DOCSLIB.ORG

Sequences and Series in the AMC and AIME

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Sequences and Series in the AMC and AIME AMC/AIME/USA(J)MO Handout Sequences and Series in the AMC and AIME Author: For: freeman66 AoPS nikenissan Date: February 1, 2021 n n 1 A proof without words of the sum of the first n consecutive integers. \Everything comes to you in perfect time, space and sequence." { Louise Hay freeman66 and nikenissan (February 1, 2021) Sequences and Series in the AMC and AIME Contents 0 Acknowledgements 3 1 Introduction 4 2 Basics of Sequences and Series4 2.1 Definitions.................................................. 4 2.2 Summation Notation............................................ 4 3 Arithmetic Sequences and Series5 4 Geometric Sequences and Series8 5 Arithmetico-Geometric Sequence 11 6 Telescoping Series 12 7 Recursive Sequences 14 8 Problems 15 A Appendix A: List of Theorems and Definitions 18 2 freeman66 and nikenissan (February 1, 2021) Sequences and Series in the AMC and AIME §0 Acknowledgements This was made for the Art of Problem Solving Community out there! I would like to thank Evan Chen for his evan.sty code. In addition, all problems in the handout were likely from the AoPS Wiki. Art of Problem Solving Community Evan Chen's Personal Sty File freeman66's Website nikenissan's Website And Evan says he would like this here for evan.sty: Boost Software License - Version 1.0 - August 17th, 2003 Copyright (c) 2020 Evan Chen [evan at evanchen.cc] https://web.evanchen.cc/ || github.com/vEnhance He also helped with the hint formatting. Evan is a LATEXgod! And finally, please do not make any copies of this document without referencing this original one. At least cite us when you are using this document. 3 freeman66 and nikenissan (February 1, 2021) Sequences and Series in the AMC and AIME §1 Introduction Sequences and series are one of the most prevalent topics in the AMC and AIME. §2 Basics of Sequences and Series §2.1 Definitions Let's start by introducing a few ideas. Definition 2.1 (Sequence) | A sequence is an ordered list of numbers. Definition 2.2 (Partial Sum) | A partial sum is the sum of a portion of the sequence. Partial sums are sometimes called finite series. Definition 2.3 (Series) | A series is the sum of the elements in a given sequence. §2.2 Summation Notation To keep things simple we introduce summation notation. Definition 2.4 (Summation Notation) | In a summation, there is an index going from a starting point to an ending point. The given element is tested through all values from the start to the end (inclusive), and then all values of the element are summed. Summation notation is also sometimes called sigma notation. An example would be b X f(a) = f(1) + f(2) + ··· + f(b − 1) + f(b): a=1 Example 2.5 Write 1 + 2 + ::: + n in summation notation. Solution. Denote x as the index of the summation. Notice the starting point of the summation is 1 and the ending point of the summation is n: Our element is x in this scenario, thus our summation can be written as n X x : x=1 Theorem 2.6 (Distributive Property of Summations) For all real constants k, X X k · f(x) = k f(x): 4 freeman66 and nikenissan (February 1, 2021) Sequences and Series in the AMC and AIME Example 2.7 Evaluate 5 X 3 · x! : (x − 1)! x=1 Solution. By the Distributive Property of Summations, 5 5 X 3 · x! X x! = 3 : (x − 1)! (x − 1)! x=1 x=1 x! Notice (x−1)! = x: Thus, 5 5 X x! X 3 = 3 x: (x − 1)! x=1 n=1 P5 x=1 x = 1 + 2 + 3 + 4 + 5 = 15: Therefore, 5 X 3 x = 3 · 15 = 45 : x=1 Theorem 2.8 (Commutative Property of Summations) For functions f(x) and g(x), X X X (f(x) ± g(x)) = f(x) ± g(x): Summations simplify an expression a lot, and make it easier and more convenient to work with. Exercise 2.9. Simplify X X (x + 1)2 − x2: Exercise 2.10. Compute the sum 1 X k2 : 2k k=1 §3 Arithmetic Sequences and Series Definition 3.1 (Arithmetic Sequence) | An arithmetic sequence is a sequence of numbers in which each term is given by adding a fixed value to the previous term. For example, −2; 1; 4; 7; 10;::: is an arithmetic sequence because each term is three more than the previous term. In this case, 3 is called the common difference of the sequence. More