DOCSLIB.ORG

Alternating Series; Absolute/Conditional Convergence

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Alternating Series; Absolute/Conditional Convergence Alternating Series; Absolute/Conditional Convergence SUGGESTED REFERENCE MATERIAL: As you work through the problems listed below, you should reference your lecture notes and the relevant chapters in a textbook/online resource. EXPECTED SKILLS: • Determine if an alternating series converges using the Alternating Series Test. • Analyze the absolute values of the terms of a series and determine if it converges. • Use any of the previously discussed convergence tests to determine if a series with negative terms converges absolutely, converges conditionally, or diverges. • Find a partial sum that approximates a convergent alternating series to some specified accuracy. PRACTICE PROBLEMS: For problems 1 { 3, show that the series converges by verifying that it satisfies the hypotheses of the Alternating Series Test, or show that the series does not satisfy the hypotheses of the Alternating Series Test. If the latter, then use some other test to determine if the series converges or diverges. 1 1 1 1 1 1 1. − + − + − + ::: 2 4 8 16 32 64 1 k X 1 This is the series (−1)k . 2 k=0 1 k ak+1 1 We have ak = 2 , so = < 1 for k ≥ 0. ak 2 +1 Thus the sequence fakgk=0 is decreasing; also, lim ak = 0. k!1 So by the Alternating Series Test the series converges. 1 [Of course this is just a geometric series with r = − 2 .] 1 X k2 2. (−1)k 2 + k3 k=2 k2 k(k3 − 4) We have a = , so a0 = − < 0 for k ≥ 2. k 2 + k3 k (2 + k3)2 +1 Thus the sequence fakgk=2 is decreasing; also, lim ak = 0. k!1 So by the Alternating Series Test the series converges. 1 1 X 1 + 2k 3. (−1)k+1 3k k=1 1 + 2k 2 lim = 6= 0, so the hypotheses of the Alternating Series Test are not satisfied. k!1 3k 3 1 + 2k Now lim (−1)k+1 DNE (half of the terms are approaching 2 , the k!1 3k 3 2 other half are approaching − 3 ). Since the limit is not zero, the series diverges by the Divergence Test. For problems 4 { 13, determine if the series converges absolutely, converges conditionally, or diverges. 1 1 1 1 1 1 1 1 1 1 4. + + − − − + + + − − ::: 2 4 8 16 32 64 128 256 512 1024 converges absolutely 1 X k2 5. (−1)k+1 2 + k3 k=1 converges conditionally 1 X 1 6. (−1)k 4k2 + 9 k=1 converges absolutely 1 X 1 7. (−1)k arcsin k k=1 converges conditionally; Detailed Solution: Here 1 X (−1)k+1 8. k! k=1 converges absolutely 1 X 9. cos(kπ)ke−k k=1 converges absolutely; Detailed Solution: Here 2 1 X 10. (−1)ke(1=k) k=1 diverges 1 p X k 11. (−1)k k − 2 k=3 converges conditionally 1 X sin k 12. k4 + 4k k=1 converges absolutely 1 X (2k + 1)! 13. (−1)k 200k k=1 diverges 1 X h π i ln k 14. sin (2k + 1) 2 k k=3 converges conditionally For problems 15 { 17, the series converge to some sum S. Find the smallest value of n so that the n-th partial sum sn will guarantee the approximation of S to the required accuracy. 1 X (−1)k+1 15. ; jS − s j < 0:001 k! n k=1 n = 6; Detailed Solution: Here 1 X (−1)k p 16. 4 ; jS − snj < 0:2 k=1 k + 7 n = 617 1 X (−1)k+1 17. ; jS − s j < 0:0005 2kk n k=1 n = 7 3 th P1 (−1)k+1 18. Is the 300 partial sum of k=1 2kk an overestimate or an underestimate of the sum of the series? Justify your answer. Let S be the sum of the series. Now the first term of the series is positive, the second is negative, the third positive, the fourth negative, and so on. Thus the sequence of odd partial sums is decreasing (and converges to S) while the sequence of even partial sums is increasing (and converges to S). So every odd partial sum is an overestimate of S and every even partial sum is an underestimate of S. Therefore the 300th partial sum of the series is an underestimate of S. 4.
