Extra Notes for Section 3.1 Calculus I - Lecture Note

Extra Notes for Section 3.1 Prior to studying these examples, you should read the material in the latter part of the Prelude to Chapter 3 in order to become more comfortable with the notation and basic ideas. Example 1: For each positive integer n, the number 1 5 n5 + 1 3 n3 + 7 15 n is an integer. Solution: We will use the Principle of Mathematical Induction. We first check the statement for the first few values of n. n = 1: n = 2: n = 3: 1 1 7 3+5+7 + + = = 1; 5 3 15 15 96 + 40 + 14 32 8 14 + + = = 10; 5 3 15 15 243 7 3 7 + 9 + = 48 + + 9 + = 59. 5 5 5 5 We now know that the expression gives an integer when n is 1, 2, or 3. Suppose that 1 5 7 k 5 + 31 k 3 + 15 k is an integer for some positive integer k; let’s call this integer m. Using the appropriate rows of Pascal’s triangle, we obtain 1 1 7 (k + 1)5 + (k + 1)3 + (k + 1) 5 3 15
1
1 7 = k 5 + 5k 4 + 10k 3 + 10k 2 + 5k + 1 + k 3 + 3k 2 + 3k + 1 + (k + 1) 5 3 15 1   1 7 1 1 7 = k5 + k3 + k + (k 4 + 2k 3 + 2k 2 + k) + (k 2 + k) + + + 5 3 15 5 3 15 = m + k 4 + 2k 3 + 3k 2 + 2k + 1, which shows that the expression is an integer when n = k + 1. By the Principle of Mathematical Induction, the number 1 5 n5 + 1 3 n3 + 7 15 n is an integer for each positive integer n. Example 2: Suppose that a0 = 3, a1 = 7, and an+1 = 5an − 6an−1 for each positive integer n ≥ 1. Find and prove a simple expression for an . Solution: Gathering some data, we find that a2 = 5a1 − 6a0 = 17, a3 = 5a2 − 6a1 = 43, a4 = 5a3 − 6a2 = 113. After spending some time looking for patterns in the numbers 3, 7, 17, 43, and 113, we discover that 3 = 2 + 1 ⇔ a0 = 21 + 30 , 7 = 4 + 3, 17 = 8 + 9, 43 = 16 + 27, 113 = 32 + 81 ⇔ a4 = 25 + 34 . It appears that an = 2n+1 + 3n for each nonnegative integer n. We can prove this using the Principle of Strong Induction. We already know that the expression for an is valid when n is 0, 1, 2, 3, or 4. Suppose that ai = 2i+1 + 3i for all integers i = 0, 1, 2, . . . , k for some positive integer k. Then (follow the steps closely) ak+1 = 5ak − 6ak−1 = 5(2k+1 + 3k ) − 6(2k + 3k−1 ) = (10 − 6) · 2k + (15 − 6) · 3k−1 = 2k+2 + 3k+1 . This shows that the expression for ak+1 is also valid. By the Principle of Strong Induction, we have shown that an = 2n+1 + 3n for each nonnegative integer n. 1 Example 3: For each positive integer n, n P ifi = nfn+2 − fn+3 + 2. i=1 Proof: We will use the Principle of Mathematical Induction. We first check the equation for a few values of n (assuming the reader is familiar with the Fibonacci sequence). For n = 1, we have 1 X ifi = f1 = 1 and 1 · f3 − f4 + 2 = 1 · 2 − 3 + 2 = 1. i=1 For n = 2, we have 2 X ifi = f1 + 2f2 = 1 + 2 · 1 = 3 and 2 · f4 − f5 + 2 = 2 · 3 − 5 + 2 = 3. i=1 For n = 3, we have 3 X ifi = f1 + 2f2 + 3f3 = 1 + 2 · 1 + 3 · 2 = 9 and 3 · f5 − f6 + 2 = 3 · 5 − 8 + 2 = 9. i=1 We now know that the equation is valid for n = 1, 2, 3. Suppose that k P ifi = kfk+2 − fk+3 + 2 for some i=1 positive integer k. We then have k+1 X i=1 ifi = k X ifi + (k + 1)fk+1 i=1
= kfk+2 − fk+3 + 2 + (k + 1)fk+1 = (k + 1)fk+2 − fk+2 − fk+3 + 2 + (k + 1)fk+1

= (k + 1) fk+1 + fk+2 − fk+2 + fk+3 + 2 = (k + 1)fk+3 − fk+4 + 2, which is the desired equality for n = k + 1. By the Principle of Mathematical Induction, the equation n P ifi = nfn+2 − fn+3 + 2 is valid for all positive integers n. i=1 Example 4: Every positive integer n ≥ 2 is either a prime number or can be factored into a product of prime numbers. Solution: We will use the Principle of Strong Induction. It is clear that 2 is a prime number. Suppose that for some positive integer k ≥ 2, each of the integers 2, 3, . . . , k is either a prime number or can be factored into a product of prime numbers. Consider the integer k + 1. If k + 1 is a prime number, there is nothing further to prove. If k + 1 is not a prime number, then k + 1 = ab, where a and b are integers between 2 and k, inclusively. By the induction hypothesis, each of the integers a and b is either a prime number or can be factored into a product of prime numbers. It follows that k + 1 = ab can be factored into a product of prime numbers. By the Principle of Strong Induction, every integer n ≥ 2 is either a prime number or can be factored into a product of prime numbers. 2 Consider the statement 1 · 4 + 2 · 7 + 3 · 10 + · · · + n · (3n + 1) = n(n + 1)2 . To claim that this statement is true for every positive integer n means that all of the following equations are true. 1 · 4 = 1 · 22 ; 1 · 4 + 2 · 7 = 2 · 32 ; 1 · 4 + 2 · 7 + 3 · 10 = 3 · 42 ; 1 · 4 + 2 · 7 + 3 · 10 + 4 · 13 = 4 · 52 ; 1 · 4 + 2 · 7 + 3 · 10 + 4 · 13 + 5 · 16 = 5 · 62 ; 1 · 4 + 2 · 7 + 3 · 10 + 4 · 13 + 5 · 16 + 6 · 19 = 6 · 72 ; 1 · 4 + 2 · 7 + 3 · 10 + 4 · 13 + 5 · 16 + 6 · 19 + 7 · 22 = 7 · 82 ; 1 · 4 + 2 · 7 + 3 · 10 + 4 · 13 + 5 · 16 + 6 · 19 + 7 · 22 + 8 · 25 = 8 · 92 ; 1 · 4 + 2 · 7 + 3 · 10 + 4 · 13 + 5 · 16 + 6 · 19 + 7 · 22 + 8 · 25 + 9 · 28 = 9 · 102 ; 1 · 4 + 2 · 7 + 3 · 10 + 4 · 13 + 5 · 16 + 6 · 19 + 7 · 22 + 8 · 25 + 9 · 28 + 10 · 31 = 10 · 112 ; .. . 1 · 4 + 2 · 7 + 3 · 10 + 4 · 13 + · · · + k · (3k + 1) = k(k + 1)2 ; 1 · 4 + 2 · 7 + 3 · 10 + 4 · 13 + · · · + k · (3k + 1) + (k + 1) · (3k + 4) = (k + 1)(k + 2)2 ; .. . It would not be all that difficult to check the first ten of these statements by hand. A computer could easily check the next million cases or so but the numbers would get rather large. However, it is impossible to check all of the equations since there are an infinite number of them. This is where the Principle of Mathematical Induction comes into play. In the set of equations above, the letter k is used to represent some generic integer that is larger than 10. If we can show that the validity of the k equation forces the validity of the k + 1 equation, then we know that all of the equations are true; each correct equation forces the next one to be true as well and so the equations are true “all the way down”. Here is an approximate template for writing solutions to induction problems. However, there will be some problems in which modifications to the template are needed. 1. We will use the Principle of Mathematical Induction (Principle of Strong Induction). 2. Let S be the set of all positive integers n such that ... 3. Since [do some simple math] ..., it follows that 1 ∈ S. 4. Now suppose that k ∈ S (suppose that 1, 2, . . . , k ∈ S) for some positive integer k. 5. Since k ∈ S, we know that [write what it means] 6. We then have [do some work here, some of it may be complicated] ... 