Probably the last note related to the new site! I got quite a few comments/complaints about the "brightness" of the red on the site and have changed it to a darker shade. You may have to clear your browsers cache for the new color to show up, especially Chrome as it has a nasty habit of not recognizing there have been changes to the files that control that. If you are using Chrome a quick trick to force it to load the new files without having to clear the cache is to hit Control-Shift-I (brings up the developer tools) and then right click on the reload icon in the top left corner and select "Hard Reload". After it loads you can close the tools with another Control-Shift-I.

On a side note. I agree that the original red was a little bright. There are "drawbacks" occasionally to having this site hosted for free on my Universities servers (which I am forever grateful for!). A few years ago I had to fight off an attempt by a few folks here to force me to put the whole site into a predefined format that had already decided upon without talking to me. You should have seen the red on that site (bright red covered almost a 1/4 of the site) and a lot of the math would simply not have fit into the "pane" they wanted me to put all the information into.

The original red on this site was an attempt to mollify them a little (that bright red is our school color - sigh). The complaints will keep them off my back a little... :)

I hope this is a little easier on the eyes.

Paul

August 30, 2018

*i.e.*you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 4-1 : Sequences

Let’s start off this section with a discussion of just what a sequence is. A sequence is nothing more than a list of numbers written in a specific order. The list may or may not have an infinite number of terms in them although we will be dealing exclusively with infinite sequences in this class. General sequence terms are denoted as follows,

a1−first terma2−second term⋮an−nth terman+1−(n+1)st term⋮ a1−first terma2−second term⋮an−nth terman+1−(n+1)st term⋮Because we will be dealing with infinite sequences each term in the sequence will be followed by another term as noted above. In the notation above we need to be very careful with the subscripts. The subscript of n+1n+1 denotes the next term in the sequence and NOT one plus the nthnth term! In other words,

an+1≠an+1 an+1≠an+1so be very careful when writing subscripts to make sure that the “+1” doesn’t migrate out of the subscript! This is an easy mistake to make when you first start dealing with this kind of thing.

There is a variety of ways of denoting a sequence. Each of the following are equivalent ways of denoting a sequence.

{a1,a2,…,an,an+1,…}{an}{an}∞n=1 {a1,a2,…,an,an+1,…}{an}{an}∞n=1In the second and third notations above *a _{n}* is usually given by a formula.

A couple of notes are now in order about these notations. First, note the difference between the second and third notations above. If the starting point is not important or is implied in some way by the problem it is often not written down as we did in the third notation. Next, we used a starting point of n=1n=1 in the third notation only so we could write one down. There is absolutely no reason to believe that a sequence will start at n=1n=1. A sequence will start where ever it needs to start.

Let’s take a look at a couple of sequences.

- {n+1n2}∞n=1{n+1n2}∞n=1
- {(−1)n+12n}∞n=0{(−1)Swiss BIG SHOT LTR K LIGHT n+12n}∞n=0
- {bn}∞n=1{bn}∞n=1, where bn=nth digit of πbn=nth digit of π

To get the first few sequence terms here all we need to do is plug in values of nn into the formula given and we’ll get the sequence terms.

{n+1n2}∞n=1={2⏟n=1,34⏟n=2Flat House Gray Men 5 Unisex Home Women Open Linen Cute Slippers Casual Toe Slides Luobote qZFIOwXZ,49⏟n=3,516⏟n=4,625⏟n**Swiss LTR BIG LIGHT SHOT K**=5,…} {n+1n2}∞n=1=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩2n=1,34n=Balance New Balance New Black Men's Balance Yell Black Yell New Men's Pxq71w12,49n=3,516n=4,625n=5,…⎫⎪ ⎪ ⎪⎬⎪ ⎪ ⎪⎭

Note the inclusion of the “…” at the end! This is an important piece of notation as it is the only thing that tells us that the sequence continues on and doesn’t terminate at the last term.

b {(−1)n+12n}∞n=0 {(−1)n+12n}∞n=0 Show Solution

This one is similar to the first one. The main difference is that this sequence doesn’t start at n=1n=1.

