# Introduction to Sequences (for Calculus)

(D4M) — Here is the video transcript for this video.

00:00
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

00:01
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

00:02
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

00:03
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

00:04
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

00:05
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

00:06
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

00:07
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

00:08
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

00:09
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.

00:10
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

00:11
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

00:12
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

00:13
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

00:14
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

00:15
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

00:16
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

00:17
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

00:18
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

00:19
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

00:20
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
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

00:21
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

00:22
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

00:23
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

00:24
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

00:25
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

00:26
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

00:27
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

00:28
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

00:29
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.

00:30
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

00:31
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

00:32
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

00:33
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

00:34
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

00:35
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

00:36
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

00:37
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

00:38
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

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
if you enjoyed this video please like and subscribe to my channel
and click the bell icon to get new video updates