formally, an arithmetic sequence an is defined recursively by a first term a1 and an = an−1 + d for n ≥ 2, where d is the common difference. Explicitly: 5 freeman66 and nikenissan (February 1, 2021) Sequences and Series in the AMC and AIME Theorem 3.2 (Terms of an Arithmetic Sequence) The nth term in an arithmetic sequence is described an = a1 + d(n − 1); th where an is the n term, a1 is the first term, and d is the difference between consecutive terms. Theorem 3.3 (Sum of an Arithmetic Sequence) The sum of the first n terms of an arithmetic sequences is n n sn = (a1 + an) = (2a1 + (n − 1)d): 2 2 Proof. We can prove this with induction, which will not be taught or explained in this handout. The following formulas are a direct consequence of the aforementioned theorems. Corollary 3.4 (Sum of First n Positive Integers) For all positive integers n, n(n + 1) 1 + 2 + 3 + ::: + n = : 2 Corollary 3.5 (Sum of First n Even Integers) For all positive integers n, 2 + 4 + ::: + 2n = n(n + 1): Corollary 3.6 (Sum of First n Odd Integers) For all positive integers n, 1 + 3 + 5 + ::: + (2n − 1) = n2: Example 3.7 (AIME I 2005/2) For each positive integer k; let Sk denote the increasing arithmetic sequence of integers whose first term is 1 and whose common difference is k: For example, S3 is the sequence 1; 4; 7; 10;:::: For how many values of k does Sk contain the term 2005? Solution. Suppose that the nth term of the sequence Sk is 2005. Then 1 + (n − 1)k = 2005 =) k(n − 1) = 2004 = 22 · 3 · 167. Because k when multiplied by an integer is equal to 2004; in order words we can say that k is a divisor/divides 2004: By the formula for the number of divisors, there are (2 + 1)(1 + 1)(1 + 1) = 12 divisors of 22 · 31 · 1671. 6 freeman66 and nikenissan (February 1, 2021) Sequences and Series in the AMC and AIME Example 3.8 (AIME II 2011/3) The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle. Solution. The average angle in an 18-gon is 160◦. In this arrangement, the average is equivalent to the middle, so the center two terms of the succession average to 160◦. In this manner for some certain (the grouping is expanding and hence non-consistent) whole number d, the center two terms are (160 − d)◦ and (160 + d)◦. Since the progression is 2d, the last term of the arrangement is (160 + 17d)◦, which is under 180◦, since the polygon is raised. This gives 17d < 20, so the lone appropriate positive whole number d is 1. The initial term is then (160 − 17)◦ = 143 : Example 3.9 (AIME I 2012/2) The terms of an arithmetic sequence add to 715. The first term of the sequence is increased by 1, the second term is increased by 3, the third term is increased by 5, and in general, the kth term is increased by the kth odd positive integer. The terms of the new sequence add to 836. Find the sum of the first, last, and middle terms of the original sequence. n X Solution. If the sum of the original sequence is ai then the sum of the new sequence can be expressed as i=1 n n X 2 X 2 ai + (2i − 1) = n + ai: Therefore, 836 = n + 715 =) n = 11: Now the middle term of the original i=1 i=1 715 sequence is simply the average of all the terms, or 11 = 65; and the first and last terms average to this middle term, so the desired sum is simply three times the middle term, or 195 : Example 3.10 (AMC 8 2015/18) An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. Each row and each column in this 5 × 5 array is an arithmetic sequence with five terms. What is the value of X? 1 25 X 17 81 (A) 21 (B) 31 (C) 36 (D) 40 (E) 42 25+1 Solution. The middle term of the first row is 2 = 13 since the middle number is just the average in an 17+81 arithmetic sequence. Similarly, the middle of the bottom row is 2 = 49. Applying this again for the middle 49+13 column, the answer is 2 = 31 . 7 freeman66 and nikenissan (February 1, 2021) Sequences and Series in the AMC and AIME Exercise 3.11 (AHSME 1969/33). Let Sn and Tn be the respective sums of the first n terms of two arithmetic series. If Sn : Tn = (7n + 1) : (4n + 27) for all n, the ratio of the eleventh term of the first series to the eleventh term of the second series is: (A) 4 : 3 (B) 3 : 2 (C) 7 : 4 (D) 78 : 71 (E) undetermined Exercise 3.12 (AIME 1999/1).
