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in this episode sequences are introduced the limit of a sequence and various

limit rules are studied including the squeeze theorem the convergence of a

bounded monotonic sequence is also explained let’s do some math

hi everyone welcome back we’re going to begin with the question what are

sequences so basically a sequence is a function

whose domain is a set of integers so remember back from earlier in

calculus when you studied uh functions when the when the input was a

real number potentially and we studied limits and derivatives of those type of

functions and in particular we talked about you know continuous functions

now we’re going to restrict our attention to sequences so a function

whose domain right the inputs are coming from the integers

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so sequence here and this will be the notation for it is a function

whose domain is a set of positive integers

um so we’ll use sometimes we’ll use this notation here a sub n

n equals one to infinity or perhaps if we want to start somewhere

else say n equals three to infinity or sometimes we’ll say um

you know like we’ll have an example like a sub n is one over n

and then we’ll say n is greater than or equal to one

and sometimes we’ll just write out a set notation where we list them all so the

functional values a1 e2 a3 a sub n are the terms of the sequence so a sub n

is called the nth term right so we have the first term the second term the nth

term of the sequence so let’s look at a couple of examples

just to make sure that we’re good so let’s look at the sequence right here

we’re going to start at 1 and so we can write it like this right here a sub n

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n equals 1 to infinity now if it’s if this is understood then we can just

write it like this and we can write it with a set notation

right here also with a1 a2 all the way to a n

and we can actually list out the values here

now that we have a formula for them so when n is one we have one over one

we have one over two one over three one over four

and so on until we have one over n and then so on

so here’s the nth term here’s the first second third and here’s the formula to

get to all of them um and so yeah we could write it uh this sequence right here

in various different ways um let’s look at another one right here

let’s look at n over n plus one and so for this one right here

we’ll just have this uh set here so when n is one we get uh one over two

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when n is two right so our domain is the is the posi

i’m starting at 1 right so when n is 2 we’re going to get 2 over 3

and then we’re going to get 3 over 4 4 over 5 and so on

and then the nth term is n over n plus 1 and then so on

so we can write it and list out the elements like this with the nth term there

or we can write it like this okay so there’s a couple of examples so um yeah so

let’s look at a couple more maybe what if we have something like

about something like um square root of n minus three

and let’s start here at three so it makes sense to naturally start

this one at three here so the this will be

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uh so you know another way to write this it would be like a to the n is

n minus 3 and then n greater than or equal to 3.

right so we can write it like this write it like this or we can just list them

all out you know when n is 3 we’re going to get square root of 0 0

and then when n is 4 we’re going to get a 1 and then when n is

3 or sorry 5 we’re going to get square root of 2

then we’ll get square root of three and so on and then we get the nth term

and then so on so you know either of these three we can write out

that example there um how about one more how about something like the sequence

cosine of n over six pi starting at i don’t know zero so

you know what’s this sequence right here right

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so we can write the terms out or we can write it out in the formula form cosine

n pi over six where n is starting at zero

we get uh when n is 0 we’re going to get cosine

uh of that and then we’re going to get a 1 and then we’re going to get square

root of 3 over 2. these are the special angles right one-half 0

and so on and then and then the nth term in pi over 6 and then so on so

there’s two more examples right there so the inputs are you know

sets of is a set of integers right so here we start at zero here we start at

three and for these two we started at one okay so

let’s reverse this now what if we’re given the list first right

so let’s look at something like this right here find the expression for the

nth term the sequence and let’s look at something like this here first let’s get

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rid of this here so i’m going to uh write it up here and say

we’re going to have one half 1 4 1 8 so this is 2 to the first 2 squared 2 to

the third 2 to the 4th and then so on so i’m going to say 1

over 2 to the n and then so on and there’s the sequence for part a right there

for part b um let’s see here for part b so this is

alternating in sine so i want to use a some kind of power of minus one

so here we’ll write it like this one half so it’s positive negative positive

and then back to negative so it’s called an alternating series

which we’ll study a lot more in detail so i’m going to put -1 to some power the

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power is determined by the index so i’m going to say something like an n let me

