, then Z 1 Z a Z 1 f(x) dx = f(x) dx + f(x) dx: −∞ −∞ a R 1 If one of the limits do not exist or is infinite, then −∞ f(x) dx diverges. 2. Infinite Discontinuity: If f is continuous on [a; b) or (a; b], the improper integral for a discon- tinuous function is defined as the respective limit of integrals over finite intervals, Z b Z t Z b Z b f(x) dx = lim f(x) dx f(x) dx = lim f(x) dx: a t!b− a a t!a+ t R c R b If f has a discontinuity at c 2 (a; b) and both a f(x) dx < 1 and c f(x) dx < 1, then Z b Z c Z b f(x) dx = f(x) dx + f(x) dx: a a c R b If one of the limits do not exist or is infinite, then a f(x) dx diverges.
    [Show full text]
  • Series: Convergence and Divergence Comparison Tests
    Series: Convergence and Divergence Here is a compilation of what we have done so far (up to the end of October) in terms of convergence and divergence. • Series that we know about: P∞ n Geometric Series: A geometric series is a series of the form n=0 ar . The series converges if |r| < 1 and 1 a1 diverges otherwise . If |r| < 1, the sum of the entire series is 1−r where a is the first term of the series and r is the common ratio. P∞ 1 2 p-Series Test: The series n=1 np converges if p1 and diverges otherwise . P∞ • Nth Term Test for Divergence: If limn→∞ an 6= 0, then the series n=1 an diverges. Note: If limn→∞ an = 0 we know nothing. It is possible that the series converges but it is possible that the series diverges. Comparison Tests: P∞ • Direct Comparison Test: If a series n=1 an has all positive terms, and all of its terms are eventually bigger than those in a series that is known to be divergent, then it is also divergent. The reverse is also true–if all the terms are eventually smaller than those of some convergent series, then the series is convergent. P P P That is, if an, bn and cn are all series with positive terms and an ≤ bn ≤ cn for all n sufficiently large, then P P if cn converges, then bn does as well P P if an diverges, then bn does as well. (This is a good test to use with rational functions.
    [Show full text]
  • Beyond Mere Convergence
    Beyond Mere Convergence James A. Sellers Department of Mathematics The Pennsylvania State University 107 Whitmore Laboratory University Park, PA 16802 [email protected] February 5, 2002 – REVISED Abstract In this article, I suggest that calculus instruction should include a wider variety of examples of convergent and divergent series than is usually demonstrated. In particular, a number of convergent series, P k3 such as 2k , are considered, and their exact values are found in a k≥1 straightforward manner. We explore and utilize a number of math- ematical topics, including manipulation of certain power series and recurrences. During my most recent spring break, I read William Dunham’s book Euler: The Master of Us All [3]. I was thoroughly intrigued by the material presented and am certainly glad I selected it as part of the week’s reading. Of special interest were Dunham’s comments on series manipulations and the power series identities developed by Euler and his contemporaries, for I had just completed teaching convergence and divergence of infinite series in my calculus class. In particular, Dunham [3, p. 47-48] presents Euler’s proof of the Basel Problem, a challenge from Jakob Bernoulli to determine the 1 P 1 exact value of the sum k2 . Euler was the first to solve this problem by k≥1 π2 proving that the sum equals 6 . I was reminded of my students’ interest in this result when I shared it with them just weeks before. I had already mentioned to them that exact values for relatively few families of convergent series could be determined.
    [Show full text]
  • Series Convergence Tests Math 121 Calculus II Spring 2015
    Series Convergence Tests Math 121 Calculus II Spring 2015 Some series converge, some diverge. Geometric series. We've already looked at these. We know when a geometric series 1 X converges and what it converges to. A geometric series arn converges when its ratio r lies n=0 a in the interval (−1; 1), and, when it does, it converges to the sum . 1 − r 1 X 1 The harmonic series. The standard harmonic series diverges to 1. Even though n n=1 1 1 its terms 1, 2 , 3 , . approach 0, the partial sums Sn approach infinity, so the series diverges. The main questions for a series. Question 1: given a series does it converge or diverge? Question 2: if it converges, what does it converge to? There are several tests that help with the first question and we'll look at those now. The term test. The only series that can converge are those whose terms approach 0. That 1 X is, if ak converges, then ak ! 0. k=1 Here's why. If the series converges, then the limit of the sequence of its partial sums n th X approaches the sum S, that is, Sn ! S where Sn is the n partial sum Sn = ak. Then k=1 lim an = lim (Sn − Sn−1) = lim Sn − lim Sn−1 = S − S = 0: n!1 n!1 n!1 n!1 The contrapositive of that statement gives a test which can tell us that some series diverge. Theorem 1 (The term test). If the terms of the series don't converge to 0, then the series diverges.