7. It follows that k + 1 ∈ S. 8. We have thus shown that the conditional statement “if k ∈ S, then k + 1 ∈ S ” (“if 1, 2, . . . , k ∈ S, then k+1 ∈ S ” is true). By the Principle of Mathematical Induction (Principle of Strong Induction), we know that S = Z+ . It follows that [repeat the statement that you are proving] for each n ∈ Z+ . 3 Here is an example to illustrate the template approach to solving such problems. It is the statement highlighted on the previous page. The second solution is the one found more often in textbooks. Problem: Prove that 1 · 4 + 2 · 7 + 3 · 10 + · · · + n · (3n + 1) = n(n + 1)2 for each positive integer n. Solution: We will use the Principle of Mathematical Induction. Let S be the set of all positive integers n such that 1 · 4 + 2 · 7 + 3 · 10 + · · · + n · (3n + 1) = n(n + 1)2 . Since 1 · 4 = 1 · 22 , it is clear that 1 ∈ S. Now suppose that k ∈ S for some positive integer k. This means that 1 · 4 + 2 · 7 + 3 · 10 + 4 · 13 + · · · + k · (3k + 1) = k(k + 1)2 . We then have 1 · 4 + 2 · 7 + 3 · 10 + · · · + (k + 1) · (3(k + 1) + 1) left-hand side of k + 1 equation = 1 · 4 + 2 · 7 + 3 · 10 + · · · + k · (3k + 1) + (k + 1) · (3(k + 1) + 1) include a term = k(k + 1)2 + (k + 1) · (3(k + 1) + 1)
= (k + 1) k(k + 1) + (3k + 4)
= (k + 1) k 2 + 4k + 4 factor out a common term = (k + 1)(k + 2)2 recognize a perfect square
= (k + 1) (k + 1) + 1 )2 , using our hypothesis combine like terms right-hand side of k + 1 equation which tells us that k + 1 ∈ S. We have thus shown that the conditional statement “if k ∈ S, then k + 1 ∈ S ” is true. By the Principle of Mathematical Induction, we know that S = Z+ . It follows that 1 · 4 + 2 · 7 + 3 · 10 + · · · + n · (3n + 1) = n(n + 1)2 for each positive integer n. Abbreviated Solution: We will use the Principle of Mathematical Induction. Since 1 · 4 = 1 · 22 , the equation is valid when n = 1. Now suppose that 1 · 4 + 2 · 7 + 3 · 10 + 4 · 13 + · · · + k · (3k + 1) = k(k + 1)2 . for some positive integer k. We then have 1 · 4 + 2 · 7 + 3 · 10 + · · · + k · (3k + 1) + (k + 1) · (3k + 4) = k(k + 1)2 + (k + 1) · (3k + 4)
= (k + 1) k(k + 1) + (3k + 4) )
= (k + 1) k 2 + 4k + 4 ) = (k + 1)(k + 2)2 , revealing that the equation is valid for n = k + 1. By the Principle of Mathematical Induction, it follows that 1 · 4 + 2 · 7 + 3 · 10 + · · · + n · (3n + 1) = n(n + 1)2 for each positive integer n. 4 Extra Notes for Section 3.2 It is important to be familiar with many examples of sequences of real numbers. Note that there are many types of sequences; any list of numbers that goes on forever is a sequence.   6. n2 = 1, 4, 9, 16, 25, . . . 1. 2n = 2, 4, 8, 16, 32, . . . n n o 1 2 3 4 5  1−n 1 1 1 1 = , , , , ,... 7. 2. 3 = 1, , , , , . . . 3n − 1 2 5 8 11 14 3 9 27 81 n 1 + (−1)n+1 o n (−1)n o 1 1 1 1 = 1, 0, 1, 0, 1, . . . 8. = −1, , − , , − , . . . 3. 2 n 2 3 4 5 √ √ √ √  √ √ √ √ √ 3 4 5 9. n n = 1, 2, 3, 4, 5, . . . 4. n = 1, 2, 3, 2, 5, . . .  5 4  11 5 n  6 n o = 7, 42 , 33 , , ,... 10. 1+ 5. (−1)n+1 n = 1, −2, 3, −4, 5, . . . n 2 5 n o 2n 2 22 23 24 25 11. = , , , , ,... 1 · 4 · 7 · · · · · (3n + 1) 1 · 4 1 · 4 · 7 1 · 4 · 7 · 10 1 · 4 · 7 · 10 · 13 1 · 4 · 7 · 10 · 13 · 16 n nX 1 1 1 1 1 1 1 1 1 1 1o = 1, 1 + , 1 + + , 1 + + + , 1 + + + + , . . . 12. k 2 2 3 2 3 4 2 3 4 5 k=1 Study these examples carefully and be absolutely certain that you understand the terms that are generated by the various formulas. Which adjectives from Definition 3.4 apply to each of these sequences? There are two ways to visualize sequences graphically; we illustrate these for the sequence {an } = {4/n}. For a one dimensional view, think of each term of the sequence as a dot on the number line; the dots are like stepping stones that must be traversed in a certain order: a1 , a2 , a3 , . . . a5 a4 a3 aa7 6 •• • • • 0 a2 a1 • 1 • 2 3 4 5 6 7 For a two dimensional view, think of an as the value of a function a(n), where the inputs are integers. We can then represent the graph of such a function as a bunch of dots in the plane. y 4 (1, a1 ) • 3 (2, a2 ) • 2 (3, a3 ) • (4, a4 ) (5, a5 ) (6, a ) • 6 (7, a7 ) • • • 1 x 1 2 3 4 5 6 7 It is very important that you be able to use these two representations of sequences to visualize the descriptive adjectives for sequences that appear in Definition 3.4. 5 n nX 1o . Example 1: Determine the first five terms of the sequence k! k=0 Solution: Substituting 1, 2, 3, 4, and 5 for n into the formula for the terms, we find that 1 X 1 1 1 = + = 1 + 1 = 2; k! 0! 1! k=0 2 X 1 1 1 1 5 1 = + + =1+1+ = ; k! 0! 1! 2! 2 2 k=0 3 X 1 1 1 1 1 1 8 1 = + + + =1+1+ + = ; k! 0! 1! 2! 3! 2 6 3 k=0 4 X 1 1 1 1 1 1 1 1 1 65 = + + + + =1+1+ + + = ; k! 0! 1! 2! 3! 4! 2 6 24 24 k=0 5 X 1 1 1 1 1 1 1 1 1 1 163 1 = + + + + + =1+1+ + + + = . k! 0! 1! 2! 3! 4! 5! 2 6 24 120 60 k=0 5 8 65 163 The first five terms of the sequence are thus 2, , , , . 2 3 24 60 p Example 2: Find the limit of the sequence n2 + 5n − n . Solution: We use the technique of multiplying by the conjugate. Since ! p  √n2 + 5n + n p
2 2 n + 5n − n = lim n + 5n − n · √ lim n→∞ n→∞ n2 + 5n + n n2 + 5n − n2 = lim √ n→∞ n2 + 5n + n  1/n  5n = lim √ · n→∞ n2 + 5n + n 1/n 5 5 = lim q = , n→∞ 2 1+ 5 +1 n (note that 1/n entered the square root as 1/n2 ), the limit of the sequence p n2 + 5n − n is 5/2. Example 3: Without finding the sum, show that the 16th term of the sequence n nX 1o is greater than 3. k k=1 Solution: We use some convenient pairing and estimation to make the sum easier to work with. 16 X 1 1 1 1 1 =1+ + + + + k 2 3 4 5 k=1 1 1 1 1 >1+ + + + + 2 4 4 8 1 2 4 8 =1+ + + + = 3. 2 4 8 16 1 1 1 1 1 1 1 1 1 1 1 + + + + + + + + + + 6 7 8 9 10 11 12 13 14 15 16 1 1 1  1 1 1 1 1 1 1 1 + + + + + + + + + + 8 8 8 16 16 16 16 16 16 16 16 It takes some practice to learn how to do estimates such as this but it is an important skill to acquire. 6 Extra Notes for Section 3.3 We begin with some examples to illustrate Theorem 3.8; study them carefully. 5. The sequences n1o n 1 o n 1 o √ √ , , and all converge to 0. 8 n4 n n n1o n 2 n o n 6 n o , − , and all converge to 0. 3 7 en n  3 n o n n 5 o n n10 o n4 − , , and all converge to 0. n 4 5 1.2n n 7 n o n 3 n o n o , − , and π n are unbounded. 4 2 n√ o n√ o n√ o n n n 7 , 100 , and 2018 all converge to 1. 6. The sequences n √ o n√ o np o k n n , k , and j j all converge to 1. 1. The sequences 2. The sequences 3. The sequences 4. The sequences 7. The sequences n 1+ n 2 n o n 1 n o 1 n o , 1+ , and 1− converge to e2 , e1/2 , and e−1/2 . n 2n 2n Combining these examples with Theorem 6, we can obtain the following results. a. The sequences n 12 o o n √ n √ and 8 100 converge to 0 and 8, respectively. 