just try an n but i also notice that the numerator is

the n and then the denominator is n plus one so i’m going to go with an n over m

plus one and then say so on so the question is does this one right here work

so when n is one we’re gonna get one over two that’s good but when n is one

this is odd so it’ll be a minus but over here it’s a positive right it’s a

positive so i actually want to change that and say m plus one

and then try out the next one when n is 2

well when n is 1 now it’s positive right here when n is 2 now it’s odd so it’ll

be negative and we’ll get this term right here so there we go right there so

there’s the sequence for part b here all right and so now let’s um

you know these sequences are really fun to work with in fact some of them are so

fun to work with they’re famous they have their own

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dedicated people who study them there’s um this sequence right here is very very

famous so this sequence starts off with one and one

the first term is one the second term is one and we’re going to get to the rest

of them by using recursion in other words so i’m

going to get the next one by looking at the two previous ones and so this is for

one and two for the first term the second term and for all the other terms

greater than two i’m going to use this formula right here

so if you’ve never seen this before let me give you a couple of examples

so a3 here that’s when n is two because i get two plus one

so for a3 is when n is two so this will be equal to a2

and then this right here will be a two so it’ll be two minus one so it’ll be a1

so in other words we already know the first two terms so to get to the next

term a3 we’re just going to add the two previous terms so this is one plus one

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which is two let’s try that out one more time a4

so to get a four here i need n to be equal to 3 now so this will be equal to a3

plus and then a3 minus 1 which is a2 so in other words to find any one of

them we add together the previous two to

find a3 i add a1 and a2 to find a4 i add a2 and a3 the one we just found and the

previous one so a3 is 2 and this one’s 1 and so this is a 3. and so this is the

way to write it out but actually it’s very intuitive a5 is just the sum of the

previous two it’s five a six is eight a seven is a 8 is 21. a 9 is 34.

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a 10 is 55 and so on one more how about a11 just add them up you get 89

and so on and so this is called the fibonacci

sequence very famous a lot of people study it has its own

journal the fibonacci journal um yeah and so this is an example of

recursion right here right in case you haven’t seen that before

so any case our domain here is a sequence right so the domain is three

four five six seven and so on all right so let’s talk about limits now

now the limit of sequence we want this to be consistent with the limit of a

function in the sense that we have already

defined what the limit of a function is when x is going to infinity

of a function of a real variable if we restrict the domain to be just a

set of a set of integers we would like that limit to be

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whatever it is equal to l and then if you restrict the domain we

would like the limit of the sequence to have the same

uh limit here and so that’s what we’re going to do we’re going to say a sequence

has a limit written like this right the limit of the sequence

if the ace of n’s the outputs can be made as close to l as we please

by taking in sufficiently large in other words as n is going to infinity a sub n

is getting closer and closer to l n is getting larger and larger and larger and

that’s forcing the ace of n’s to get closer and closer to l

and so if this happens then we say the limit converges

and if it doesn’t happen we say that the sequence diverges

and so here’s the theorem that we i was talking about that we want if you know

that the limit of the function of a real variable is equal to l

and then if you use this function right here to define a sequence

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then when you restrict the domain of that function in other words using a

sub n is f of n then that should also be equal to l

and so i’d like to just kind of give you a simple diagram to help you

to help us understand that to make sure that our intuition on that is good

so let’s look at a function like f of x equals 1 over x

and so what does that look like right so i’m just going to restrict this to

looking at the positive part here but it looks like this right here

um in fact maybe i can make it drawn out a little bit better and make it look

something like that um and then what happens if we have f of n or let’s say a

sub n which is f of n which is one over n in other words what

if our sequence is one over n and and so here i’m going to say n is

one two three right just a set of integers right here

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i’m here and here x is just you know greater than or equal to one

right so what if we look at these two right here what’s the difference

so here i’ll still have the same one over n here but this time

when i have one i have a height when i have a two i have a height whatever three

and so on and over here though this is discrete

so it’ll look it’ll have the same shape but instead it’ll just be a set of

points here this is one and this height here is one right

and this height here is one half and this is two is one half and so on and so

you know as we’re going to as x is going to infinity

as x is getting larger and larger the height here is going to zero

well if you restrict it in other words you take out all this continuous part

here and you just look at those points right there

what is the height approaching the height still approaching zero even

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though we’re missing these pieces right here in terms of the overall behavior