Load more

Recommended publications
  • Riemann Sums Fundamental Theorem of Calculus Indefinite
    Riemann Sums Partition P = {x0, x1, . , xn} of an interval [a, b]. ck ∈ [xk−1, xk] Pn R(f, P, a, b) = k=1 f(ck)∆xk As the widths ∆xk of the subintervals approach 0, the Riemann Sums hopefully approach a limit, the integral of f from a to b, written R b a f(x) dx. Fundamental Theorem of Calculus Theorem 1 (FTC-Part I). If f is continuous on [a, b], then F (x) = R x 0 a f(t) dt is defined on [a, b] and F (x) = f(x). Theorem 2 (FTC-Part II). If f is continuous on [a, b] and F (x) = R R b b f(x) dx on [a, b], then a f(x) dx = F (x) a = F (b) − F (a). Indefinite Integrals Indefinite Integral: R f(x) dx = F (x) if and only if F 0(x) = f(x). In other words, the terms indefinite integral and antiderivative are synonymous. Every differentiation formula yields an integration formula. Substitution Rule For Indefinite Integrals: If u = g(x), then R f(g(x))g0(x) dx = R f(u) du. For Definite Integrals: R b 0 R g(b) If u = g(x), then a f(g(x))g (x) dx = g(a) f( u) du. Steps in Mechanically Applying the Substitution Rule Note: The variables do not have to be called x and u. (1) Choose a substitution u = g(x). du (2) Calculate = g0(x). dx du (3) Treat as if it were a fraction, a quotient of differentials, and dx du solve for dx, obtaining dx = .
    [Show full text]
  • The Radius of Convergence of the Lam\'{E} Function
    The radius of convergence of the Lame´ function Yoon-Seok Choun Department of Physics, Hanyang University, Seoul, 133-791, South Korea Abstract We consider the radius of convergence of a Lame´ function, and we show why Poincare-Perron´ theorem is not applicable to the Lame´ equation. Keywords: Lame´ equation; 3-term recursive relation; Poincare-Perron´ theorem 2000 MSC: 33E05, 33E10, 34A25, 34A30 1. Introduction A sphere is a geometric perfect shape, a collection of points that are all at the same distance from the center in the three-dimensional space. In contrast, the ellipsoid is an imperfect shape, and all of the plane sections are elliptical or circular surfaces. The set of points is no longer the same distance from the center of the ellipsoid. As we all know, the nature is nonlinear and geometrically imperfect. For simplicity, we usually linearize such systems to take a step toward the future with good numerical approximation. Indeed, many geometric spherical objects are not perfectly spherical in nature. The shape of those objects are better interpreted by the ellipsoid because they rotates themselves. For instance, ellipsoidal harmonics are shown in the calculation of gravitational potential [11]. But generally spherical harmonics are preferable to mathematically complex ellipsoid harmonics. Lame´ functions (ellipsoidal harmonic fuctions) [7] are applicable to diverse areas such as boundary value problems in ellipsoidal geometry, Schrodinger¨ equations for various periodic and anharmonic potentials, the theory of Bose-Einstein condensates, group theory to quantum mechanical bound states and band structures of the Lame´ Hamiltonian, etc. As we apply a power series into the Lame´ equation in the algebraic form or Weierstrass’s form, the 3-term recurrence relation starts to arise and currently, there are no general analytic solutions in closed forms for the 3-term recursive relation of a series solution because of their mathematical complexity [1, 5, 13].
    [Show full text]
  • 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]
  • Absolute and Conditional Convergence, and Talk a Little Bit About the Mathematical Constant E
    MATH 8, SECTION 1, WEEK 3 - RECITATION NOTES TA: PADRAIC BARTLETT Abstract. These are the notes from Friday, Oct. 15th's lecture. In this talk, we study absolute and conditional convergence, and talk a little bit about the mathematical constant e. 1. Random Question Question 1.1. Suppose that fnkg is the sequence consisting of all natural numbers that don't have a 9 anywhere in their digits. Does the corresponding series 1 X 1 nk k=1 converge? 2. Absolute and Conditional Convergence: Definitions and Theorems For review's sake, we repeat the definitions of absolute and conditional conver- gence here: P1 P1 Definition 2.1. A series n=1 an converges absolutely iff the series n=1 janj P1 converges; it converges conditionally iff the series n=1 an converges but the P1 series of absolute values n=1 janj diverges. P1 (−1)n+1 Example 2.2. The alternating harmonic series n=1 n converges condition- ally. This follows from our work in earlier classes, where we proved that the series itself converged (by looking at partial sums;) conversely, the absolute values of this series is just the harmonic series, which we know to diverge. As we saw in class last time, most of our theorems aren't set up to deal with series whose terms alternate in sign, or even that have terms that occasionally switch sign. As a result, we developed the following pair of theorems to help us manipulate series with positive and negative terms: 1 Theorem 2.3. (Alternating Series Test / Leibniz's Theorem) If fangn=0 is a se- quence of positive numbers that monotonically decreases to 0, the series 1 X n (−1) an n=1 converges.