8 n o n n5 o √ √ n n n and + 2018 converge to 2 and 1, respectively. 5n n 1 n√  6 n o √ o n c. The sequences √ − n n and 100 − − converge to −1 and 1, respectively. 7 n n√  n1  2 n o 1 n o d. The sequences n n 1 + · 1 + and converge to e2 and 0, respectively. n en 2n n n√ n √ n √ no 100 o n converge to 1 and e, respectively. e. The sequences √ and  n 1 7 1 − 2n b. The sequences n√ n 7+ n √  2 n o n f. The sequence 3 7n − 2 1 + converges to 3 − 2e2 . n Finally, to illustrate the Squeeze Theorem, consider the sequence 2n ≤ 5n2 + 2n ≤ 7n2 and thus √ n 2n ≤ np o n 5n2 + 2n . Note that p √ n n 5n2 + 2n ≤ 7n2 n√ o n√ o n n for all positive integers n. Since the sequences 2n and 7n2 both converge to 1, the sequence np o n 5n2 + 2n must also converge to 1. (If you are uncertain about the limits of the two new sequences, n√ √ o n√ √ √ o n n note that they can be written as 2 · n n and 7 · n n · n n , respectively.) 7 Although Exercise 1 in the textbook asks you to write out a detailed proof for the limit of a sequence (in order to help you better understand the basic properties of convergent sequences), usually we do not go into this much detail. The following examples illustrate this. Example 1: Find the limit of the sequence n 2 n + n5 o . 3n + n Solution: Using parts (2) and (3) of Theorem 3.8, we find that n5 + n 3 3 = 0 + 0 = 0. n 1+0 1+ n 3  2 n n lim n→∞ 5  2 n + n5 n 2 +n 1/3 = lim · n→∞ 3n + n 3n + n 1/3n The limit of the sequence n 2n + n5 o 3n + n denominator that grows most quickly. = lim n→∞ is thus 0. We chose the multiplier 1/3n because 3n is the term in the Example 2: Find the limit of the sequence n 32n−1 + 4n n o . 7n+1 + 9n−1 Solution: Using parts (2) and (3) of Theorem 3.8, we find that 2n−1 lim n→∞ n  32n · 3−1 + 4n n n 3 +4 n 1/9 = lim · n→∞ 7 · 7n + 9−1 · 9n 1/9n 7n+1 + 9n−1 Therefore, the limit of the sequence  4 n 1 1 +n +0 3 9 = 3 = 3. = lim   n 7 1 1 n→∞ 7· 0+ + 9 9 9 n 32n−1 + 4n n o is 3. (We multiplied numerator and denominator by 7n+1 + 9n−1 1/9 since it is the term in the denominator that grows most rapidly.) n Example 3: Find the limit of the sequence n√ n o 3n + π n . Solution: The idea here is that for really large values of n, the number π n will be much bigger than 3n (since π > 3). Ignoring the number 3n , we end up with the nth root of π n , which is just π. It thus appears that the limit of the sequence will be π. To prove this, we use the Squeeze Theorem. Note that π n ≤ 3n + π n ≤ 2π n and thus π≤ √ n 3n + π n ≤ π √ n 2   √ n for all positive integers n. Since the sequences π (a constant sequence) and π 2 both converge to π √ (using part (6) of Theorem 3.8), the sequence n 3n + π n must also converge to π by the Squeeze Theorem. Example 4: Find the limit of the sequence n 3n n o . 3n + 1 Solution: Using some basic algebra, we find that −1  3n n  3n + 1 −n  1/3 n = = 1+ . 3n + 1 3n n n 1/3 n o By part (7) of Theorem 3.8, the sequence 1+ converges to e1/3 . It follows that (see part (e) of n n 3n n o Theorem 3.6) the sequence converges to e−1/3 . 3n + 1 8 Extra Notes for Section 3.4 Example 1: Consider the sequence {an } defined by a1 = 2 and an+1 = 5 − 2 for each positive integer n. an Prove that {an } converges and find its limit. Solution: Using the recursive definition of the sequence, we find that 2 =5− a1 2 a3 = 5 − =5− a2 a2 = 5 − 2 2 41 =5− = ≈ 4.5556; a3 9/2 9 2 2 187 a5 = 5 − = ≈ 4.5610. =5− a4 41/9 41 2 = 4; 2 2 9 = = 4.5; 4 2 a4 = 5 − From these first few terms of the sequence, we conjecture that the sequence {an } is increasing and bounded by 5. To verify this, we will prove the following two statements using mathematical induction: I. the inequality 2 ≤ an < 5 is valid for each positive integer n; II. the inequality an < an+1 is valid for each positive integer n. To prove I, we already know that the inequality is true for n = 1 so assume that 2 ≤ ak < 5 for some positive integer k. To determine the next term ak+1 in the sequence, we refer to the recursive definition of the sequence and see that we need to take the reciprocal of ak , multiply it by −2, and then add 5. Performing these steps (and being careful with the inequalities) yields 2 ≤ ak < 5 1 1 1 ≥ > 2 ak 5 ⇒ ⇒ −1 ≤ − 2 2 <− ak 5 ⇒ 4 ≤ 5− 2 2 < 5− ak 5 ⇒ 4 ≤ ak+1 < 4.6. It then follows that 2 ≤ ak+1 < 5. By the Principle of Mathematical Induction, the inequality 2 ≤ an < 5 is valid for each positive integer n. We conclude that the sequence {an } is bounded. To prove II, we once again know that the inequality is true for n = 1 since a1 < a2 . Assume that ak < ak+1 for some positive integer k. Proceeding in much the same way as in the previous paragraph, we find that (for the record, we are using the fact that all of the terms of the sequence are positive) ak < ak+1 ⇒ 1 1 > ak ak+1 ⇒ − 2 2 <− ak ak+1 ⇒ 5− 2 2 <5− ak ak+1 ⇒ ak+1 < ak+2 , which is the appropriate inequality for n = k + 1. By the Principle of Mathematical Induction, the inequality an < an+1 is valid for each positive integer n. It follows that the sequence {an } is increasing. Since the sequence {an } is bounded and monotone, it converges by the Completeness Axiom. Let L be the limit of the sequence. As n goes to infinity, the recurrence relation an+1 = 5 − 2 an becomes L = 5 − 2 L since the an terms approach L as n gets larger. Solving for L (using the quadratic formula) gives √ √ 5 ± 25 − 8 5 ± 17 2 2 L = 5L − 2 ⇒ L − 5L + 2 = 0 ⇒ L = = . 2 2 Since the sequence {an } is increasing, we know that the limit has to be larger than 2 so the L value √ 5 + 17 corresponding to the minus sign is too small. Hence, the limit of the sequence {an } is ≈ 4.5615528. 2 9 n nX 1 o converges. Example 2: Prove that the sequence k2k k=1 Solution: For each positive integer n, let bn = n X 1 ; we want to show that the sequence {bn } converges. k2k k=1 Using the formula for the terms, we find that the first few terms of this sequence are b1 = 1 X 1 1 = ; k2k 1 · 21 b4 = k=1 4 X 1 1 1 1 1 = + + + ; k 1 2 3 k2 1·2 2·2 3·2 4 · 24 k=1 2 X 1 1 1 b2 = = + ; k 1 k2 1·2 2 · 22 b5 = k=1 5 X 1 1 1 1 1 1 = + + + + ; k2k 1 · 21 2 · 22 3 · 23 4 · 24 5 · 25 k=1 3 X 1 1 1 1 b3 = = + + ; k 1 2 k2 1·2 2·2 3 · 23 b6 = b5 + k=1 1 . 