it’s still going to zero right here so this limit right here as x goes to

infinity of f of x is zero and so sin and this what this theorem is

saying is that if this happens and i use the sequence defined like this then

that implies that the limit as n goes to infinity of

one over n is zero so that’s what this theorem right here is saying i hope that

shed some light on that for you all right so let’s look at some limit laws

so you remember limit laws from calculus one

we have uh two sequences now instead of two functions so we have two sequences

um instead of functions of real variables now we have functions of of

integers we have two sequences and we’re going to assume that they’re convergent

so in other words i’m assuming that this limit right here exists

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and it’s equal to some number let’s call it l and this limit right here

of this one right here the sequence right here b sub n

also exists and it’s equal to some let’s call it m

and let’s move that l down just a tad uh right and so these both exist that’s

what this is saying it’s convergent and c is a constant

so here’s our limit laws um that you might remember from calculus one but now

it’s worded in terms of sequences so how do you find the limit of a sum of two

sequences well you can break it up and add up the limits um

find this limit that’s equal to l find this limit is equal to m and just say l

plus m right so this is l plus m so we can find this limit right here now keep

in mind though you can only use equality

right here you only know these are equal whenever you know that these limits

right here are actually equal to uh exist right here

now we have the same thing for subtraction

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and just like limits of real variables we can pull out constants right so c is

a constant it doesn’t depend upon n we can pull it outside the limit and

then we have the same for products so the to find the limit of a product of

two sequences we can find the limits of them individually and then multiply them

and we can do the same thing for quotient

of course providing that we don’t divide by zero and then the last one here is

we can uh pull the limit inside of a power as long as the power is positive

and your sequence is positive here so we

can just slip this limit inside and then take the power and that simplifies

things a lot because if you know your your terms uh have a power on them

then we can just bring the limit inside and then raise the power after the fact

after finding the limit okay so let’s look at a couple of examples now

so for our first example we’re going to look at the sequence here uh n over

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n plus one two n plus one um and so you know what’s our feeling

that it’s going to converge or diverge so we can start looking at some values

right here so what what are some numbers in the sequence right so when n is one

what are we going to get we’re going to get one over and this will be 3

and when n is 2 we’re going to get 2 over 5

when n is 3 we’re going to get 3 over 7 and so this may not give you much of a

feeling for the sequence right here so sometimes writing out terms will help

you sometimes not necessarily um but we were trying to decide if it’s

convergent or not so writing out the terms like that let’s just take the limit

so what’s the limit as n goes to infinity of a sub n here so that limit is

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right how do we take this limit here now what i’m hoping is that you remember

um how to do this right here when we have a real variable

so how do you find this limit right here right think back to previous calculus

sorry limit as x goes to infinity of x over two x plus one right think

back how do we find this limit one way is to say oh this is infinity over

infinity we can use lavato’s rule or you may think back even further

and say oh i’m going to divide by the highest power

so we’ll divide the numerator by x we’ll divide the denominator by x and we’ll

work out some algebra well let’s take that approach here so

i’m going to divide the uh sorry the numerator by an n and the denominator by

an n so remember n is just some really large

uh integer for example n is 200 billion right so then i divide by that right so

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in other words n is not zero it’s it’s just a really large and so and so what

happens here um so now if we like to we can

apply our limit laws step by step so we can say this is the limit as n goes to

infinity of the numerator divided by the limit in the denominator

2 n plus 1 over n and we can simplify this now that’s just one

and we can take this limit right here out and say this is two over n

two n over n which is two and this will be the limit right here as

n goes to infinity of one over n and so this will be just one over two plus

zero or so differently one half so you know depending on how many steps

you want to show there are you required to show every limit law uh usually not

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you do enough examples out maybe you just need to step here

the limit as it goes to infinity of n over two n plus one is obviously one half

check it’s convergent right so you know you can work out the limit

laws and get the same answer or you know you do enough examples you

get your intuition and you just got the limit in any case

this is convergent here and let’s see another one

what about this one right here now so this one is just like that previous one

the only difference is we have the uh alternating sign here

so if we were to uh work out some values here

when say n is one then we’re going to get a positive we’re going to get a one

over uh sorry one over when n is one we get a three when n is two

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now we’re going to get a minus sign here so i get two over five