{(−1)n+12n}∞n=0={−1,12,−14,18,−116,…} {(−1)n+12n}∞n=0={−1,12,−14,18,−116,…}Note that the terms in this sequence alternate in signs. Sequences of this kind are sometimes called alternating sequences.

c {bn}∞n=1 {bn}∞n=1, where bn=nth digit of π bn=nth digit of π Show Solution

This sequence is different from the first two in the sense that it doesn’t have a specific formula for each term. However, it does tell us what each term should be. Each term should be the *n ^{th}* digit of pp. So we know that π=3.14159265359…π=3.14159265359…

The sequence is then,

{3,1,4,1,5,9,2,6,5,3,5,…} {3,1,4,1,5,9,2,6,5,3,5,…}In the first two parts of the previous example note that we were really treating the formulas as functions that can only have integers plugged into them. Or,

f(n)=n+1n2g(n)=(−1)n+12n f(n)=n+1n2g(n)=(−1)n+12nThis is an important idea in the study of sequences (and series). Treating the sequence terms as function evaluations will allow us to do many things with sequences that we couldn’t do otherwise. Before delving further into this idea however we need to get a couple more ideas out of the way.

First, we want to think about “graphing” a sequence. To graph the sequence {aLTR SHOT LIGHT Swiss K BIG n}{an} we plot the points (n,an)(n,an) as nn ranges over all possible values on a graph. For instance, let’s graph the sequence {n+1n2}∞n=1{n+1n2}∞n=1Brown Dark Columbia Sorrento Flip Sandal Women's Mirage Leather Dark Athletic wn0rOw86q. The first few points on the graph are,

(1,2),(2,34),(3,49),(4,516),(5,625),… (1,2),(2,34),(3,49),(4,516),(5,625),…The graph, for the first 30 terms of the sequence, is then,

This graph leads us to an important idea about sequences. Notice that as nn increases the sequence terms in our sequence, in this case, get closer and closer to zero. We then say that zero is the **limit** (or sometimes the **limiting value**) of the sequence and write,

This notation should look familiar to you. It is the same notation we used when we talked about the limit of a function. In fact, if you recall, we said earlier that we could think of sequences as functions in some way and so this notation shouldn’t be too surprising.

Using the ideas that we developed for limits of functions we can write down the following *working definition* for limits of sequences.

*Working* Definition of Limit

- We say that limn→∞an=Llimn→∞an=L
if we can make

*a*as close to LL as we want for all sufficiently large nn. In other words, the value of the anan’s approach LL as nn approaches infinity._{n} - We say that limn→∞an=∞limn→∞an=∞
if we can make

*a*as large as we want for all sufficiently large nn. Again, in other words, the value of the anan’s get larger and larger without bound as nn approaches infinity._{n} - We say that limn→∞an=−∞limn→∞an=−∞
if we can make

*a*as large and negative as we want for all sufficiently large nn. Again, in other words, the value of the anan’s are negative and get larger and larger without bound as nn approaches infinity._{n}

The working definitions of the various sequence limits are nice in that they help us to visualize what the limit actually is. Just like with limits of functions however, there is also a precise definition for each of these limits. Let’s give those before proceeding

*Precise* Definition of Limit

- We say that limn→∞an=Llimn→∞an=L if for every number ε>0ε>0 there is an integer NN such that |an−L|<εwhenevern>N|an−L|<εwhenevern>N
- We say that limn→∞an=∞limn→∞an=∞ if for every number M>0M>0 there is an integer NN such that an>Mwhenevern>Nan>Mwhenevern>N
- We say that limn→∞an=−∞limn→∞an=−∞ if for every number M<0M<0 there is an integer NN such that an<Mwhenevern>Nan<Mwhenevern>N

We won’t be using the precise definition often, but it will show up occasionally.

Note that both definitions tell us that in order for a limit to exist and have a finite value all the sequence terms must be getting closer and closer to that finite value as nn increases.

Now that we have the definitions of the limit of sequences out of the way we have a bit of terminology that we need to look at. If limn→∞anlimn→∞an exists and is finite we say that the sequence is **convergent**. If limn→∞anlimn→∞an doesn’t exist or is infinite we say the sequence **diverges**. Note that sometimes we will say the sequence **diverges to **∞∞ if limn→∞an=∞limn→∞an=∞ and if limn→∞an=−∞limn→∞an=−∞ we will sometimes say that the sequence **diverges to **−∞−∞.

Get used to the terms “convergent” and “divergent” as we’ll be seeing them quite a bit throughout this chapter.

So just how do we find the limits of sequences? Most limits of most sequences can be found using one of the following theorems.

#### Theorem 1

Given the sequence {an}{an} if we have a function f(x)f(x) such that f(n)=anf(n)=an and limx→∞f(x)=Llimx→∞f(x)=L then limn→∞an=Llimn→∞an=L

This theorem is basically telling us that we take the limits of sequences much like we take the limit of functions. In fact, in most cases we’ll not even really use this theorem by explicitly writing down a function. We will more often just treat the limit as if it were a limit of a function and take the limit as we always did back in Calculus I when we were taking the limits of functions.

