DOCSLIB.ORG

11.3 the Integral Test, 11.4 Comparison Tests, 11.5 Alternating Series

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
11.3 the Integral Test, 11.4 Comparison Tests, 11.5 Alternating Series Math 115 Calculus II 11.3 The Integral Test, 11.4 Comparison Tests, 11.5 Alternating Series Mikey Chow March 23, 2021 Table of Contents Integral Test Basic and Limit Comparison Tests Alternating Series p-series 1 ( X 1 converges for = np n=1 diverges for 1 X Z 1 an converges if and only if f (x)dx converges. n=1 1 The Integral Test Theorem (The Integral Test) Suppose the sequence an is given by f (n) where f is a continuous, positive, decreasing sequence on [1; 1). Then Example Determine the convergence of the series of the sequence e−1; 2e−2; 3e−3;::: Table of Contents Integral Test Basic and Limit Comparison Tests Alternating Series 1 1 X X 0 ≤ an ≤ bn n=1 n=1 P1 P1 n=1 an n=1 bn P1 P1 n=1 bn n=1 an The (Basic) Comparison Test Theorem (The (Basic) Comaprison Test) Suppose fang and fbng are two sequences such that 0 ≤ an ≤ bn for all n. Then In particular, If converges, then converges also. If diverges, then diverges also. (What does this remind you of?) Remark We can still apply the basic comparison test if we know that a lim n = L n!1 bn where L is a finite number and L 6= 0. 1 1 X X an converges if and only if bn converges. n=1 n=1 The Limit Comparison Test Theorem (The Limit Comaprison Test) Suppose fang and fbng are two sequences such that Then Example P1 k sin2 k Determine the convergence of k=2 k3−1 Example P1 logpn Determine the convergence of n=1 n n Group Problem Solving Instructions: For each pair of series below, determine the convergence of one of them without using a comparison test. Then use a comparison test with these two series to determine the convergence of the other. 1 1 X 1 X 1 2n 2n + 1 n=1 n=1 1 1 X 1 X 1 2n 2n − 1 n=1 n=1 1 1 X 1 X n2 + 1 n2 n4 − n2 + 1 n=1 n=1 1 1 p X 1 X i 2 + 1(i 3 + 3) n i 5 − i 4 + i 3 n=1 i=1 Group Problem Solving Instructions: For each pair of series below, determine the convergence of one of them without using a comparison test. Then use a comparison test with these two series to determine the convergence of the other. 1 1 X sin n + 2 X 1 n2 n2 n=1 n=1 1 1 X n sin2 n + cos n + 3 X 1 n4 + 5 − 4 cos n n3 n=1 n=1 1 1 X X 12e−n e−n log(n + 1) n=1 n=1 1 1 X e10−n X 1 n5 N5 n=1 N=1 Rough Work Rough Work Rough Work Table of Contents Integral Test Basic and Limit Comparison Tests Alternating Series stuff Alternating Sequences An alternating sequence is a sequence (an) of the form where Alternating Series Test Example P1 (−1)n+3 Determine the convergence of n=1 n . Example P1 (−1)n+53n2 Determine the convergence of n=1 4n2+1 ..
Load more

Recommended publications
  • Calculus and Differential Equations II
    Calculus and Differential Equations II MATH 250 B Sequences and series Sequences and series Calculus and Differential Equations II Sequences A sequence is an infinite list of numbers, s1; s2;:::; sn;::: , indexed by integers. 1n Example 1: Find the first five terms of s = (−1)n , n 3 n ≥ 1. Example 2: Find a formula for sn, n ≥ 1, given that its first five terms are 0; 2; 6; 14; 30. Some sequences are defined recursively. For instance, sn = 2 sn−1 + 3, n > 1, with s1 = 1. If lim sn = L, where L is a number, we say that the sequence n!1 (sn) converges to L. If such a limit does not exist or if L = ±∞, one says that the sequence diverges. Sequences and series Calculus and Differential Equations II Sequences (continued) 2n Example 3: Does the sequence converge? 5n 1 Yes 2 No n 5 Example 4: Does the sequence + converge? 2 n 1 Yes 2 No sin(2n) Example 5: Does the sequence converge? n Remarks: 1 A convergent sequence is bounded, i.e. one can find two numbers M and N such that M < sn < N, for all n's. 2 If a sequence is bounded and monotone, then it converges. Sequences and series Calculus and Differential Equations II Series A series is a pair of sequences, (Sn) and (un) such that n X Sn = uk : k=1 A geometric series is of the form 2 3 n−1 k−1 Sn = a + ax + ax + ax + ··· + ax ; uk = ax 1 − xn One can show that if x 6= 1, S = a .
