NCSSM Online AP Calculus (BC) Distance Education & Extended Programs Distance Education & Extended Programs    Lesson  3.1  ...  Sequences       INTRODUCTION This first lesson on sequences reviews the general concepts that you learned in Algebra and/or Pre-Calculus. In those classes you may have worked with arithmetic and geometric sequences. We will expand on your knowledge in this lesson.

Suppose you were the first person to arrive at a meeting. How many handshakes are possible between you and…you? (We will define a handshake as shaking a right hand with another right hand.) Bazaar question, isn’t it? This will make sense really soon. The answer to the question is 0, right? You can’t shake your own right hand with your right hand! How many handshakes are possible between you and the next person who arrives at the meeting? The answer is 1, correct? Let’s organize this information in a table and define the pattern we notice.

Number of People in the Room

Number of Handshakes

1 0

2 1

3 2

4 3

We can see that the number of handshakes is always exactly 1 less than the number of people involved. So, a general rule we can assign to the number of handshakes is n - 1, where n represents the number of people involved. As we move along in this unit, we will be identifying patterns and expressing them mathematically.

NCSSM Online AP Calculus (BC) Distance Education & Extended Programs Distance Education & Extended Programs SEQUENCES We begin with a definition:

Definition: A sequence is an ordered list of numbers whose terms follow a certain rule or function. It is a function whose domain is a set of integers. Examples of sequences are the following:

Integers ℤ : {−2,−1, 0, 1, 2,… } Fibonacci Sequence: {1, 1, 2, 3, 5, 8, 13,… }

The terms in a sequence can be written as follows:

𝑎!  ,𝑎!  ,𝑎!  ,𝑎!  ,… ,𝑎!  ,… Sequences can either have a finite number of terms (like a list of monthly interest payments on a loan that is eventually paid off) or an infinite number (like the Fibonacci sequence). In the case of infinite sequences, we can determine whether the terms in the sequence increase or decrease indefinitely (diverge), oscillate between values that never reach a limit (diverge), or approach a certain finite value (converge). Note: As noted in the definition, the domain of any sequence is a set of integers. The value of n + 1 is the position of the term 𝑎! in the order of the list, where n being 0 refers to the initial term in the sequence. For example, the first term in a sequence is given by 𝑎!, the second term by 𝑎!, and so on. DISPLAYING SEQUENCES ON OUR CALCULATORS

(Keystrokes and calculator instructions adapted from http://www.prenhall.com/divisions/esm/app/calc_v2/index.html by Prentice Hall)

In order to work with sequences on our calculators, we need to set the function mode to “Sequence.” Press to access the mode menu. In the fourth line of the mode menu (the “Func” line), highlight Seq and then press . It’s also a good idea to set the graph style to Dot (highlight Dot and press

). (See Figure 1 below.)

Figure 1

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Now, press Y= to enter the equation editor. Enter the value 0 for “nMin” and the sequence 3𝑛 − 2 in for 𝑢(𝑛), and press . Notice that the graph style icon to the left of 𝑢(𝑛) consists of three dots in a diagonal line, indicating that the graph style is in dot mode. Press to access the Zoom menu, and select 6: ZStandard to view a graph. The graph using this Window, displayed in the middle screen below, is not too helpful. For a better view, press to access the Window menu. Figure 2 below at the right uses the Window settings Xmin = 0, Xmax = 10, Xscl = 1, Ymin = 0, Ymax = 30, Yscl = 1. Then press to view a plot of the sequential points.

Figure 2

Our graph looks quite linear. In fact, it is! If we look at the definition of the sequence, we notice that each successive term is 3 times the term’s position minus two. So, the function that mimics the behavior of this sequence is 𝑓 𝑥 = 3𝑥 − 2.

Another problem … In order to find the first three terms in the sequence 𝑎! =

!!!!

, we substitute 𝑛 = 0, 1,nd  2 into the general term. So the first three terms in the sequence are as follows:

𝑎! =0

0+ 1 = 0              𝑎! =1

1+ 1 =12              𝑎! =

22+ 1 =

23              𝑎!" =

4949+ 1 =

4950

As before, we will type the general formula into the calculator’s equation editor. Remember to revise your Window so that you get a good view of your sequence. Figure 3 below displays the following Window: Xmin = 0, Xmax = 10, Xscl = 1, Ymin = 0, Ymax = 1, Yscl = 0.1. Figure 3

ZoomStandard Window Revised Window

NCSSM Online AP Calculus (BC) Distance Education & Extended Programs Distance Education & Extended Programs This scatterplot acts like a portion of a Rational function, and in fact, if we were to connect the dots, that function could be described by the function.

𝑓 𝑥 =𝑥

𝑥 + 1  

Do you see the relationship between this function and the general term? (Recall, though, that the graph of a sequence will be disconnected dots, since it is a discrete function.) CONVERGENCE & DIVERGENCE OF SEQUENCES As noted above, we can determine whether the terms in infinite sequences increase or decrease indefinitely (diverge), oscillate between values that never reach a limit (diverge), or approach a certain finite value (converge). We do this by examining the limit of the general term, 𝑎!, that define sequences. If lim

n→∞an = L for some value L, then the sequence defined by 𝑎! converges. Otherwise, the sequence

diverges. See Examples 3-7 on text Pages 601-602 for examples of determining convergence or divergence of a sequence. Now try some yourself! Answers are included at the end of this document. Determine whether each sequence associated with the following general terms converges or diverges. If the sequence converges, find the limit.

#1) 𝑎! =!!

#2) 𝑎! = 3

#3) 𝑎! =!!

!

#4) 𝑎! = cos !"!

#5) 𝑎! =!

!!!

#6) 𝑎! =!!

!

You will have an opportunity to practice identifying whether sequences converge or diverge in your assignment for this lesson.

NCSSM Online AP Calculus (BC) Distance Education & Extended Programs Distance Education & Extended Programs Problem Answers:

#1) Since limn→∞

1n= 0 , this sequence converges to 0.

#2) Since lim

n→∞3= 3 , this sequence converges to 3.

#3) Since limn→∞

12

⎛⎝⎜

⎞⎠⎟n= limn→∞

12n

= 0 , this sequence converges to 0.

#4) The terms in this sequence alternate between 0, -1, and 1. Hence, since limn→∞

cos πn2

⎛⎝⎜

⎞⎠⎟ does not

exist, the sequence diverges.

#5) Since limn→∞

nn +1

= 1 , this sequence converges to 1.

#6) We find that limn→∞

2n

n= limn→∞

2n ln21

by application of L’Hopital’s Rule. So, since limn→∞

2n ln21

DNE but approaches +∞ , this sequence diverges.