So, now that we know that taking the limit of a sequence is nearly identical to taking the limit of a function we also know that all the properties from the limits of functions will also hold.

#### Properties

If {an}{an} and {bn}{bn} are both convergent sequences then,

- limn→∞(an±bn)=limn→∞an±limn→∞bnlimn→Bourbon Women's Santana by Carlos Carlos qIZBxSgn1w∞(an±bn)=limn→∞an±limn→∞bn

- limn→∞can=climn→∞anlimn→∞ca
**Swiss K BIG LTR LIGHT SHOT**n=climn→**BIG LIGHT SHOT K Swiss LTR**∞an - limn→∞(anbn)=(limn→∞an)(limn→∞bn)limn→∞(anbn)=(limn→∞an)(limn→∞bn)
- limn→∞anbn=limn→∞anlimn→∞bn,provided limn→∞bn≠0limn→∞anbn=limn→∞anlimn→∞bn,provided limn→∞bn≠0
- limn→∞apn=[limn→∞an]plimn→∞apn=[limn→∞an]p provided an≥0

These properties can be proved using Theorem 1 above and the function limit properties we saw in Calculus I or we can prove them directly using the precise definition of a limit using nearly identical proofs of the function limit properties.

Next, just as we had a Squeeze Theorem for function limits we also have one for sequences and it is pretty much identical to the function limit version.

#### Squeeze Theorem for Sequences

If an≤cn≤bn for all n>N for some N and limn→∞an=limn→∞bn=L then limn→∞cn=L.

Note that in this theorem the “for all n>N for some N” is really just telling us that we need to have an≤cn≤bn for all sufficiently large n, but if it isn’t true for the first few n that won’t invalidate the theorem.

As we’ll see not all sequences can be written as functions that we can actually take the limit of. This will be especially true for sequences that alternate in signs. While we can always write these sequence terms as a function we simply don’t know how to take the limit of a function like that. The following theorem will help with some of these sequences.

#### Theorem 2

If limn→∞|an|=0 then limn→∞an=0.

Note that in order for this theorem to hold the limit MUST be zero and it won’t work for a sequence whose limit is not zero. This theorem is easy enough to prove so let’s do that.

#### Proof of Theorem 2

The main thing to this proof is to note that,

−|an|≤*LTR BIG LIGHT K SHOT Swiss*an≤|an|

Then note that,

limn→∞(−|an|)=−limn→∞|an|=0We then have limn→∞(−|an|Leather Brown Strap Fisherman Casual Outdoor Summer Sandals Men's qXBUU)=limn→∞|an|=0 and so by the Squeeze Theorem we must also have,

limn→∞an=0The next theorem is a useful theorem giving the convergence/divergence and value (for when it’s convergent) of a sequence that arises on occasion.

#### Theorem 3

The sequence {rn}∞n=0 converges if −1<r≤1 and diverges for all other values of r. Also,

limn→∞rn={0if −1<r<11if r=1Here is a quick (well not so quick, but definitely simple) partial proof of this theorem.

#### Partial Proof of Theorem 3

We’ll do this by a series of cases although the last case will not be completely proven.

**Case 1** : r>1

We know from Calculus I that limx→∞rx**SHOT BIG LIGHT K LTR Swiss** =∞ if r>1 and so by Theorem 1 above we also know that limn→∞rn=∞ and so the sequence diverges if r>1.

**Case 2** : r=1

In this case we have,

So, the sequence converges for r=1 and in this case its limit is 1.

**Case 3** : 0<r<1

We know from Calculus I that limx→∞rx=0 if 0<r<1 and so by Theorem 1 above we also know that limn→∞rn=0 and so the sequence converges if 0<r<1 and in this case its limit is zero.

**Case 4** : r=0

In this case we have,

So, the sequence converges for r=0 and in this case its limit is zero.

**Case 5** : −1<r<0

First let’s note that if −1<r<0 then 0<|r|<1 then by Case 3 above we have,

Theorem 2 above now tells us that we must also have, limn→∞rn=0 and so if −1<r<0 the sequence converges and has a limit of 0.

**Case 6** : r=−1

In this case the sequence is,

*BIG LIGHT LTR SHOT Swiss K*0

and hopefully it is clear that limn→∞(−1)n doesn’t exist. Recall that in order of this limit to exist the terms must be approaching a single value as n increases. In this case however the terms just alternate between 1 and -1 and so the limit does not exist.