6 · 26 It should be obvious that the sequence {bn } is increasing. (It is very important that you see this; you get bn+1 from bn by adding a positive number to bn .) To prove that {bn } converges, we need to prove that it is bounded. Using some simple overestimates (when the denominator of a fraction is smaller the fraction is larger) and the formula from Exercise 7 in Section 2.1, we obtain bn = n X 1 1 1 1 1 = + + + ··· + k 1 2 3 k2 1·2 2·2 3·2 n · 2n k=1 ≤  1 n 1 1 1 1 (1/2)n − 1 1 + + + · · · + = · = 1 − <1 21 22 23 2n 2 (1/2) − 1 2 for each positive integer n. This shows that the sequence {bn } is bounded by 1. Since the sequence {bn } is bounded and monotone, it converges by the Completeness Axiom. For the record, the limit of this sequence is not 1; see Example 5 in the next set of notes. For ease of reference, we record five ways to prove that a sequence {an } is increasing; there are analogous ways to prove that a sequence is decreasing. 1. Show algebraically that an+1 ≥ an for each n. 2. Show algebraically that an+1 − an ≥ 0 for each n. 3. Show algebraically that an+1 /an ≥ 1 for each n, ASSUMING all the an terms are positive. 4. Show that an+1 ≥ an for each n using the Principle of Mathematical Induction. 5. Convert the formula for an to a function f (x) and show that f 0 (x) ≥ 0 for all x ≥ 1. Often the form of the sequence will provide an indication of which of these five approaches is best for a particular sequence. 10 Extra Notes for Section 3.5 Example 1: Find an explicit formula for the sequence of partial sums for the series ∞ X 6 . 4k k=1 Solution: Using the formula from Exercise 7 in Section 2.1, we find that the nth partial sum of this series is n n  k  X X 6 1 1 1 − (1/4)n 1 = 6 = 6 · . · = 2 1 − 4k 4 4 1 − (1/4) 4n k=1 k=1 Example 2: Prove that the series ∞ X 3+ k=1 5 √ k k2 diverges. Solution: We will use the Divergence Test. Referring to a previous result (part (6) of Theorem 3.8), we find that lim k→∞ 3+ 5 √ k k2 = lim k→∞ 5 5 5 √ √ = = . k k 3 + 1 · 1 4 3+ k· k Since the terms of the series do not converge to 0, the series diverges by the Divergence Test. ∞ X Example 3: Find the sum of the series k=2 3k 22k−1 . Solution: Rewriting the series as ∞ X k=2 3k 22k−1 = ∞ X k=2 ∞ ∞ k=2 k=2 X 2 · 3k X  3 k 3k 2 = = , 22k · 2−1 4k 4 we recognize that this is a geometric series with r = 3/4. Since |r| < 1, the series converges. The sum of a convergent geometric series is the first term of the series divided by the quantity 1 − r, where r is the common ratio. The first term in our case corresponds to k = 2 so we have ∞ X k=2 3k 22k−1 Example 4: Find the sum of the series = ∞  k X 3 2(9/16) 9 2 = = . 4 1 − (3/4) 2 k=2 ∞ X 4k−1 + 7k+1 9k k=0 . Solution: Using part (b) of Theorem 3.10 and the formula for the sum of a geometric series, we find that ∞ X 4k−1 + 7k+1 k=0 9k = ∞ X 4k−1 k=0 9k + ∞ X 7k+1 k=0 9k = 1/4 7 9 63 + = + = 31.95. 1 − (4/9) 1 − (7/9) 20 2 As in this example, you should be able to determine the first term and the common ratio of a geometric series without doing a lot of extra work. 11 ∞ ∞ X X 3k + 5 1 = ln 2, find the sum of the series . Example 5: Given that k2k k2k k=3 k=1 Solution: We need to be careful here because the given series and the series in question start at different k values. We begin by noting that ln 2 = ∞ ∞ X 1 1 1 X 1 = + + k2k 2 8 k2k k=1 and thus ∞ X 5 1 = ln 2 − . k2k 8 k=3 k=3 (We have just pulled the first two terms out of the series.) Using part (b) of Theorem 3.10 and the formula for the sum of a geometric series, we obtain ∞ X 3k + 5 k=3 k2k = ∞ ∞  X X 3/8 3 5 5 3 25 19 + = = + 5 ln 2 − + 5 ln 2 − = 5 ln 2 − ≈ 1.090736. 2k k2k 1 − (1/2) 8 4 8 8 k=3 k=3 16 32 64 + − + · · ·. 3 9 27 Example 6: Find the sum of the series 18 − 12 + 8 − Solution: Whenever a problem asks you to find the sum of a series, you should first check to see whether or not the series happens to be geometric. Since the ratios −12/18 and 8/(−12) of consecutive terms are both equal to −2/3, we see (after perhaps checking the next couple of terms) that the given series is a geometric series with common ratio −2/3. It follows that 18 − 12 + 8 − 16 32 64 18 54 + − + ··· = = . 3 9 27 1 − (−2/3) 5 (Do be careful when the common ratio r is negative.) Example 7: Suppose that the sequence of partial sums of the series ∞ P ak is given by {sn } = k=1 n 5n o . 2n + 1 Find a27 and the sum of the series. Solution: By definition, the sum of a series is the limit of its sequence of partial sums. Thus ∞ X k=1 ak = lim sn = lim n→∞ n→∞ 5n 5 = . 2n + 1 2 To find a27 , we simply compute (make certain you understand why this works) a27 = s27 − s26 = 5 · 27 5 · 26 27 130 1 − = − = . 55 53 11 53 583 In general, we can use this idea to find an for any n > 1: an = sn − sn−1 =  n(2n − 1) − (n − 1)(2n + 1)  5n 5n − 5 5 − =5 = . 2n + 1 2n − 1 (2n − 1)(2n + 1) (2n − 1)(2n + 1) Finally, note that a1 = s1 for any series of the form ∞ P k=1 12 ak . Extra Notes for Section 3.6 Example 1: Use the Integral Test to determine whether or not the series ∞ X 1 converges. 5/2 k k=1 1 . This function is continuous and decreasing on x5/2 the interval [1, ∞). Computing the improper integral of f on [1, ∞) yields Z ∞ Z b 1 1 dx = lim dx 5/2 5/2 b→∞ x x 1 1 Solution: Consider the function f defined by f (x) = ∞ Z Since the improper integral 1 b 2 = lim − x−3/2 b→∞ 3 1  2 2 1 = lim − − 1 = . b→∞ 3 b3/2 3 ∞ X 1 1 dx converges, the series converges by the Integral Test. We 5/2 5/2 x k k=1 can use the equations next to the figure in Section 3.6 to estimate the value of the sum of this series. Referring to the last inequality that is adjacent to the figure and letting n go to infinity, we see that (noting that a1 = 1 for our particular series) Z ∞ Z ∞ ∞ X 1 1 1 dx ≤ ≤1+ dx 5/2 5/2 5/2 x k x 1 1 k=1 ∞ and thus 2 X 1 5 ≤ ≤ . 5/2 3 3 k k=1 With the aid of technology (many options), we can find some partial sums of this series and get a better approximation to the sum. For example, s1000 = 1000 X k=1 1 ≈ 1.3414662. k 5/2 The one potential difficulty with this calculation is that we do not know how accurate it is. Another calculation gives s2000 ≈ 1.3414798. We can thus (somewhat cautiously) conclude that the sum of this series is about 1.3415. Example 2: Use the Integral Test to determine whether or not the series ∞ X k=3 √ 3 4 converges. 