and then now when n is three now we’re back to an even so it’s a positive so

when n is three i’m gonna get three over seven and we can just keep going and

you know if we keep going i guess the point is is that um

these are the same numbers that we’re getting before only difference is some

of them are evens and some of them are odds so this is the same sequence in the

previous example except for the factor of minus one to

the n plus one that’s that’s what’s different right there so the

sequence is going to oscillate oscillate between

the value that we got last time the month of the one-half and the minus one-half

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so you know if we look at the even terms being uh negatives so that’s going to

give us the minus one half so we got minus two fifths we got the next one the

next one the next one and so we have um a sequence there that’s going to minus

one half and then we have another one that’s going to one half if you just

look at the even and odd terms right so this is coming from the um odd terms

the first term the third term that’s going to give us the positive

it’s going to go to the positive one half so these are the odd terms

and these this is what’s coming from the even terms

and so the limit is going back and forth between these two values right here so

here’s here’s an example of one where we’re going to have to say we diverge

right here so if that’s confusing to you let me

just kind of maybe um refresh your memory in terms of maybe you saw

something like this in in calculus one or earlier in calculus say you have

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something like a sine x and you have something like x times sine x

how are these different right here right and so the sine x is just going up

and down and so as x goes to infinity what’s

happening to this right here does this converge to anything

right so as x is going farther and farther out it’s just going to 1 and

then back down to minus one back up to one back up to minus one back down to

minus one it’s just going up and down it doesn’t ever converge to something this

one right here is different and it’s you know as you’re getting closer um

yeah so this one is just going to fly off here it’s just going to it’s going

to come in here like this like that but it’s it’s going to go up like this

so it’s x is going to infinity this one’s just going to fly off right there

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so very different type of behavior um this one’s blowing up whereas this

one’s diverging because of oscillating behavior in any case

let’s look at another example another one here that’s oscillating here

between positive and negative and so when we look at this right here we’re

going to say 1 over n goes to goes to 0 as n goes to infinity

and so what is this product here minus one to the n plus one

one over n that right here oscillates between positive and negative values

and so the odd numbers are approaching zero odd terms approach zero

and the even terms also approach zero the even terms approach zero also

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so this one the uh what we’re approaching even and

odd is the same so in other words as we go to infinity here

this minus one to the n plus one over one over n

is zero right here so in this case the sequence converges and it converges to

zero and i i’d like to take a look at this in

more detail right here so that says approaches two p’s

even terms approach zero the odd terms approach zero unlike the previous

example where it was one half and minus one half so it’s oscillating between two

different heights here on this one the even terms are coming down to zero and

the odd terms are going to zero also so the whole limit right here is zero so

the sequence converges to zero right here

now let’s compare that to this one right here

if we take the limit right here on this one as n goes to infinity of 3n minus 2n

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and you know this right here is blowing up or

down if you want let’s go into minus infinity right here so this one right

here the sequence diverges it’s certainly going to diverge

and so there’s really nothing else to say here this diverges here

what about the sequence right here so let’s look at the uh limit as n goes

to infinity and we’re going to look at n over 2 n plus 2 2 to the n

right and so when we try to take this limit here how would we do that

so here i’m not going to divide by the highest power

instead this reminds me of something that we would do

when we were studying functions of a real variable here

and you know if we had something like this right here how would we try to

solve this problem here here so you know you try to use the hopital’s

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rule right you try to take derivatives we have an indeterminate form infinity

over infinity so what we’re going to do here is we’re

going to say and we know if this limit exists let’s say it’s equal to l

if we can find this l then we know that this limit right here

would be equal to l by the by the first theorem that we talked about

so i want to try to find this and using the habitats rule

and so here’s what we’re going to do i’m going to say f of x is equal to x

and that’s for the numerator and g of x is 2 to the x

so now i have the limit as x goes to infinity of x over 2 to the x

and that’s the limit as uh x goes to infinity you know using the hopital’s rule

1 over and then we’re going to have 2 to the x and then times right so the base

is 2 so natural log of two and so what’s happening is this is now fixed at one

and this is getting larger and larger and larger right ln2 is a fixed constant

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but this is going larger and larger right here so this whole limit here is