    [Show full text]
  • 1 Mean Value Theorem 1 1.1 Applications of the Mean Value Theorem
    Seunghee Ye Ma 8: Week 5 Oct 28 Week 5 Summary In Section 1, we go over the Mean Value Theorem and its applications. In Section 2, we will recap what we have covered so far this term. Topics Page 1 Mean Value Theorem 1 1.1 Applications of the Mean Value Theorem . .1 2 Midterm Review 5 2.1 Proof Techniques . .5 2.2 Sequences . .6 2.3 Series . .7 2.4 Continuity and Differentiability of Functions . .9 1 Mean Value Theorem The Mean Value Theorem is the following result: Theorem 1.1 (Mean Value Theorem). Let f be a continuous function on [a; b], which is differentiable on (a; b). Then, there exists some value c 2 (a; b) such that f(b) − f(a) f 0(c) = b − a Intuitively, the Mean Value Theorem is quite trivial. Say we want to drive to San Francisco, which is 380 miles from Caltech according to Google Map. If we start driving at 8am and arrive at 12pm, we know that we were driving over the speed limit at least once during the drive. This is exactly what the Mean Value Theorem tells us. Since the distance travelled is a continuous function of time, we know that there is a point in time when our speed was ≥ 380=4 >>> speed limit. As we can see from this example, the Mean Value Theorem is usually not a tough theorem to understand. The tricky thing is realizing when you should try to use it. Roughly speaking, we use the Mean Value Theorem when we want to turn the information about a function into information about its derivative, or vice-versa.
    [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]
  • What Is on Today 1 Alternating Series
    MA 124 (Calculus II) Lecture 17: March 28, 2019 Section A3 Professor Jennifer Balakrishnan, [email protected] What is on today 1 Alternating series1 1 Alternating series Briggs-Cochran-Gillett x8:6 pp. 649 - 656 We begin by reviewing the Alternating Series Test: P k+1 Theorem 1 (Alternating Series Test). The alternating series (−1) ak converges if 1. the terms of the series are nonincreasing in magnitude (0 < ak+1 ≤ ak, for k greater than some index N) and 2. limk!1 ak = 0. For series of positive terms, limk!1 ak = 0 does NOT imply convergence. For alter- nating series with nonincreasing terms, limk!1 ak = 0 DOES imply convergence. Example 2 (x8.6 Ex 16, 20, 24). Determine whether the following series converge. P1 (−1)k 1. k=0 k2+10 P1 1 k 2. k=0 − 5 P1 (−1)k 3. k=2 k ln2 k 1 MA 124 (Calculus II) Lecture 17: March 28, 2019 Section A3 Recall that if a series converges to a value S, then the remainder is Rn = S − Sn, where Sn is the sum of the first n terms of the series. An upper bound on the magnitude of the remainder (the absolute error) in an alternating series arises form the following observation: when the terms are nonincreasing in magnitude, the value of the series is always trapped between successive terms of the sequence of partial sums. Thus we have jRnj = jS − Snj ≤ jSn+1 − Snj = an+1: This justifies the following theorem: P1 k+1 Theorem 3 (Remainder in Alternating Series).
    [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]
  • 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]
  • Math 6B Notes Written by Victoria Kala [email protected] SH 6432U Office Hours: R 12:30 − 1:30Pm Last Updated 5/31/2016
    Math 6B Notes Written by Victoria Kala [email protected] SH 6432u Office Hours: R 12:30 − 1:30pm Last updated 5/31/2016 Green's Theorem We say that a closed curve is positively oriented if it is oriented counterclockwise. It is negatively oriented if it is oriented clockwise. Theorem (Green's Theorem). Let C be a positively oriented, piecewise smooth, simple closed curve in the plane and let D be the region bounded by C. If F(x; y) = (P (x; y);Q(x; y)) with P and Q having continuous partial derivatives on an open region that contains D, then Z Z ZZ @Q @P F · ds = P dx + Qdy = − dA: C C D @x @y In other words, Green's Theorem allows us to change from a complicated line integral over a curve C to a less complicated double integral over the region bounded by C. The Operator r We define the vector differential operator r (called \del") as @ @ @ r = i + j + k: @x @y @z The gradient of a function f is given by @f @f @f rf = i + j + k: @x @y @z The curl of a function F = P i + Qj + Rk is defined as curl F = r × F: Written out more explicitly: i j k @ @ @ curl F = r × F = @x @y @z PQR @ @ @ @ @ @ = @y @z i − @x @z j + @x @y k QR PR PQ @R @Q @P @R @Q @P = − i + − j + − k @y @z @z @x @x @y 1 The divergence of function F = P i + Qj + Rk is defined as @P @Q @R div F = r · F = + + : @x @y @z Notice that the curl of a function is a vector and the divergence of a function is a scalar (just a number).