2k − 5 4 Solution: Consider the function f defined by f (x) = √ . This function is continuous and decreasing 3 2x − 5 on the interval [3, ∞). (We can still apply the Integral Test but we need to use [3, ∞) rather than [1, ∞) to be certain that the values are positive.) Computing the appropriate improper integral in this case yields Z ∞ Z b 4 4 √ dx = lim dx 3 b→∞ (2x − 5)1/3 2x − 5 3 3 b = lim 3(2x − 5)2/3 b→∞ 3   = lim 3(2b − 5)2/3 − 3 = ∞. b→∞ Z Since the improper integral 3 ∞ √ 3 ∞ X 4 4 √ diverges by the Integral Test. dx diverges, the series 3 2x − 5 2k −5 k=3 13 Example 3: Estimate a value of n for which n X 1 √ > 100. k k=1 Solution: Since the given series is a p-series with p ≤ 1, we know that the series diverges. According to the inequalities listed by the figure in Section 3.6, we find that Z n n X √ 1 dx √ ≥ √ =2 x x k 1 k=1 n √ = 2 n − 2. 1 √ It is thus sufficient to find a value of n for which 2 n − 2 > 100. It is easy to see that n > (51)2 = 2601 will work. In other words, we know that 2602 X k=1 1 √ > 100. k Using some technology, we can actually compute these values more carefully and find that 2602 X k=1 1 √ ≈ 100.5691 k 2573 X while k=1 1 √ ≈ 99.9989971 k and 2574 X k=1 1 √ ≈ 100.0187075. k The estimates provided by the Integral Test are important for both practical and theoretical purposes, but we will not spend much time on this aspect of series. 1 1 1 Example 4: For each positive integer n, let yn = 1 + √ + √ + · · · + √ − n 2 3 a decreasing sequence. Z n 1 dx √ . Prove that {yn } is x Solution: We want to prove that yn+1 ≤ yn for each positive integer n. This is equivalent to proving that yn − yn+1 ≥ 0. Writing out this expression and using properties of integrals, we find that yn − yn+1 Z n Z n+1 1 1 1 dx   1 1 1 dx  √ √ = 1 + √ + √ + ··· + √ − − 1 + √ + √ + ··· + √ − n x x n+1 2 3 2 3 1 1 Z n+1 Z n 1 dx dx √ −√ √ + =− x x n + 1 1 1 Z n+1 dx 1 √ −√ = . x n +1 n  √ Since the function 1/ x is decreasing, we can use property (5) of integrals (see Section 2.5) to find that Z n+1 yn − yn+1 = n dx 1 1 1 √ −√ ≥√ −√ = 0. x n+1 n+1 n+1 (Alternatively, we can draw a picture and look at areas as in the text; the area under the curve is larger than the area of the inscribed rectangle.) We have thus shown that yn ≥ yn+1 for each positive integer n. It follows that {yn } is a decreasing sequence. Note that, even though we proved that an equation was valid for all positive integers n, there was no need to use the Principle of Mathematical Induction. We chose the subtraction form for a decreasing sequence since the expression for yn involved a sum. Finally, referring once again to the inequalities listed by the figure in Section 3.6, we see that each of the terms yn is positive. By the Completeness Axiom, the sequence {yn } converges. 14 Extra Notes for Section 3.7 Example 1: Use the Comparison Test to determine whether or not the series ∞ X k=1 Solution: Looking at just the highest powers of k, this series looks a bit like k converges. 7k 2 − 6 ∞ X 1 , which is a divergent k k=1 p-series. This leads us to believe that the series diverges so we need to show that its terms are larger than the terms of a divergent series of positive numbers. We can make fractions smaller by making their denominators larger. In this case, we note that k 1 k 1 1 > 2 = = · −6 7k 7k 7 k 7k 2 for all k ≥ 1. Since the series ∞ ∞ X X 1 k diverges (it is a multiple of a divergent p-series), the series 7k 7k 2 − 6 k=1 k=1 diverges by the Comparison Test. Example 2: Use the Comparison Test to determine whether or not the series ∞ X k=1 k converges. 7k 3 − 6 ∞ X 1 , which is a convergent Solution: Looking at just the highest powers of k, this series looks a bit like k2 k=1 p-series. This leads us to believe that the series converges so we need to show that its terms are smaller than the terms of a convergent series. We can make positive fractions larger by making their denominators smaller. We also want to do this in a way that makes it possible to combine like terms. In this case, we note that k k k 1 k k k 1 < 3 = 3 = 2 or perhaps < 3 = 3 = 2. 7k 3 − 6 7k − 6k 3 k k 7k 3 − 6 7k − k 3 6k 6k The first inequality is valid for all k ≥ 1, while the second is valid for all k ≥ 2 (since we need k 3 to be larger ∞ ∞ X X k 1 than 6). In either case, since the series (or one of its multiples) converges, the series 2 3 k 7k − 6 k=1 k=1 converges by the Comparison Test. Example 3: Use the Comparison Test to determine whether or not the series ∞ X 4k + 5k k=1 3k + 6k Solution: Looking at the exponential terms with the larger bases, this series looks a bit like converges. ∞  k X 5 k=1 6 , which is a convergent geometric series. Noting that  5 k 5k + 5 k 5k + 5 k 4k + 5k 5k < < = 2 · = 2 3k + 6k 3k + 6 k 6k 6k 6 ∞ X  5 k for all k ≥ 1 and that the series 2 is a convergent geometric series (the absolute value of the common 6 k=1 ratio is less than 1), the series ∞ X 4k + 5k k=1 3k + 6 k converges by the Comparison Test. 15 Example 4: Use the Limit Comparison Test to determine if the series ∞ X k=1 2k + 3 converges. k 2 + 4k − 2 Solution: For large values of k, the terms of this series resemble those of the series ∞ X 2 . This series k k=1 diverges (since it is a multiple of a divergent p-series) and we find that 2k + 3 2 + 4k − 2 2k 2 + 3k k = lim α = lim = 1. 2 k→∞ 2k 2 + 8k − 4 k→∞ k Since α > 0, the series ∞ X k=1 k2 2k + 3 diverges by the Limit Comparison Test. (We can certainly use 1/k + 4k − 2 instead of 2/k and reach the same conclusion; obtaining just a slightly different positive value for α.) Example 5: Determine whether or not the series ∞ X 7k 2 + 3k − 4 converges. 2k 4 + 4k 3 − 2 k=1 Solution: For large values of k, the terms of this series resemble those of a multiple of the series ∞ X 1 . We k2 k=1 thus predict that this series converges. (It is important for you to be able to make such conjectures quickly AND it is necessary that you know how to back up your conjectures with a proof if necessary.) To prove this fact, we can use either the Comparison Test or the Limit Comparison Test. To use the LCT, we compute the limit (skipping straight to the invert and multiply step):  7k 2 + 3k − 4 k 2  7k 4 + 3k 3 − 4k 2 7 α = lim · = lim = . k→∞ 2k 4 + 4k 3 − 2 k→∞ 2k 4 + 4k 3 − 2 1 2 Since α < ∞ and the series ∞ ∞ X X 1 7k 2 + 3k − 4 converges, the series converges by the Limit Comparison 2 k 2k 4 + 4k 3 − 2 k=1 k=1 Test. On the other hand, we can be very careful with inequalities and note that 7k 2 + k 2 7k 2 + k 2 8k 2 4 7k 2 + 3k − 4 < 4 < = 4 = 2 4 3 3 4 2k + 4k − 2 2k + 4k − 2 2k 2k k for all k ≥ 1. Since ∞ ∞ X X 4 7k 2 + 3k − 4 converges, the series converges by the Comparison Test. k2 2k 4 + 4k 3 − 2 k=1 k=1 Example 6: Estimate (without technology) the sum of the series ∞ X k=1 2k . 