CHAPTER 11. SEQUENCES AND SERIES 114. a 2 = 2 p 3 a 3 = 3 p 4 a 4 = 4 p 5 a 5 = 5 p 6. n +1. 2n p 2n +1

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1 CHAPTER. SEQUENCES AND SERIES.2 Series Example. Let a n = n p. (a) Find the first 5 terms of the sequence. Find a formula for a n+. (c) Find a formula for a 2n. (a) a = 2 a 2 = 2 p 3 a 3 = 3 p a = p 5 a 5 = 5 p 6 We simply plug n + into the formula has an n: a n+ = p = p. + n +2 (c) This is like part, except that we plug 2n into a 2n = 2n p 2. n p everywhere that n p : Example 2. List the first 5 terms of the sequence, simplify if possible (a) a n =( ) n n 2 a n = n! 2 n (a) a =( ) 2 = a 2 =( ) =() = a 3 =( ) 3 =( ) 32 9 = 9 a =( ) 2 =() 6 = 6 a 5 =( ) =( ) 25 = 25

2 CHAPTER. SEQUENCES AND SERIES 5 a =! 2 = 2 a 2 = 2! 2 2 = 2 = 2 a 3 = 3! 2 = = 3 a =! 2 = = 3 2 a 5 = 5! 2 = = 5 Example 3. Let a n =. Write down (just write them, don t calculate n2 them) the following sums (or at least enough of them to see the pattern): 5 (a) (c) a n 25 a n n=2 n=2 a n (a) (c) Example. Find 0 0 (9n ( ) n )= 9n 0 =9 n (9n ( ) n ) ( ) n ( ) n =9( ) ( =9(55) 0

3 CHAPTER. SEQUENCES AND SERIES 6 Example 5. Zeno s paradox is a problem posed by the ancient Greek philosopher Zeno, who was arguing against the position that space and time are infinitely divisible. Here s one version of it. Suppose we want to walk one city block. We must first walk half the distance, and then half the remaining distance, and half the remaining distance, etc. Since we must go through an infinite number of half-way points, we can never finish the task: we can never walk one block. Translate this problem into a series and solve it. The distance traveled after n steps is n. We can write this using notation as n i= We can start by giving a formula for how far we have walked after n half-way points: n =: n =3: n =2: 2 i 2 + = 3 or = 7 8 or = 5 6 or 6... general n n = 2n 2 n or Assuming that this formula is correct, then after n half-way steps, the total distance traveled is 2 n Taking the limit as n!gives a total distance of lim = 0= n! 2n Thus, we get to the end of the block after all! Another way to summarize this: (/2) n = 2 n Example 6. Re-calculate the example about Zeno s paradox, using our formula.

4 CHAPTER. SEQUENCES AND SERIES 7 The distance traveled in Zeno s paradox is (/2) n. Thus, the total distance is first # common ratio = /2 /2 = /2 /2 =. Example 7. Find the exact formula for each of the following: n 2 (a) n+ n= n 5 (a) It probably helps to write out the first few terms of this series, to make it easier to see how it s a geometric series. n 2 = 5 2 n n=5 = It does not matter that this series is not written explicitly in the form ar n. The main work is to identify what r is. Whatever each term successive term gets multiplied by, that s r. Look at how each term changes when we go to t. Both the power of and the power of increase. How much to they increase by? Each one increases by. Thus, each term equals the previous one, times.inotherwords,we can rewrite the given series this way n 2 3 = n n=5 r = < Now we can see exactly what a and r equal, we can write down the sum instantly, n 2 = 3 / 6 n+ / n=5 As before, it probably helps to write out a few terms of this series n = = ar n

5 CHAPTER. SEQUENCES AND SERIES 8 Now we apply Equation (??) toget = 25 5 (26 ) Example 8. Write the following number as a fraction of integers, equals The definition of decimal notation means that this number where as soon as we get to... the numerators repeat. If we get a common denominator, we can write this as follows We can write this more compactly, and see what the pattern is, if we use powers of {z 8 } geometric Now we find a formula for the infinite repeating part Now we add this to 3: geometric = first # common ratio 3+ 5 = 5/0 /0 = 5 0 = 5 = = 32 Example 9. Show that the series 0000n diverges. The terms in this series look small, because of the factors of 0000 and 0 9 on the bottom, but, they do not keep getting smaller, and smaller, approaching 0. To be more precise: lim n! = =0

6 CHAPTER. SEQUENCES AND SERIES 9 Since this limit is not 0, the series diverges, by the Divergence Test (just imagine adding to itself an infinite number of times: eventually 0000 this will get pretty big!). Example 0. [Harmonic Series] Show that n diverges. I ll give a direct proof now, but later we give a better proof using the Integral test. We add groups of terms in this series and show that no matter how far out we go in the terms, we can always group enough of them together to add up to /2. Let s write out the first 30 terms (or at least some of them) and group the first term, the next 2, the next, the next 8, the next 6, etc {z 3} {z 7} {z 5} = = 2 = {z 3} No matter how big n gets, we can always group together enough terms to get another sum. Thus, as n goes to, wegetaninfinite 2 number of s. Well, that s infinite! 2 Example. Find the sum First we factor n 2 + n n 2 + n = n(). by using partial fractions. Then we split the fraction up using partial fractions n() = A n + B =A()+Bn n = ) =A 0 B ) B = n =0) =A +B 0 ) A = The harmonic series is a really important example. It shows that you can have terms that are approaching 0, and yet adding them together doesn t give a finite answer.