    [Show full text]
  • Calculus Online Textbook Chapter 10
    Contents CHAPTER 9 Polar Coordinates and Complex Numbers 9.1 Polar Coordinates 348 9.2 Polar Equations and Graphs 351 9.3 Slope, Length, and Area for Polar Curves 356 9.4 Complex Numbers 360 CHAPTER 10 Infinite Series 10.1 The Geometric Series 10.2 Convergence Tests: Positive Series 10.3 Convergence Tests: All Series 10.4 The Taylor Series for ex, sin x, and cos x 10.5 Power Series CHAPTER 11 Vectors and Matrices 11.1 Vectors and Dot Products 11.2 Planes and Projections 11.3 Cross Products and Determinants 11.4 Matrices and Linear Equations 11.5 Linear Algebra in Three Dimensions CHAPTER 12 Motion along a Curve 12.1 The Position Vector 446 12.2 Plane Motion: Projectiles and Cycloids 453 12.3 Tangent Vector and Normal Vector 459 12.4 Polar Coordinates and Planetary Motion 464 CHAPTER 13 Partial Derivatives 13.1 Surfaces and Level Curves 472 13.2 Partial Derivatives 475 13.3 Tangent Planes and Linear Approximations 480 13.4 Directional Derivatives and Gradients 490 13.5 The Chain Rule 497 13.6 Maxima, Minima, and Saddle Points 504 13.7 Constraints and Lagrange Multipliers 514 CHAPTER Infinite Series Infinite series can be a pleasure (sometimes). They throw a beautiful light on sin x and cos x. They give famous numbers like n and e. Usually they produce totally unknown functions-which might be good. But on the painful side is the fact that an infinite series has infinitely many terms. It is not easy to know the sum of those terms.
    [Show full text]
  • MATH 409 Advanced Calculus I Lecture 16: Tests for Convergence (Continued)
    MATH 409 Advanced Calculus I Lecture 16: Tests for convergence (continued). Alternating series. Tests for convergence [Condensation Test] Suppose an is a nonincreasing { } ∞ sequence of positive numbers. Then the series n=1 an ∞ k converges if and only if the “condensed” series Pk=1 2 a2k converges. P [Ratio Test] Let an be a sequence of positive numbers. { } Suppose that the limit r = lim an+1/an exists (finite or n→∞ ∞ infinite). (i) If r < 1, then the series n=1 an converges. ∞ (ii) If r > 1, then the series n=1 an Pdiverges. P [Root Test] Let an be a sequence of positive numbers and { } n r = lim sup √an. n→∞ ∞ (i) If r < 1, then the series n=1 an converges. ∞ (ii) If r > 1, then the series P n=1 an diverges. P Refined ratio tests [Raabe’s Test] Let an be a sequence of positive numbers. { } a Suppose that the limit L = lim n n 1 exists (finite →∞ n an+1 − ∞ or infinite). Then the series n=1 an converges if L > 1, and diverges if L < 1. P Remark. If < L < then an+1/an 1 as n . −∞ ∞ → →∞ [Gauss’s Test] Let an be a sequence of positive numbers. { } an L γn Suppose that =1+ + 1+ε , where L is a constant, an+1 n n ε> 0, and γn is a bounded sequence. Then the series ∞ { } an converges if L > 1, and diverges if L 1. n=1 ≤ P Remark. Under the assumptions of the Gauss Test, n(an/an+1 1) L as n .