going to zero right here so um by by the first theorem that we

talked about by theorem uh we can conclude that the limit as n goes to infinity

of n over two to the m is in fact zero so you know we can use the habitat’s rule

um we find this limit right here by looking at this right here and we found

it l to be zero there all right so is there anything else

besides limit laws well you might remember previous calculus

um that we had something called the squeeze theorem

and these right here were functions of a real variable um

and so now we’re going to try to squeeze these two sequences right here or

they’re going to try to squeeze the b this limit right here

so if i know that i have this inequalities here

and i pass the limit then this limit right here is equal to l

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and this limit of this right here is also equal to l then that will force

the limit of the b sub n’s to also be equal to l

and so let’s see that example on something like this a cosine where we

have some factor right here so the way to do that is to say

or to start with something with cosine right so because we need a way to bound

so i’m going to start off with -1 cosine’s between minus one and one

and this is for all in actually and so we just need this to be true

starting at some number and so you know i can use a one here

for all n greater than or equal to one we don’t really even need to say that we

could just say for all in but you know to try to match the squeeze theorem

because cosine no matter what you input whether you input integers or real

numbers it’s going to be bounded between -1 and 1.

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now i don’t have minus 1 here though you know i have

i don’t have cosine in so i need to involve this somehow so i’m going to

take this right here and divide it up all three of them so here we’re going to

get minus 1 over n squared plus 1 and cosine n over n squared plus 1.

now the reason why we can do this is because this is positive right

when n is greater than or equal to one the n squared is positive plus one add a

little bit more positive to it this is positive so right so the inequalities

are not going to change sign and so now we have what we need we have

an ace of n right here and we have a b sub n right here that

we’re interested in the middle one is what we’re trying to trap

and the c sub n is this one right here and now we need to know that the limit

of these two are equal to each other and we’ll find out what that is so

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here we’re going to say that the limit as n goes to infinity of the a sub n

is that equal to question mark the limit of this one right here

of the c sub n here this one on the right here

and yeah they are both equal to what they’re both equal to each other right n

is getting larger and larger and larger this is a fixed number up here so this

is going to zero this one’s going to zero they’re both going to zero

and so by the squeeze theorem so let’s say by the squeeze theorem

the limit of the middle one here that’s squeezed which is this one right here

which is the one that we’re interested in has to be zero also

and so yeah we can use the squeeze theorem just like we did for functions

of real variables we can find the limit of something that we’re squeezing

all right um in fact let’s look at another um example here because i wanted

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to talk about when we have the positive and negative terms before

so let’s look at another example here let me get this out of the way real quick

all right so squeeze theorem again uh but now i’m going to show that if the

limit of the absolute value is zero then the limit of the ace of ends is zero

um and so this says if i put the absolute value on

and that’s zero then we can just conclude that without the absolute value

it’s also zero right here and so the way that we’re going to do that is by using

the squeeze theorem and you might guess that that’s true

because remember absolute value is based in terms of inequalities so i’m

going to start off by saying that minus the absolute value of a sub n

is less than or equal to a sub n and the reason why is because

if i put absolute value on it i’m guaranteeing it’s positive and then i

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put a minus so this is guaranteeing that this is minus

um and then that’s going to be less than or equal to i’m really guaranteed that

these are positive right here these are positive right here who knows about the

ace of n’s but these are the negative ace of n’s these are the positive ace of

n’s right and i can take the limit of the these

two right here so as n goes to infinity of minus these right here

is that equal to this limit right here of the positive ones here so

we’re assuming if this right here is 0 well

remember the limit laws this right here is the limit as n goes to infinity the

minus one right here is a constant i can pull it out and say minus

minus one times the limit and this is minus zero which of course is 0. so all

these are equal to each other there’s no question anymore

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right so all these are equal this is 0 this minus 0 that’s minus