3k + 4k Solution: Using some simple over and under estimates, we see that 2k 2k 2k 1  1 k 2k 1  2 k < < or < < 4k + 4k 3k + 4k 3k + 3k 2 2 3k + 4k 2 3 for all k ≥ 1. Using the formula for the sum of a geometric series, we find that ∞ ∞ ∞ k=1 k=1 k=1 X 1  2 k 1 X 1  1 k X 2k = < < = 1. k k 2 2 2 3 +4 2 3 Hence, the sum of the series is between 0.5 and 1. (Technology gives the actual sum as about 0.633754.) 16 Extra Notes for Section 3.8 Recall that the main goal of the last few sections is to determine whether or not the series ∞ P ak k=1 converges. That is, we want to know if there is a number represented by the infinite sum a1 + a2 + a3 + a4 + a5 + a6 + · · · . We have seen that this occurs in some cases and not in others. The tests that we have developed thus far (with paraphrased informal synopses) to help us decide are the Divergence Test (if the terms ak do not have a limit of 0, then the series diverges), the Integral Test (the series and the corresponding improper integral behave in the same way), the Comparison Test (less than convergent is convergent and greater than divergent is divergent), and the Limit Comparison Test (if the limit of the ratios of the terms exists, then the series do the same thing). Using these ideas, we have learned that ∞ X i. geometric series ark converge when |r| < 1 and diverge when |r| ≥ 1; k=0 ∞ X 1 ii. p-series converge when p > 1 and diverge when 0 < p ≤ 1. kp k=1 Furthermore, for geometric series, we know that the sum of the series is a/(1 − r), that is, the sum is the first term divided by 1 minus the common ratio. For p-series, there are no simple formulas for the sum when the series converges (although some of the sums are known); we simply know that the series converges and we can then use technology to find an approximation for the sum. The latter three tests require series with positive terms and this can be a serious drawback for certain ∞ ∞ P P series. The concept of absolute convergence is helpful in some cases: if |ak | converges then ak must k=1 also converge. Since the terms of ∞ P k=1 |ak | are positive (or at least nonnegative), we can use one of the k=1 comparison tests. However, the next three tests that we consider (Alternating Series Test, Ratio Test, and Root Test) allow us to look at series with infinitely many positive and negative terms. Example 1: Prove that the series ∞ X sin(3k) k=1 k 4/3 converges. Solution: Using basic properties of the sine function, we see that sin(3k) | sin(3k)| 1 = ≤ 4/3 4/3 4/3 k k k ∞ ∞ X X 1 sin(3k) for all positive integers k. Since is a convergent p-series, the series converges by the 4/3 k k 4/3 k=1 k=1 Comparison Test. It follows that the series ∞ X sin(3k) k=1 k 4/3 17 converges. Example 2: Determine whether or not the series ∞ X (−1)k+1 k=1 ∞ X (−1)k and converge. 2k − 1 2k − 1 k=1 Solution: These two series behave in the same way since one is just a constant multiple of the other. It is easy to see that the sequence {1/(2k − 1)} is decreasing (the denominator increases with k while the ∞ X (−1)k+1 numerator stays the same) and converges to 0. We conclude that converges by the Alternating 2k − 1 k=1 ∞ ∞ X X (−1)k (−1)k+1 Series Test. It then follows that the series = (−1) converges as well. 2k − 1 2k − 1 k=1 k=1 This is perhaps a good time to recall various ways to prove that a sequence is monotone. Suppose we want to prove that a sequence {ak } is decreasing. There are several ways to do so. a. Use algebra to prove that ak ≥ ak+1 for all positive integers k. b. Use algebra to prove that ak − ak+1 ≥ 0 for all positive integers k. c. Use algebra to prove that ak+1 /ak ≤ 1 for all positive integers k, ASSUMING all terms are positive. d. Use induction to prove that an ≥ an+1 for all positive integers n. e. Find a function f such that f (k) = ak for each k and show that f 0 (x) ≤ 0 for all x ≥ 1. Methods for proving that a sequence is increasing are analogous (see the extra notes for Section 3.4). Example 3: Prove that the series ∞ X (−1)k+1 k k=1 Solution: We first consider the series 7k 2 − 6 ∞ X k=1 is conditionally convergent. k , which consists of the absolute value of each term. −6 7k 2 Referring to Example 1 in the extra notes for Section 3.7, we see that this series diverges. (If you are confronted with a new series, you would need to include the details here; these solutions do get a bit ∞ X (−1)k+1 k converges, we use the Alternating Series Test. Ignoring the lengthy.) To verify that the series 7k 2 − 6 k=1 n k o (−1)k+1 term, we consider the sequence . This sequence clearly converges to 0 and has positive 7k 2 − 6 terms. To prove that it is decreasing, let f be the function defined by f (x) = x −6 7x2 and note that f 0 (x) = 7x2 − 6 − x(14x) −7x2 − 6 = . 2 2 (7x − 6) (7x2 − 6)2 Since f 0 (x) < 0 for all x ≥ 1, the function f is decreasing on [1, ∞). It follows that decreasing sequence. By the Alternating Series Test, the series ∞ X (−1)k+1 k k=1 the series ∞ X k=1 7k 2 − 6 n k o is a 7k 2 − 6 converges. Finally, since ∞ ∞ X X k (−1)k+1 k (−1)k+1 k diverges and the series converges, we conclude that is a 2 2 7k − 6 7k − 6 7k 2 − 6 k=1 k=1 conditionally convergent series. 18 Extra Notes for Section 3.9 Example 1: Determine whether or not the series ∞  X 2k + 17 k converges. 3k − 100 k=1 Solution: Since the form of the terms involves k as part of the exponent, the Root Test seems like a good choice. Note that r 2k + 17 2 2k + 17 k k = . ` = lim = lim k→∞ k→∞ 3k − 100 3k − 100 3 ∞  X 2k + 17 k converges absolutely by the Root Test. Since ` < 1, the series 3k − 100 k=1 Example 2: Determine whether or not the series ∞  2 X 2k − 1 k converges. 2k k=1 Solution: As with Example 1, the Root Test seems like a good choice. In this case, we find that r  2k − 1 k2  2k − 1 k  −1/2 k k ` = lim = lim = lim 1 + = e−1/2 k→∞ k→∞ k→∞ 2k 2k k (see part 7 of Theorem 3.8). Since ` < 1, the series ∞  2 X 2k − 1 k k=1 Example 3: Determine whether or not the series 2k ∞ X (−1)k (2k)! 3k (k!)2 k=0 converges absolutely by the Root Test. converges. Solution: Since the terms of this series involve factorials, it is best to use the Ratio Test. The kth term and (k + 1)st term of the sequence of terms are (−1)k (2k)! 3k (k!)2 and (−1)k+1 (2k + 2)! , 3k+1 ((k + 1)!)2 respectively. Be certain that you notice what happens when k + 1 is substituted for k and be very careful with parentheses as, for instance, the number 2k! is very different from the number (2k)!. Omitting the invert and multiply step, we find that ` = lim k→∞ Since ` > 1, the series 3k (k!)2 4 (−1)k+1 (2k + 2)! (2k + 2)(2k + 1) · = lim = . k+1 2 k→∞ 3 ((k + 1)!) (−1)k (2k)! 