Load more

Recommended publications
  • 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]
  • On a Series of Goldbach and Euler Llu´Is Bibiloni, Pelegr´I Viader, and Jaume Parad´Is
    On a Series of Goldbach and Euler Llu´ıs Bibiloni, Pelegr´ı Viader, and Jaume Parad´ıs 1. INTRODUCTION. Euler’s paper Variae observationes circa series infinitas [6] ought to be considered important for several reasons. It contains the first printed ver- sion of Euler’s product for the Riemann zeta-function; it definitely establishes the use of the symbol π to denote the perimeter of the circle of diameter one; and it introduces a legion of interesting infinite products and series. The first of these is Theorem 1, which Euler says was communicated to him and proved by Goldbach in a letter (now lost): 1 = 1. n − m,n≥2 m 1 (One must avoid repetitions in this sum.) We refer to this result as the “Goldbach-Euler Theorem.” Goldbach and Euler’s proof is a typical example of what some historians consider a misuse of divergent series, for it starts by assigning a “value” to the harmonic series 1/n and proceeds by manipulating it by substraction and replacement of other series until the desired result is reached. This unchecked use of divergent series to obtain valid results was a standard procedure in the late seventeenth and early eighteenth centuries. It has provoked quite a lot of criticism, correction, and, why not, praise of the audacity of the mathematicians of the time. They were led by Euler, the “Master of Us All,” as Laplace christened him. We present the original proof of the Goldbach- Euler theorem in section 2. Euler was obviously familiar with other instances of proofs that used divergent se- ries.
    [Show full text]
  • Series, Cont'd
    Jim Lambers MAT 169 Fall Semester 2009-10 Lecture 5 Notes These notes correspond to Section 8.2 in the text. Series, cont'd In the previous lecture, we defined the concept of an infinite series, and what it means for a series to converge to a finite sum, or to diverge. We also worked with one particular type of series, a geometric series, for which it is particularly easy to determine whether it converges, and to compute its limit when it does exist. Now, we consider other types of series and investigate their behavior. Telescoping Series Consider the series 1 X 1 1 − : n n + 1 n=1 If we write out the first few terms, we obtain 1 X 1 1 1 1 1 1 1 1 1 − = 1 − + − + − + − + ··· n n + 1 2 2 3 3 4 4 5 n=1 1 1 1 1 1 1 = 1 + − + − + − + ··· 2 2 3 3 4 4 = 1: We see that nearly all of the fractions cancel one another, revealing the limit. This is an example of a telescoping series. It turns out that many series have this property, even though it is not immediately obvious. Example The series 1 X 1 n(n + 2) n=1 is also a telescoping series. To see this, we compute the partial fraction decomposition of each term. This decomposition has the form 1 A B = + : n(n + 2) n n + 2 1 To compute A and B, we multipy both sides by the common denominator n(n + 2) and obtain 1 = A(n + 2) + Bn: Substituting n = 0 yields A = 1=2, and substituting n = −2 yields B = −1=2.