and i can bring the limit side so i have this limit right here this

left side is zero and the limit of the right side is zero and both of these are

zeros because of the assumption and so now we can say by squeeze theorem

squeeze theorem we have the limit of the middle one right here

got squeezed we found the limit of this one right here we found the limit of

this one right here and so the middle one here just a sub n

without absolute value that has been squeezed to zero so this is a nice uh

example right here because it’s saying that if you put absolute value on and

you get zero then it’s zero without the absolute

value and the reason why this is nice is because you know when you have um you

know sequence with positive and negative terms in it it’s a lot easier to just

think about in terms of absolute value right there

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all right so um the next question is do sequences

and continuous functions play nice with each other

and so here’s an example right here or here’s the theorem right here

so if we know that the sequence is l right so we can find this limit and we

know it’s l and if we know the function is continuous

then we can actually apply the limit of the function and it’s going to be f of l

and so this is nice right here because what it’s saying is that we can find the

limit of the sequence separately and then we can just take the function

of both sides and we have the limit right there

and so i’m going to use this right here to show that the

converse of what we just showed is actually also true

that if you have the limit of the ace of n’s and they’re zero

then the limit of the absolute value one sequence right there also must be zero

so let’s see how to do that so first off i’m going to assume let’s

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write it right here assume that this right here is true right that’s our

if so assume that this limit as n goes to infinity of a sub n is zero

and use i’m going to use this f of x is absolute value of x that’s going to be

my continuous function so remember that’s continuous at zero

in case you don’t just visualize it right there’s absolute value coming

right through there right just absolute value of x

and yeah it’s continuous at zero and so therefore by this theorem right here

the limit as n goes to infinity of the absolute value of the ace of n’s

is equal to the absolute value of the l and the l is

zero right here right so the limit is equal to l so the l is right here so

this is going to be absolute value of 0 which we know is 0 right here so if we

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know this is true then we get the limit of the absolute values is in fact 0

right there so now we have the complete understanding between the absolute value

of the sequence and the sequence when we’re looking around zero

all right so now we’re going to talk about monotonic sequences

so a sequence of natural numbers one two

three four and so on is is is an example of a strictly increasing sequence it’s

just getting larger and larger so this is also an example of a strictly

increasing sequence we have point five at 0.75 and so on

so this is a constant sequence it’s just four four four four

and this is an example of a non-decreasing sequence in particular

and so we’re going to say more generally

a strict strictly increasing sequence is

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when the terms are strictly greater than the previous one so a2 is strictly