    [Show full text]
  • ∑ (E)-N N=1 This Is a Geometric Series with First Term 1/E and Constant Ratio 1/E (Whose Magnitude Is Less Than 1), So the Series 1 1 Converges to E =
    Lab 2 Convergence/Divergence Testing - Some Guidelines Sample Solutions 1. If a series is expressed using sigma notation, write out the first few terms to get a feel for what the series looks like. Try: ∞ ∑ (-1)n (n/[n+1]) = -1/2 + 2/3 – 3/4 + 4/5 – 5/6 + … n=1 2. Is the series a geometric series? If so, it’s easy to see whether it converges or diverges just by determining the magnitude of the constant ratio. If it converges, you can calculate the sum by a formula, page 408. Try: ∞ a. ∑ (e)-n n=1 This is a geometric series with first term 1/e and constant ratio 1/e (whose magnitude is less than 1), so the series 1 1 converges to e = . 1 e 1 1− − e ∞ b. ∑ (1/2)-n n=1 This is a geometric series with first term 2 and constant ratio 2 (whose magnitude is greater than 1), so the series diverges. 3. Is the limit of the nth term of the series equal to zero? If not, the series diverges (page 413, Thm 9.2 #3). Try: ∞ ∑ [n+2]/ [5n+17] n=1 2 1 n + 2 + 1 The limit of the nth term, [n+2]/ [5n+17], is lim = lim n = . n→∞ 5n 17 n→∞ 17 5 + 5 + n Since 1/5 is not equal to 0, the series diverges. 4. Is each term of the series positive? If so, a comparison test (page 417) may be used. Try: ∞ a. ∑ 1 / [2n + 1] n=1 The nth term, 1 / [2n + 1], is less than 1/2n , the nth term of a ∞ convergent geometric series, so the series ∑ 1/[2n+1] n=1 converges.
    [Show full text]
  • Theorem the Geometric Series ∑ N=0 Arn = ∑ Ar + Ar 2 + ··· Is Convergent
    P∞ n Pn n−1 P Theorem The geometric series n=0 ar = i=1 ar = a + Definition A series an is called absolutely convergent if the 2 P∞ n P ar + ar + ··· is convergent if |r| < 1 and its sum is n=0 ar = series of absolute values |an| is convergent. a P 1−r |r| < 1. If |r| ≥ 1, the geometric series is divergent. Definition A series an is called conditionally convergent if Example Find the exact value of the repeating decimal 0.156 = it is convergent but not absolutely convergent. 0.156156156 ... as a rational number. Theorem If a series converges absolutely, then is also converges. Solution Note that The Ratio Test. 156 156 1 156 1 2 an+1 P∞ 0.156 = 1000 + 1000 · 1000 + 1000 · 1000 + ··· (i) If limn→∞ = L < 1, then the series an is abso- an n=1 P∞ 156 1 n 156 = n=0 1000 1000 . This is a geometric series with a = 1000 lutely convergent (and therefore convergent). and r = 1 and has the value: 1000 an+1 an+1 156 156 (ii) If limn→∞ = L > 1 or limn→∞ = ∞, then the 1000 1000 156 an an 1 = 999 = 999 . ∞ (1− ) 1000 P 1000 series n=1 an is divergent. The Test For Divergence. If limn→∞ an does not exist or if P∞ an+1 limn→∞ an 6= 0, then the series an is divergent. (iii) If limn→∞ = 1, the Ratio Test is inconclusive. n=1 an The Integral Test. Suppose f is a continuous, positive, de- The Root Test. creasing function on [1, ∞) and let an = f(n).
    [Show full text]
  • 10.6 Alternating Series Contemporary Calculus 1
    10.6 Alternating Series Contemporary Calculus 1 10.6 ALTERNATING SERIES In the last two sections we considered tests for the convergence of series whose terms were all positive. In this section we examine series whose terms change signs in a special way: they alternate between positive and negative. And we present a very easy–to–use test to determine if these alternating series converge. An alternating series is a series whose terms alternate between positive and negative. For example, the following are alternating series: ∞ 1 1 1 1 k+1 1 k+1 1 (1) 1 – 2 + 3 – 4 + 5 – . + (–1) k + . = ∑ (–1) k (alternating harmonic series) k=1 ∞ 1 2 3 4 5 k k k k (2) – 3 + 4 – 5 + 6 – 7 + . + (–1) k+2 + . = ∑ (–1) k+2 k=1 1 1 1 1 1 1 ∞ 1 (3) – + – + + – ... + (–1)k + . = (–1) k . 3 5 7 9 11 2k+1 ∑ 2k+1 k=1 Figures 1, 2 and 3 show graphs and tables of values of several partial sums sn for each of these series. As we move from left to right in each graph (as n increases), the partial sums alternately get larger and smaller, a common pattern for the partial sums of alternating series. The same pattern appears in the tables. n s n sn n n k k n s = (–1) 1 –0.3333 k+1 1 1.0 n ! k+2 s = (–1) 1 k=1 2 0.1667 1 n ! k 2 0.5 k=1 3 – 0.4333 3 0.8333 0.3 0.693 4 0.5833 4 0.2333 5 0.7833 5 –0.4810 6 0.6166 1 2 3 4 5 10 15 6 0.2690 7 0.7595 7 – 0.5087 8 0.2913 1 2 3 4 5 10 15 8 0.6345 –0.6 9 0.7456 9 – 0.5269 10 0.6456 10 0.3064 Fig.
    [Show full text]