    [Show full text]
  • 3.2 Introduction to Infinite Series
    3.2 Introduction to Infinite Series Many of our infinite sequences, for the remainder of the course, will be defined by sums. For example, the sequence m X 1 S := : (1) m 2n n=1 is defined by a sum. Its terms (partial sums) are 1 ; 2 1 1 3 + = ; 2 4 4 1 1 1 7 + + = ; 2 4 8 8 1 1 1 1 15 + + + = ; 2 4 8 16 16 ::: These infinite sequences defined by sums are called infinite series. Review of sigma notation The Greek letter Σ used in this notation indicates that we are adding (\summing") elements of a certain pattern. (We used this notation back in Calculus 1, when we first looked at integrals.) Here our sums may be “infinite”; when this occurs, we are really looking at a limit. Resources An introduction to sequences a standard part of single variable calculus. It is covered in every calculus textbook. For example, one might look at * section 11.3 (Integral test), 11.4, (Comparison tests) , 11.5 (Ratio & Root tests), 11.6 (Alternating, abs. conv & cond. conv) in Calculus, Early Transcendentals (11th ed., 2006) by Thomas, Weir, Hass, Giordano (Pearson) * section 11.3 (Integral test), 11.4, (comparison tests), 11.5 (alternating series), 11.6, (Absolute conv, ratio and root), 11.7 (summary) in Calculus, Early Transcendentals (6th ed., 2008) by Stewart (Cengage) * sections 8.3 (Integral), 8.4 (Comparison), 8.5 (alternating), 8.6, Absolute conv, ratio and root, in Calculus, Early Transcendentals (1st ed., 2011) by Tan (Cengage) Integral tests, comparison tests, ratio & root tests.
    [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]
  • INFINITE SERIES SERIES and PARTIAL SUMS What If We Wanted
    INFINITE SERIES SERIES AND PARTIAL SUMS What if we wanted to sum up the terms of this sequence, how many terms would I have to use? 1, 2, 3, . 10, . ? Well, we could start creating sums of a finite number of terms, called partial sums, and determine if the sequence of partial sums converge to a number. What do you think that this sequence of partial sums is converging to? It is approaching the value of 2. Therefore, we can conclude that the sum of all the terms of this sequence is 2. To discuss this topic fully, let us define some terms used in this and the following sets of supplemental notes. DEFINITION: Given a sequence of numbers {a n }, the sum of the terms of this sequence, a 1 + a 2 + a 3 + . + a n + . , is called an infinite series. DEFINITION: FACT: If the sequence of partial sums converge to a limit L, then we can say that the series converges and its sum is L. FACT: If the sequence of partial sums of the series does not converge, then the series diverges. There are many different types of series, but we going to start with series that we might of seen in Algebra. GEOMETRIC SERIES DEFINITION: FACT: FACT: If | r | 1, then the geometric series will diverge. EXAMPLE 1: Find the nth partial sum and determine if the series converges or diverges. SOLUTION: Now to calculate the sum for this series. EXAMPLE 2: Find the nth partial sum and determine if the series converges or diverges. 1 - 3 + 9 - 27 + .
    [Show full text]
  • A SIMPLE COMPUTATION of Ζ(2K) 1. Introduction in the Mathematical Literature, One Finds Many Ways of Obtaining the Formula
    A SIMPLE COMPUTATION OF ζ(2k) OSCAR´ CIAURRI, LUIS M. NAVAS, FRANCISCO J. RUIZ, AND JUAN L. VARONA Abstract. We present a new simple proof of Euler's formulas for ζ(2k), where k = 1; 2; 3;::: . The computation is done using only the defining properties of the Bernoulli polynomials and summing a telescoping series, and the same method also yields integral formulas for ζ(2k + 1). 1. Introduction In the mathematical literature, one finds many ways of obtaining the formula 1 X 1 (−1)k−122k−1π2k ζ(2k) := = B ; k = 1; 2; 3;:::; (1) n2k (2k)! 2k n=1 where Bk is the kth Bernoulli number, a result first published by Euler in 1740. For example, the recent paper [2] contains quite a complete list of references; among them, the articles [3, 11, 12, 14] published in this Monthly. The aim of this paper is to give a new proof of (1) which is simple and elementary, in the sense that it involves only basic one variable Calculus, the Bernoulli polynomials, and a telescoping series. As a bonus, it also yields integral formulas for ζ(2k + 1) and the harmonic numbers. 1.1. The Bernoulli polynomials | necessary facts. For completeness, we be- gin by recalling the definition of the Bernoulli polynomials Bk(t) and their basic properties. There are of course multiple approaches one can take (see [7], which shows seven ways of defining these polynomials). A frequent starting point is the generating function 1 xext X xk = B (t) ; ex − 1 k k! k=0 from which, by the uniqueness of power series expansions, one can quickly obtain many of their basic properties.