greater than a1 a3 is strictly greater than a3 and so on

and we’re going to say it’s increasing when these could be potentially equal

here right so increasing is potentially strictly increasing

but it could also be constant they could all be equal here

similarly a sequence is called strictly decreasing

when the terms are getting strictly smaller than their previous ones

and it’s called decreasing when it’s potentially got some equal signs in here

and the sequence is called a monotonic and let me see if i get out of the way

here it’s called monotonic when its terms are non-decreasing

which is this one right here in other words increasing

so it’s usually said non-decreasing right so you could have all equals in

here you could have like four four four four that would be monotonic and when

00:39

it’s terms or when it’s terms are non-increasing right there so if one of

these two are holding we’re going to call it monotonic right there

all right and so let’s look at an example here

so show that the sequence right here is a strictly increasing sequence

so i believe i said that on the previous screen here

this right here is a strictly increasing sequence so let’s prove it

how do we prove this right here so to be a strictly increasing sequence

we need to to prove this right here and how can we do that right

so what we’re going to do is we’re going to let uh let me move down here we’re

going to let i’ll start over here i guess so let a sub n be equal to

n over n plus one right that’s we have right here

and i’m going to look at n plus one and that’s going to be n plus one over n

00:40

plus two right if that’s a sub n our nth term then our next one is

n plus one over n plus one plus one right and so what i’m gonna do is i’m gonna

look at a sub n plus 1 minus a sub n and i’m going to decide what that is

right so here’s n plus 1 [Music] over n plus 2 minus

and then the a sub n was n over m plus 1

and i’m going to look at this right here and just get a common denominator and

simplify and what we’re going to get is 1 over right we’re going to get some

cancellations in the numerator and we’re

going to get m plus 1 times n plus 2 the common denominator and if you notice

that this is always positive right this is 1

and remember the ends are like 1 2 3 4 and so on so this is always positive

this is always strictly greater than zero and so this is a really nice way of

00:41

thinking about the difference right there this difference right here is

always greater than zero so what that means hence

a sub n plus one minus a sub n is greater than zero

right that’s just connecting the dots so a sub n

is greater than a sub n a sub n plus 1 is greater than a sub n so in other

words we have strictly increasing right there

that’s just exactly what we needed or if you know if you want to put it a sub n

is strictly less than the a sub n plus one so we’re strictly increasing

another way is to perhaps think about this uh f of x as x over x plus one

as a function of a real variable and you know use the first derivative test

um you know something like that might be um interesting to look at but any case

00:42

um you know here’s here’s all you need right here let’s look at um

what are bounded sequences now what are bounded sequences

um so before we get to balance sequences let me just say that

this video is part of the series uh sequences and series

infinite sequences and series complete in depth tutorials for calculus so look

below uh in the description for the link to the entire series

and in the next episode we’re going to go over in more detail bounded and

monotonic uh sequences um and so this video is just kind of an introduction to

sequences all right so the sequence of natural

numbers has no upper bound it’s just going to keep getting larger and larger

and larger so this one is certainly not bounded

the sequence right here which we keep looking at right here is bounded above

by one so no matter how far you go out every term is less than uh one

00:43

or equal to one right or less than one so in fact one is the least upper bound

in other words if you take any other upper bound

one will be less than or equal to that upper bound

so the constant sequence right so this sequence this sequence hey what about

this sequence the constant sequence is bounded above and below by

four you know it’s just four four four four

all right so let’s uh make a definition here so we have a sequence

and a sequence is bounded above by some number m

if every term in the sequence is less than or equal to m

so every term in the sequence for all in and then m is called the upper bound of

the sequence right so every so m is a upper bound

and then we also have bounded below if and let me move out of the way here

so if you can bound the sequence below by some real number

00:44

so notice that that the m and n are real numbers

like 0.5 one half whatever real number and but this is true not just for sum of

the ace of n’s but for all of the terms of the sequence then it is called the

lower bound so you know you may have a sequence right here

and it’s just jumping all over the place perhaps who knows what it’s doing

but it’s staying less than or equal to all the terms are

staying less than or equal to m maybe it’s getting really close to m

maybe in fact equals to m but it does not get any larger than m

and then over here all the terms are staying greater than or equal to n

and so you know it may actually hit the n

but it does not get any less than the n so there it’s bounded it’s stuck between

the n and the m it’s trapped and if you have such a case

00:45

the sequence is called bounded when it’s bounded above and bounded below

and so this is what we’re going to mean by bounded i like to think of it as

being trapped right the sequence could be doing all kinds of things

now what we’re going to look at though is a sequence is actually headed in one

direction but it’s bounded then that’s particular type interesting

type of sequence if it’s headed a certain direction but it’s bounded right so

um we’re going to have something called the bounded convergence theorem

if a sequence is bounded it’s trapped it has an imminent end

and it’s monotonic it’s headed a certain direction or it’s stuck

as a constant one then it converges and this is a huge theorem

and yeah so and this is just kind of an introduction to it we’re going to talk