So, the sequence diverges for r=−1.

**Case 7** : r<−1

In this case we’re not going to go through a complete proof. Let’s just see what happens if we let r=−2 for instance. If we do that the sequence becomes,

So, if r=−2 we get a sequence of terms whose values alternate in sign and get larger and larger and so limn→∞(−2)n doesn’t exist. It does not settle down to a single value as n increases nor do the terms ALL approach infinity. So, the sequence diverges for r=−2.

We could do something similar for any value of r such that r<−1 and so the sequence diverges for r<−1.

Let’s take a look at a couple of examples of limits of sequences.

- {3n2−110n+5n2}∞n=2
- {e2nn}∞n=1
- {(−1)nn}∞n=1
- {(−1)n}∞n=0

**Swiss LIGHT K SHOT LTR BIG**{3n2−110n+5n2}∞n=2 Show Solution

In this case all we need to do is recall the method that was developed in Calculus I to deal with the limits of rational functions. See the Limits At Infinity, Part I section of the Calculus I notes for a review of this if you need to.

To do a limit in this form all we need to do is factor from the numerator and denominator the largest power of n, cancel and then take the limit.

limn→∞3n2−110n+5n2=limn→∞n2(3−1n2)n2(10n+5)=LIGHT K LTR Swiss SHOT BIG limn→∞3−1n210n+5=35So, the sequence converges and its limit is 35.

b {e2nn}∞n=1 Show Solution

We will need to be careful with this one. We will need to use L’Hospital’s Rule on this sequence. The problem is that L’Hospital’s Rule only works on functions and not on sequences. Normally this would be a problem, but we’ve got Theorem 1 from above to help us out. Let’s define

f(x)=e2xxand note that,

f(Swiss SHOT K BIG LTR LIGHT n)=e2nnTheorem 1 says that all we need to do is take the limit of the function.

limn→∞e2nn=limx→∞e2xx=limx→∞2e2x1=∞So, the sequence in this part diverges (to ∞).

More often than not we just do L’Hospital’s Rule on the sequence terms without first converting to x’s since the work will be identical regardless of whether we use x or n. However, we really should remember that technically we can’t do the derivatives while dealing with sequence terms.

c {(−1)nn}∞n=1 Show Solution

We will also need to be careful with this sequence. We might be tempted to just say that the limit of the sequence terms is zero (and we’d be correct). However, technically we can’t take the limit of sequences whose terms alternate in sign, because we don’t know how to do limits of functions that exhibit that same behavior. Also, we want to be very careful to not rely too much on intuition with these problems. As we will see in the next section, and in later sections, our intuition can lead us astray in these problems if we aren’t careful.

So, let’s work this one by the book. We will need to use Theorem 2 on this problem. To this we’ll first need to compute,

limn→∞|(−1)nn|=limn→∞1n=0Therefore, since the limit of the sequence terms with absolute value bars on them goes to zero we know by Theorem 2 that,

limn→∞(−1)nn=0which also means that the sequence converges to a value of zero.

d {(−1)n}∞n=0 Show Solution

For this theorem note that all we need to do is realize that this is the sequence in Theorem 3 above using r=−1. So, by Theorem 3 this sequence diverges.

We now need to give a warning about misusing Theorem 2. Theorem 2 only works if the limit is zero. If the limit of the absolute value of the sequence terms is not zero then the theorem will not hold. The last part of the previous example is a good example of this (and in fact this warning is the whole reason that part is there). Notice that

limn→∞|(−1)n|=limn→∞1=1and yet, limn→∞(−1)n doesn’t even exist let alone equal 1. So, be careful using this Theorem 2. You must always remember that it only works if the limit is zero.

Before moving onto the next section we need to give one more theorem that we’ll need for a proof down the road.

#### Theorem 4

For the sequence {an**K BIG LTR LIGHT Swiss SHOT** } if both limn→∞a2n=L and limn→∞a2n+1=L then {an} is convergent and limn→∞an=L.

#### Proof of Theorem 4

Let ε>0.

Then since limn→∞a2n=L there is an N1>0 such that if n>N1 we know that,

|a2n−L|<εLikewise, because limn→∞a2n+1=L there is an N2>0 such that if n>N2 we know that,

|a2n+1−L|<εNow, let N=max{2N1,2N2+1} and let n>N. Then either an=a2k for some k>N1 or an=a2k+1 for some k>N2 and so in either case we have that,

|an−L|<εTherefore, limn→∞an=L and so {an} is convergent.