    [Show full text]
  • In Praise of an Elementary Identity of Euler Arxiv:1102.0659V3 [Math.CO] 12 Jun 2011
    In Praise of an Elementary Identity of Euler Gaurav Bhatnagar Educomp Solutions Ltd. [email protected] June 7, 2011 Dedicated to S. B. Ekhad and D. Zeilberger Abstract We survey the applications of an elementary identity used by Euler in one of his proofs of the Pentagonal Number Theorem. Using a suitably reformulated ver- sion of this identity that we call Euler's Telescoping Lemma, we give alternate proofs of all the key summation theorems for terminating Hypergeometric Series and Basic Hypergeometric Series, including the terminating Binomial Theorem, the Chu{Vandermonde sum, the Pfaff–Saalsch¨utzsum, and their q-analogues. We also give a proof of Jackson's q-analog of Dougall's sum, the sum of a terminating, bal- anced, very-well-poised 8φ7 sum. Our proofs are conceptually the same as those obtained by the WZ method, but done without using a computer. We survey iden- tities for Generalized Hypergeometric Series given by Macdonald, and prove several identities for q-analogs of Fibonacci numbers and polynomials and Pell numbers that have appeared in combinatorial contexts. Some of these identities appear to be new. arXiv:1102.0659v3 [math.CO] 12 Jun 2011 Keywords: Telescoping, Fibonacci Numbers, Pell Numbers, Derangements, Hy- pergeometric Series, Fibonacci Polynomials, q-Fibonacci Numbers, q-Pell numbers, Basic Hypergeometric Series, q-series, Binomial Theorem, q-Binomial Theorem, Chu{Vandermonde sum, q-Chu{Vandermonde sum, Pfaff–Saalsch¨utzsum, q-Pfaff– Saalsch¨utzsum, q-Dougall summation, very-well-poised 6φ5 sum, Generalized Hy- pergeometric Series, WZ Method. MSC2010: Primary 33D15; Secondary 11B39, 33C20, 33F10, 33D65 1 1 Introduction One of the first results in q-series is Euler's 1740 expansion of the product (1 − q)(1 − q2)(1 − q3)(1 − q4) ··· into a power series in q.
    [Show full text]
  • MATH 341: TAKEAWAYS We Used a Variety of Results and Techniques from 103 and 104: (1) Standard Integration Theory: for Us, the M
    MATH 341: TAKEAWAYS STEVEN J. MILLER ABSTRACT. Below we summarize some items to take away from the class (as well as previous classes!). In particular, what are one time tricks and methods, and what are general techniques to solve a variety of problems, as well as what have we used from various classes. Comments and additions welcome! 1. CALCULUS I AND II (MATH 103 AND 104) We used a variety of results and techniques from 103 and 104: (1) Standard integration theory: For us, the most important technique is integration by parts; one of many places we used this was in computing the moments of the Gaussian. Integration by parts is a very powerful technique, and is frequently used. While most of the time it is clear how to choose the functions u and dv, sometimes we need to beR a bit clever. For exam- ¡1=2 1 2 2 ple, consider the second moment of the standard normal: (2¼) ¡1 x exp(¡x =2)dx. The natural choices are to take u = x2 or u = exp(¡x2=2), but neither of these work as they lead to choices for dv that do not have a closed form integral. What we need to do is split the two ‘natural’ functions up, and let u = x and dv = exp(¡x2=2)xdx. The reason is that while there is no closed form expression for the anti-derivative of the standard normal, once we have xdx instead of dx then we can obtain nice integrals. One final remark on integrating by parts: it is a key ingredient in the ‘Bring it over’ method (which will be discussed below).