about this in the next episode more but let’s work out one more example here in

00:46

this video and just kind of get a feeling for how this works right so we

need to know that it’s bounded and we need to know that it’s monotonic

then we’ll have that it converges now this is a huge theorem but the downside

of this theorem is it doesn’t actually help us find the limit it just says it

has to have a limit sometimes it can be very difficult to actually find the

limit but for this one right here we’ll be able to actually

know that it has a limit and we’ll actually be able to find it too so here we go

so the first thing i’m going to do is i’m going to let a sub n be equal to my

sequence i’m just going to declare what a sub n is

so it’s going to be e to the n over n factorial right there’s our

there’s our sequence and now what i’m going to do is i’m

going to try to show that it’s monotonic first

um in fact i’m going to get a nice result here so i’m going to look at the

ratio of the terms so a to the n plus 1 over a to the n

i’m going to look at the ratio of the terms right here let me scoot that over

00:47

a to the n plus 1 a to the n and so what will happen if i plug in n

plus 1 here so i get n plus 1 n plus 1. so this will be e to the n plus 1 over

n plus 1 factorial and that’s just for the numerator right and now divided by

the denominator and i wish i wrote smaller font size

i’ll just do it again n plus 1 over n plus 1 factorial

and then over and now the a sub n is just e to the n over n factorial

there we go so now let’s just write this out a

little bit differently um m plus 1 factorial times n factorial over

and e to the n right here and maybe i’ll move down here now all right and so

you know it’s just dividing by the reciprocal or multiplying by the

reciprocal i mean but in any case um we’re going to get n plus 1 factorial

00:48

that’s 1 times 2 times 3 times n times n plus 1

in other words you know when we factor this i’ll just write this out here real

quick i don’t want to lose anybody here um so this is just 1 times 2 times

all the way to n and this is 1 times 2 times delta all

the way to n but this has an n plus 1 so times n plus one

point is is that when you cancel everything

everything cancels and we’re left with just a one on top and an n plus one down

here so when i look at this over this i’m getting an n stuck down here

so i’ll write that as 1 over n and then what about e to the m plus 1

over e to the n e to the n plus 1 think of that as just e to the n times e

and so if i divide by e to the n you see the e’s to the n’s cancel and so

00:49

this is just times e or said differently this is just e to the

oh i forgot my n plus 1 didn’t i i explained how it was m plus 1

left over but i forgot to write it and so this is m plus 1 here

all right so the ratio of a to the n plus 1 over a to the n is e over n plus one

so what we can do with this is you know we can write um [Music] so

we can write this as a to the m plus one is equal to e

let me scoot this down here a little bit 8 to the n plus 1 is equal to e

n plus 1 and then 8 to the n um and so this is a nice formula right

00:50

here i’m going to put a box around this right here so all i did was say 8 to the

a a sub n plus 1 over 8 to the n is equal to this right so multiply both

sides by 8 to the n a sub n in any case we get this nice formula right here

now notice that this tells us that a sub n plus 1 is less than a sub n

you know if you compare what these two are right here

and this is going to be true for all n greater or equal to one

and so what we have is that um you know this is going to be bounded above

rounded above by e and how do i get that well when n is one

for example we’re going to get e to the one

over one in other words we just get out e

in other words the sequence starts at e right if we start writing it out it’s

00:51

going to be e and then and then but each of the terms are getting smaller

so you know each of the terms are getting smaller than the previous this

is getting smaller and smaller and smaller so it’s bounded above by e

and it’s bounded below by can you guess bounded below by

notice all the terms in this sequence are all positive right e to the n over n

is positive n factorial that’s positive so they’re all positive so if they’re

bounded below by zero so what’s happening on this sequence

right here if you want to think about it on the real line

is we got zero right here we got e right here and it starts off at e is as the

first output and then it keeps getting you know it’s it keeps getting smaller

and smaller and smaller but they’re all positive and so is this is bounded

it’s monotonic right because it’s it’s strictly decreasing

00:52

that’s we got right here you know if you want to write it like a sub n is less

than a to n plus one right n plus one it’s strictly you know

um or greater than sorry right so the next term is strictly

smaller than so it’s bounded it’s strictly decreasing which means it’s

also monotonic so this theorem right here plays an important role as bounded

monotonic it has to converge so now the question is what does it converge to

um it looks like it converges to zero because

um no matter how far out we go it’s gonna get smaller and smaller and

smaller but it’s not gonna get past zero

right so but we kind of need a way to to figure that out so let’s move away from

intuition and actually just pass a limit right here so

let’s say the limit as n goes to infinity of a to the in a sub n plus one

00:53

and that’ll be the limit as n goes to infinity of e n plus one a sub n

and this right here is now we can break it up you see saying limit laws

so e n plus one and then times the limit as we n goes to infinity of a sub n

and now equals this limit right here is 0 times this limit right here

now we don’t actually know what this limit right here is except that it exists

and you know that’s what’s important right here it exists

bounded convergence theorem and so exists times zero is zero

and so this sequence right here the limit of the sequence right here has

to be zero right here and so yeah this is our first example here in this series

where the bounded convergence theorem is used and actually even more we actually

00:54

were able to find the limit right here so i want to say thank you for watching

and i’ll see you in the next episode have a great day

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