    [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]
  • Dirichlet’S Test
    4. CONDITIONALLY CONVERGENT SERIES 23 4. Conditionally convergent series Here basic tests capable of detecting conditional convergence are studied. 4.1. Dirichlet’s test. Theorem 4.1. (Dirichlet’s test) n Let a complex sequence {An}, An = k=1 ak, be bounded, and the real sequence {bn} is decreasing monotonically, b1 ≥ b2 ≥ b3 ≥··· , so that P limn→∞ bn = 0. Then the series anbn converges. A proof is based on the identity:P m m m m−1 anbn = (An − An−1)bn = Anbn − Anbn+1 n=k n=k n=k n=k−1 X X X X m−1 = An(bn − bn+1)+ Akbk − Am−1bm n=k X Since {An} is bounded, there is a number M such that |An|≤ M for all n. Therefore it follows from the above identity and non-negativity and monotonicity of bn that m m−1 anbn ≤ An(bn − bn+1) + |Akbk| + |Am−1bm| n=k n=k X X m−1 ≤ M (bn − bn+1)+ bk + bm n=k ! X ≤ 2Mbk By the hypothesis, bk → 0 as k → ∞. Therefore the right side of the above inequality can be made arbitrary small for all sufficiently large k and m ≥ k. By the Cauchy criterion the series in question converges. This completes the proof. Example. By the root test, the series ∞ zn n n=1 X converges absolutely in the disk |z| < 1 in the complex plane. Let us investigate the convergence on the boundary of the disk. Put z = eiθ 24 1. THE THEORY OF CONVERGENCE ikθ so that ak = e and bn = 1/n → 0 monotonically as n →∞.
    [Show full text]
  • Chapter 3: Infinite Series
    Chapter 3: Infinite series Introduction Definition: Let (an : n ∈ N) be a sequence. For each n ∈ N, we define the nth partial sum of (an : n ∈ N) to be: sn := a1 + a2 + . an. By the series generated by (an : n ∈ N) we mean the sequence (sn : n ∈ N) of partial sums of (an : n ∈ N) (ie: a series is a sequence). We say that the series (sn : n ∈ N) generated by a sequence (an : n ∈ N) is convergent if the sequence (sn : n ∈ N) of partial sums converges. If the series (sn : n ∈ N) is convergent then we call its limit the sum of the series and we write: P∞ lim sn = an. In detail: n→∞ n=1 s1 = a1 s2 = a1 + a + 2 = s1 + a2 s3 = a + 1 + a + 2 + a3 = s2 + a3 ... = ... sn = a + 1 + ... + an = sn−1 + an. Definition: A series that is not convergent is called divergent, ie: each series is ei- P∞ ther convergent or divergent. It has become traditional to use the notation n=1 an to represent both the series (sn : n ∈ N) generated by the sequence (an : n ∈ N) and the sum limn→∞ sn. However, this ambiguity in the notation should not lead to any confu- sion, provided that it is always made clear that the convergence of the series must be established. Important: Although traditionally series have been (and still are) represented by expres- P∞ sions such as n=1 an they are, technically, sequences of partial sums. So if somebody comes up to you in a supermarket and asks you what a series is - you answer “it is a P∞ sequence of partial sums” not “it is an expression of the form: n=1 an.” n−1 Example: (1) For 0 < r < ∞, we may define the sequence (an : n ∈ N) by, an := r for each n ∈ N.
    [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]
  • Chapter 11 Outline Math 236 Summer 2002
    Chapter 11 Outline Math 236 Summer 2002 11.1 Infinite Series Sigma notation: Be able to go from a written-out sum to sigma notation and back. • Know how to change indicies and algebraically manipulate a sum in sigma notation. Infinite series as sequences of partial sums: What does it mean to say that an • infinite series converges? See Definition 11.1.1 which describes a series as the limit of its sequence of partial sums. Do not confuse the series with the corresponding sequence! Telescoping series: Sometimes the terms of a series can be rewritten using partial • fractions so that you can find a formula for the nth partial sum for the series. Then you can take the limit of this sequence of partial sums to get the sum of the series (if it exists). Be able to do this and be able to explain what you are doing (including notation) while you do it. Not all series are \telescoping" like this. The geometric series: What is the general form of a geometric series? When • does it converge, and to what? Notice that the formula for the sum of a convergent geometric series only works if you start from k = 0. Be able to prove (in a particular 1 case like x = 4 or in general) the formula for the sum of a convergent geometric series. The proof involves finding a formula for the nth partial sum of the series (start by computing sn xsn) and then taking the limit of that formula to get the sum of the series.
    [Show full text]