    [Show full text]
  • 6.1 Sigma Notation & Convergence / Divergence
    Math 123 - Shields Infinite Series Week 6 Zeno of Elea posed some philosophical problems in the 400s BC. His actual reasons for doing so are somewhat murky, but they are troubling regardless. In his Dichotomy paradox he argues that any objects in motion must arrive at the halfway point before they arrive at their goal. However, we can iterate the reasoning, and an object must also arrive at the halfway point to the halfway point first, and so on :::. If the goal is m meters away and you travel at a speed of v, then the time needed to traverse the distance is x x x x t = + + + + ::: 2v 4v 8v 16v x1 1 1 1 = + + + + ::: v 2 4 8 16 The parenthetical expression is an infinite sum. Zeno said that the time for the motion then must be infinite since if we add infinitely many positive numbers together then this must result in a positive number. As beings of logic we are forced to conclude that motion is an illusion. We have all been duped. On a completely unrelated note recall from Calc I that for small angles θ we can use a local linear approxi- mation to eθ. That is eθ ≈ 1 + θ This approximation is pretty good, but it's not perfect because the approximation is a line and eθ bends. Well in a similar way we can make our estimate better by taking a quadratic approximation and say 1 eθ ≈ 1 + θ + θ2 2 This is better, but still ultimately an approximation with some built in error.
    [Show full text]
  • On Series Involving Fibonacci and Lucas Numbers I 1 Introduction
    On series involving Fibonacci and Lucas numbers I Daniel Duverney and Iekata Shiokawa 1 Introduction This paper is the first of two survey papers devoted to the study of series involving Fibonacci and Lucas numbers. There has been, up to now, a great deal of works on this subject, and it seems necessary to summarize and classify it. To begin with, we observe that series involving Fibonacci and Lucas numbers can be divided, roughly, into two large classes ; first, the subscript of Fibonacci and Lucas numbers appear in arithmetic progression ; second, they appear in geometrie progression, We give here two classical examples of both cases. For these two exmples, the sum of the series can be explicitly computed. Example 1.1. We have (1.1) 1+V5 . where <I>= 2 lS the Golden Number. It seems difficult to find the first mention of this result, which is a direct conse- quence of the well-known continued fraction expansion of 'If (see for example [15], Exercise 3.11). But it is undoubtedly very old. Example 1.2. Lucas found in 1878 [27] that ~_1 = 7-V5. (1.2) ~F2n 2 n=O This result can be easily obtained by taking x = 'If in de Morgan's series (cf. [9]) x ---, (1.3) 1-x which is readily deduced from the identity 2 x X x --- (1.4) 1- x 1- x2· 1 There are three natural questions about senes involving Fibonacci and Lucas numbers. First question: Given a series, can we compute its sum? That means, can we express this series in a closed form as in Examples 1.1 and 1'.2, or alternatively, can we express it by using classical functions? These will include, as will see later, q-hypergeometric functions, Lambert series, elliptic integrals, and theta functions.
    [Show full text]
  • Review Solutions: Chapter 11
    Review Solutions: Chapter 11 1. What does it mean to say that a series converges? th SOLUTION: We define the n partial sum sn as follows: n X S1 = a1 S2 = a1 + a2 S3 = a1 + a2 + a3 ··· Sn = an k=1 The partial sums Sn form a sequence (of numbers). The (infinite) series is said to converge to a sum S if and only if the limit of the partial sums is S. That is, if the limit lim Sn = S n!1 then the series is said to converge, and the sum is said to be equal to S: 1 X an = S n=1 2. Does the given sequence or series converge or diverge? 1 1 (a) X p n=2 n − n P 1 SOLUTION: Using the dominating terms, this looks a lot like n , so we use the limit comparison (note that both series, the one given and the template, have all positive terms) 1p n 1 lim n− n = lim p = lim = 1 n!1 1 n!1 n!1 p1 n n − n 1 − n Therefore, both series diverge together (by the limit comparison test). (b) Take the limit: n + 2 n + 2 lim ln = ln lim = ln(1) = 0 n!1 n n!1 n n o np (c) 1+ n SOLUTION: Take the limit; You can use L'Hospital's rule if you like. To be precise, we ought to change notation to x (since you cannot formally take the derivative of a discrete sequence): x 1 p lim p = lim p = lim 2 x = 1 x!1 1 + x x!1 1=2 x x!1 Therefore, the sequence diverges.
    [Show full text]