3(k + 1)2 3 ∞ X (−1)k (2k)! k=0 3k (k!)2 diverges by the Ratio Test. It is very important that you understand how the factorial terms cancel. For instance, (2k + 2)! (2k + 2)(2k + 1)(2k)(2k − 1) · · · 3 · 2 · 1 = = (2k + 2)(2k + 1) (2k)! (2k)(2k − 1) · · · 3 · 2 · 1 or preferably (indicating a better understanding) (2k + 2)! (2k + 2)(2k + 1)(2k)! = = (2k + 2)(2k + 1). (2k)! (2k)! 19 Example 4: Determine whether or not the series ∞ X (−1)k (k 2 + 3k) k=1 k 4 − 5k + 7 converges absolutely, converges con- ditionally, or diverges. Solution: Since the numerator and denominator of the fraction that makes up the terms of this series involve polynomials, the Ratio Test and Root Test will give no information. To see why this happens, note that (k + 1)2 + 3(k + 1) k 4 − 5k + 7 ` = lim · k→∞ (k + 1)4 − 5(k + 1) + 7 k 2 + 3k √ k k 2 + 3k ` = lim √ = 1. k k→∞ k 4 − 5k + 7   = 1; The first limit is 1 due to the fact that the polynomials in the fraction are both of degree 6 with equal leading coefficients. The second limit is 1 as a consequence of part 6 of Theorem 3.8. It is very important that you recognize this in advance and that you do not then waste your time doing these tests when you know they will give you no information. We can, however, apply the Limit Comparison Test to the series ∞ X k 2 + 3k . The terms of this series are similar to 1/k 2 and we find that k 4 − 5k + 7 k=1  k 2 + 3k k2  · = 1. k→∞ k 4 − 5k + 7 1 lim Since ∞ ∞ X X 1 (−1)k (k 2 + 3k) is a convergent p-series, the series converges by the Limit Comparison 2 k k 4 − 5k + 7 k=1 k=1 Test. It follows that the series ∞ X (−1)k (k 2 + 3k) k=1 k 4 − 5k + 7 is absolutely convergent. Example 5: Find all the values of x for which the series ∞ X (−1)k+1 k 3 k=1 3k (x + 1)k converges. Solution: We apply the Ratio Test and find that ` = lim k→∞  (k + 1)3  |x + 1| (−1)k+2 (k + 1)3 (x + 1)k+1 3k · = lim |x + 1| = . k→∞ 3k+1 (−1)k+1 k 3 (x + 1)k 3k 3 3 We know that ` < 1 is needed to guarantee convergence so we want values of x that satisfy |x + 1| < 3, which is equivalent to −3 < x + 1 < 3 or −4 < x < 2. Hence, the series converges for all x in the interval (−4, 2). When considering series like this one, be very careful to include the absolute value of the terms when performing either the Ratio Test or the Root Test. 20 Extra Notes for Section 3.10 Example 1: Find the radius of convergence of the power series ∞ X 4k + 2 k=0 3k (x − 1)k . Solution: We apply the Ratio Test and find that ` = lim k→∞  4k + 6  |x − 1| (4k + 6)(x − 1)k+1 3k = lim · |x − 1| = . k→∞ 3(4k + 2) 3k+1 (4k + 2)(x − 1)k 3 We know the power series converges for those values of x that make ` < 1. In this case, we find that |x − 1| < 3. This means that the series converges for all values of x that are within three units of the center 1. It follows that the radius of convergence of the power series is 3. Example 2: Find the radius of convergence of the power series ∞ X (−1)k 5k k=0 (2k)! xk . Solution: We apply the Ratio Test and find that ` = lim k→∞ 5k+1 xk+1 (2k)! 5|x| · k k = lim =0 k→∞ (2k + 2)(2k + 1) (2k + 2)! 5 x for any value of x. It follows that the radius of convergence of the power series is ∞. Example 3: Find the interval of convergence for the power series ∞ X k=0 (−1)k (x − 1)k . + 1) 2k (3k Solution: We apply the Ratio Test and find that ` = lim k→∞  |x − 1|  3k + 1 (x − 1)k+1 2k (3k + 1) · |x − 1| = . = lim k→∞ 2(3k + 4) 2k+1 (3k + 4) (x − 1)k 2 It follows that the power series converges for those values of x that satisfy |x − 1| < 2 or −1 < x < 3. However, the Ratio Test does not give us any information about the behavior of the power series at the endpoints (where the value of ` is 1). Checking these values separately, we find that x = −1 gives ∞ X k=0 x=3 gives ∞ X k=0 ∞ X 1 (−1)k (−2)k = , k 2 (3k + 1) 3k + 1 which diverges by the Comparison Test; k=0 ∞ X (−1)k (−1)k (2)k = , k 2 (3k + 1) 3k + 1 which converges by the Alternating Series Test. k=0 The interval of convergence for this power series is thus (−1, 3]. As mentioned in the textbook discussion of power series, this means that the domain of the function f defined by f (x) = ∞ X k=0 (−1)k (x − 1)k 2k (3k + 1) is the set of all x values that satisfy −1 < x ≤ 3. 21 Example 4: Determine in more familiar terms the function represented by f (x) = ∞ X (−1)k+1 k=1 7k (x − 2)k . Solution: Since we are asked to find the sum of a series, we need to think about which series we have previously learned that allow us to find the sum. A little thought (or page turning through the textbook) reveals that the only series we have seen thus far that come with a formula for the sum are the geometric series. So, we hope that this series is actually a geometric series. Given the form of the terms (all the k values appear as exponents), we see that the series is indeed geometric. The formula for the sum is the first term divided by 1 minus the common ratio. It follows that f (x) = ∞ X (−1)k+1 7k k=1 x−2 x−2 (x − 2) = .  7x − 2  = x+5 1− − 7 k Recall that a geometric series converges when |r| < 1. In our case, we have used −(x − 2)/7 as the value for r. It follows that the sum of the series is only valid for those values of x that satisfy − x−2 <1 7 ⇔ |x − 2| < 7. This fact indicates that the radius of convergence of the power series must be 7. It also shows that the domain of the function f defined by the power series is −5 < x < 9. As a final point, notice that the simplified formula for f (x) is not defined at x = −5 and the distance from this point to the center x = 2 of the power series is 7, the same as the radius of convergence of the power series. As we will see, this is no accident. 22 Extra Notes for Section 3.11 Example 1: Use a geometric series to find a power series expression centered at 0 for f (x) = 4 . 3x + 1 Solution: Using the familiar formula for the sum of a geometric series (first term divided by 1 minus the common ratio as long as the absolute value of the common ratio is less than 1), we find that ∞ X 4 4 = = 4(−3x)k 3x + 1 1 − (−3x) k=0 assuming that | − 3x| < 1. It follows that a power series expression centered at 0 for the function f is f (x) = ∞ X 4(−1)k 3k xk , k=0 which is valid for all x that satisfy |x| < 1/3. Example 2: Use a geometric series to find a power series expression centered at 0 for f (x) = x . 