    [Show full text]
  • Calculus II Chapter 9 Review Name______
    Calculus II Chapter 9 Review Name___________ ________________________ SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. A recursion formula and the initial term(s) of a sequence are given. Write out the first five terms of the sequence. an 1) a = 1, a = 1) 1 n+1 n + 2 Find a formula for the nth term of the sequence. 1 1 1 1 2) 1, - , , - , 2) 4 9 16 25 Find the limit of the sequence if it converges; otherwise indicate divergence. 9 + 8n 3) a = 3) n 9 + 5n 3 4) (-1)n 1 - 4) n Determine if the sequence is bounded. Δn 5) 5) 4n Find a formula for the nth partial sum of the series and use it to find the series' sum if the series converges. 3 3 3 3 6) 3 - + - + ... + (-1)n-1 + ... 6) 8 64 512 8n-1 7 7 7 7 7) + + + ... + + ... 7) 1·3 2·4 3·5 n(n + 2) Find the sum of the series. Q n 9 8) _ (-1) 8) 4n n=0 Use partial fractions to find the sum of the series. 3 9) 9) _ (4n - 1)(4n + 3) 1 Determine if the series converges or diverges; if the series converges, find its sum. 3n+1 10) _ 10) 7n-1 1 cos nΔ 11) _ 11) 7n Find the values of x for which the geometric series converges. n n 12) _ -3 x 12) Find the sum of the geometric series for those x for which the series converges.
    [Show full text]
  • Calculus Terminology
    AP Calculus BC Calculus Terminology Absolute Convergence Asymptote Continued Sum Absolute Maximum Average Rate of Change Continuous Function Absolute Minimum Average Value of a Function Continuously Differentiable Function Absolutely Convergent Axis of Rotation Converge Acceleration Boundary Value Problem Converge Absolutely Alternating Series Bounded Function Converge Conditionally Alternating Series Remainder Bounded Sequence Convergence Tests Alternating Series Test Bounds of Integration Convergent Sequence Analytic Methods Calculus Convergent Series Annulus Cartesian Form Critical Number Antiderivative of a Function Cavalieri’s Principle Critical Point Approximation by Differentials Center of Mass Formula Critical Value Arc Length of a Curve Centroid Curly d Area below a Curve Chain Rule Curve Area between Curves Comparison Test Curve Sketching Area of an Ellipse Concave Cusp Area of a Parabolic Segment Concave Down Cylindrical Shell Method Area under a Curve Concave Up Decreasing Function Area Using Parametric Equations Conditional Convergence Definite Integral Area Using Polar Coordinates Constant Term Definite Integral Rules Degenerate Divergent Series Function Operations Del Operator e Fundamental Theorem of Calculus Deleted Neighborhood Ellipsoid GLB Derivative End Behavior Global Maximum Derivative of a Power Series Essential Discontinuity Global Minimum Derivative Rules Explicit Differentiation Golden Spiral Difference Quotient Explicit Function Graphic Methods Differentiable Exponential Decay Greatest Lower Bound Differential
    [Show full text]
  • Learning Goals: Power Series • Master the Art of Using the Ratio Test
    Learning Goals: Power Series • Master the art of using the ratio test to find the radius of convergence of a power series. • Learn how to find the series associated with the end points of the interval of convergence. 1 X xn • Become familiar with the series : n! n=0 xn • Become familiar with the limit lim : n!1 n! • Know the Power series representation of ex • Learn how to manipulate the power series representation of ex using tools previously mastered such as substitution, integration and differentiation. • Learn how to use power series to calculate limits. 1 Power Series, Stewart, Section 11.8 In this section, we will use the ratio test to determine the Radius of Convergence of a given power series. We will also determine the interval of convergence of the given power series if this is possible using the tests for convergence that we have already developed, namely, recognition as a geometric series, harmonic series or alternating harmonic series, telescoping series, the divergence test along with the fact that a series is absolutely convergent converges. The ratio and root tests will be inconclusive at the end points of the interval of convergence in the examples below since we use them to determine the radius of convergence. We may not be able to decide on convergence or divergence for some of the end points with our current tools. We will expand our methods for testing individual series for convergence in subsequent sections and tie up any loose ends as we proceed. We recall the definition of the Radius of Convergence and Interval of Convergence of a power series below.
    [Show full text]