3x + 2 Solution: Proceeding as in the previous example (but with the recognition that a 1 is needed in the denominator), we find that ∞ ∞ k=0 k=0 X x  3x k X (−1)k 3k x/2 x = = = xk+1 − 3x + 2 1 − (−3x/2) 2 2 2k+1 assuming that |x| < 2/3. Since the final form of the above power series involves the k + 1 exponent, we can shift the indices to obtain ∞ X (−1)k 3k k=0 2k+1 xk+1 = ∞ X (−1)k−1 3k−1 k=1 2k xk . The exponents are reduced by 1 to compensate for the sum beginning at a higher number. If you do not believe that the two series are the same, write out the first three terms of each of them and see that they are indeed the same sum. It follows that a power series expression centered at 0 for the function f is f (x) = ∞ X (−1)k+1 3k−1 2k k=1 xk . which is valid for all x that satisfy |x| < 2/3. (Note that (−1)k−1 = (−1)k+1 for all values of k.) Example 3: Use a geometric series to find a power series expression centered at 0 for f (x) = 2 . (3x + 1)2 Solution: Because of the square in the denominator, we must do some work to see how a geometric series comes into play. As with one of the examples in the textbook, we take the derivative of the appropriate function to find that f (x) = 2 2 d  1  =− · . 2 (3x + 1) 3 dx 3x + 1 23 Using the geometric series, we find that ∞ ∞ k=0 k=0 X X 1 1 = = (−3x)k = (−1)k 3k xk 3x + 1 1 − (−3x) and thus ∞ ∞ ∞ k=0 k=1 k=1 X 2X 2 d X (−1)k 3k xk = − (−1)k 3k kxk−1 = 2(−1)k+1 3k−1 kxk−1 . f (x) = − · 3 dx 3 (The starting index changed from k = 0 to k = 1 because the derivative of the first term is 0.) As in Example 2, we can shift the index to obtain f (x) = ∞ X 2(−1)k 3k (k + 1)xk . k=0 This power series expression for the function f is valid for all x that satisfy |x| < 1/3. Example 4: Use a geometric series to find a power series expression centered at 2 for f (x) = 1 . 3x + 1 Solution: Performing some creative algebra, we can make a multiple of x − 2 appear as the r value for the geometric series. Doing so yields 1 1 f (x) = = = 3x + 1 3(x − 2) + 7 1 ∞ ∞ X 1  −3(x − 2) k X (−1)k 3k 7 = (x − 2)k .  −3(x − 2)  = 7 7 7k+1 k=0 k=0 1− 7 It is important that you learn how to read these sorts of equations. Look at each equals sign and convince yourself how each step was carried out. Note that the series converges for all values of x that satisfy |x − 2| < 7/3 and that the distance from the center 2 to the undefined point −1/3 of the function is also 7/3. Example 5: Find the sum of the series ∞ X 5k − 3 k=1 4k . Solution: We begin by using the linearity properties of series and some simplification to obtain ∞ X 5k − 3 k=1 4k =5 ∞ ∞ ∞ ∞  k  1 k X X X X 1 k 1 − 3 . − 3 = 5 k k k 4 4 4 4 k=1 k=1 k=1 k=1 The second series is a geometric series so we know how to find its sum. The first series is of the form ∞ X k=1 kxk = x , (1 − x)2 whose formula was derived in this section. It follows that ∞ X 5k − 3 k=1 4k =5 ∞ ∞  k  1 k  1/4   1/4  20 X X 1 11 k −3 =5 − 3 = −1= . 2 4 4 (3/4) 3/4 9 9 k=1 k=1 24 Extra Notes for Section 3.12 Example 1: Use known Maclaurin series to determine in more familiar terms the functions represented by ∞ ∞ X X (−6)k (−1)k k x and x2k . the power series 2k+1 k! (2k + 1)! k=0 k=0 Solution: Since the first series involves k!, we conjecture that the relevant function is related to ex . Writing the series in the appropriate form, we find that ∞ ∞ ∞ X (−1)k k 1 X (−1/2)k k 1 X 1  x k 1 − x = x = = e−x/2 . k+1 2 k! 2 k! 2 k! 2 2 k=0 k=0 k=0 The second series has the factorials of odd integers and this leads us to look for an appropriate sine function. Performing some algebra yields ∞ ∞ ∞ X X √
(−1)k √
2k 1 X (−1)k √
2k+1 1 (−6)k x2k = 6x =√ 6x =√ sin 6 x . (2k + 1)! (2k + 1)! 6 x k=0 (2k + 1)! 6x k=0 k=0 Pay attention to the steps required to express the sum in the appropriate form. Ask questions, either in class or during office hours, for any steps that you do not understand. Example 2: Find f (101) (0) for the function f defined by f (x) = xe−x 2 /2 . Solution: We start by finding the Maclaurin series for f . Using the known Maclaurin series for ex , we find that 2 f (x) = xe−x /2 =x ∞ ∞ ∞ X X 1  x2 k (−1)k 2k X (−1)k 2k+1 x = x . − =x k! 2 2k k! 2k k! k=0 k=0 k=0 We know from the theory of power series (for the case when the center is 0) that f (k) (0) = k! ck , where ck represents the coefficient on the term xk . For our situation, the term x101 appears when k = 50. It follows that the coefficient on this power of x, which we denote by c101 , is given by c101 = (−1)50 1 = 50 . 250 50! 2 50! Therefore, we find that f (101) (0) = 101! c101 = 101! . 250 50! Pay particular attention to how the k values were used in this calculation. Since these numbers are so large (the final integer involves 81 digits), we do not simplify them any further. 25 √ Example 3: Find the Taylor series centered at 4 for the function f (x) = 1/ x. Solution: Since the given function is not one that has been considered thus far, we have to return to the definition and actually find the coefficients. The first step is to write out some derivatives and try to spot a pattern. f (x) = x−1/2 1 f 0 (x) = − x−3/2 2 3 1 f 00 (x) = · x−5/2 2 2 5 3 1 f (3) (x) = − · · x−7/2 2 2 2 7 5 3 1 f (4) (x) = · · · x−9/2 2 2 2 2 9 7 5 3 1 f (5) (x) = − · · · · x−11/2 2 2 2 2 2 for k ≥ 1, 1 · 3 · 5 · · · · · (2k − 1) −(2k+1)/2 x 2k 1 · 3 · 5 · · · · · (2k − 1) −(2k+1)/2 f (k) (4) = (−1)k · 4 2k 1 · 3 · 5 · · · · · (2k − 1) f (k) (4) = (−1)k · 2k 22k+1 f (k) (x) = (−1)k · As with the example in the textbook, when searching for a pattern, it is best not to multiply out terms. We also used the fact that 41/2 = 2. Now recall that the coefficients of the power series are given by ck = f (k) (4) 1 · 3 · 5 · · · · · (2k − 1) = (−1)k · k! 23k+1 k! for k ≥ 1. When k = 0, we find that c0 = f (4)/0! = 1/2. Since the first term of the Taylor series does not fit the pattern, it is pulled out separately. We thus obtain ∞ 1 X 1 · 3 · 5 · · · · · (2k − 1) 1 √ = + (−1)k · (x − 4)k . 2 23k+1 k! x k=1 Remember to write the series using powers of x − a, where a is the center of the power series. In this case, we need powers of x − 4. As another observation, we can multiply by a messy version of 1 to obtain 1 · 3 · 5 · · · · · (2k − 1) 1 · 2 · 3 · 4 · 5 · 6 · 7 · · · · (2k − 1) · (2k) (2k)! (2k)! = = k = k , k! 2 · 4 · 6 · · · · · (2k) k! 2 k! · k! 2 (k!)2 from which it follows that ck = (−1)k (2k)! . 24k+1 (k!)2 Note that this formula conveniently gives c0 = 1/2 so we no longer need to pull out the first term. We thus find that ∞ X (−1)k (2k)! 1 √ = (x − 4)k . 24k+1 (k!)2 x k=0 Formulas such as this one may help explain some of the power series exercises in Section 3.10 that involved factorials or other similar products. 26