BAM! Mr. Tarrou. In this calculus lesson we
are going to be taking a look at four examples
involving the integral test. The first example
is going to be one where we do the nth term
test and make sure that the limit as n approaches
infinity of a sub n is actually zero opening
the door for the possibility of the series
to converge. Turns out it's going to diverge.
It's not too difficult of a question. the
second example though is going to be a series
that, we are going to prove with the integral
test, does converge and give us a little bit
of a challenging review and with our examples
that involve integration by parts and a little
bit of L'Hopital's rule, so it will be somewhat
involved example that will help you along
with your homework and then the last two example
are going to go very very quickly because
we are going to just remember that before
we try to apply an integral test or any of
these tests that we are going to learn in
this chapter and when we are looking for whether
the series converges or diverges, let’s
make sure that we check the nth term test,
make sure that the limit of a sub n is equal
to zero as n goes to infinity. right.. save
yourself some work!
This section of our textbook also includes
the p series test. I am going to have a link
to that lesson in the description just as
soon as it is done, well hopefully by the
time you are watching this, it will be. The
integral test. If f is positive, continuous
and decreasing for x is greater than or equal
to zero and a sub n is equal to f(n), basically
what we are saying here is that we are going
to be an integral test so we are going to
be setting up, sort of like a let f(x) equal
a sub n so that we have some kind of function
that we could lay down on graph on top of
the graph of the sequence and everywhere that
n, the positive integer, the domain of our
sequence is, the function is going to have
the same value or agree with the terms of
our sequence. So if that is the case, the
series of a sub n or the summation of a sub
n from n going from 1 to infinity and the
definite integral from 1 to infinity of f(x)
dx those are both going to either converge
or diverge well because they are very closely
related, we have a sum of a series and we
have a limit of sums. Or looking over here
again with our picture, I have got some blue
dots showing values that we have in a sequence
that ultimately, adding up, letting that end
go to infinity or talking about that series
right? And does that series converge or diverge.
Well when we lay that function f(x) on top
of it, it agrees that our f(n) equal a sub
n, we can look at the integral and we have
some inscribed rectangles and some circumscribed
rectangles that I deal way back in, well in
my textbook that I teach from, in chapter
four when we first start learning about integration
and estimating areas bound between a curve
and the x axis using a finite number of rectangles,
partial sums with our sequences, we have the
idea that we have summation from, i starting
at 2 going to n of f(i) is equal to this inscribed
area, we have the summation from i starting
at 1 to n-1 of f(i) is our circumscribed area
and somewhere between the lower sum and the
upper sum, we have the definite integral from
1 to n of f(x) dx so just sort of tying the
idea that yeah we can look at determining
whether a series converges or diverges and
actually do it with an integral. You might
be bothered a little bit here when you are,
I am not giving a full proof, I am just trying
to give you a visual on these statements to
help you relate the series and the definite
integral together that when we talked about
estimating areas with a finite number or rectangles,
we had some kind of f(i) notation and then
there was delta n with the rectangles cause
of course our function is going to give us
the height but with the domain of our sequences
and series being positive integers the widths
of these rectangles are always going to be
one and that’s why maybe you don’t see
a delta x here, not a delta i, delta n with
these notations that you might have been remembering
that you saw when you first started looking
at Riemann sums and estimating areas when
we first built, sort of started building the
foundation of definite integrals which of
course doesn’t always actually, often doesn’t
apply to area but that’s how we first started
building that understanding. So here we go,
let’s get to our first example right now!
So for our first example and I must apologize
for the sound quality of the first part of
this video, I am actually shooting every example
over. I don’t know what happened with my
camera abd its working with my wireless mic
system, but hopefully this works with my backup
camera. We are going to, let me read the directions
in some kind of, now sort of reshooting all
this, confirm that the integral test can be
applied to these series. If yes then use the
integral test to determine the convergence
or divergence of this series. So in this example,
I didn’t actually really give us a notation
for the series, I have just simply given us
sort of an expanded version of the series
where the first term is 1/4 plus 2/7 plus
1/4 plus 2/19 so on and so on with the little
dot dot dot, we are talking about adding forth
basically forever, an infinite number of terms,
thus it’s a series. So that means we need
to take the n values of 1, 2,3,4,5 and so
on and come up with an nth term just like
we did when we started reviewing sequences.
Okay, well it looks like the numerators are
all matching n except for this third term,
well that one term, that third term seems
to be standing out so if that numerator and
n are matching for every single one of the
first five given terms, then maybe we can
just come in here and multiply this fraction
by an equivalent value of 1 and make it 3/12
and just to save some space even though that
was not much room because well I had a big
set of directions written up here and now
I am just hoping that this video works, we
have 3/12. I don’t know if you are a regular
watcher of my videos, but I have been posting
a lot of videos about sequences and series,
this is my older camcorder that I used to
use make my YouTube videos and now I use a
Nikon DSLR which.. Maybe it’s starting to
not work very well. If you happen to like
the quality of this video than the one I just
posted last week, give me a feedback below!
Okay so it seems that we have an a sub n term
of well the numerator now is matched n, and
by the way when you come up with an nth term
expression doesn’t mean it’s the only
answer, someone else can come up with another
a sub n expression that gives those values
in the sequence or in this case because it
is infinite addition, a series and it still
works so this is a, a version or one of the
possible terms. Now how do we get the n values
of 1,2,3,4 and so on to give us the values
in the denominator if you want to give that
a shot, just pause the video and try it yourself,
it does take practice to be able to get these
a sub n terms but just to not languish you
too much, 1 squared is 1, 2 squared is 4 , 3
squared is 9 , 4 squared is 16, every time
I take these one of these n values, I am three
units or I need to add 3 to it to get our
denominators. So we have a sub n is equal
to n over n squared plus 3. so that means
our series is going be summation where n starts
at 1 and goes to infinity of n over n squared
plus 3. Great! Well can we apply the integral
test? Let’s not worry about trying to apply
the integral test and let’s first make sure
to take the nth term test. I mean can we just
let n approach infinity and prove that a sub
n is not equal to zero thus this thing absolutely
positively diverges? Well this is... hopefully
by now not expecting too much work with these
problems, the numerator has a degree of one,
the denominator has a degree of two so this
is going to approach, as n approaches infinity,
this expression, the a sub n term is going
to approach zero. That doesn’t mean the
sequence, excuse me, that doesn’t mean the
series converges but it definitely, if this
were not equal to zero would guarantee that
it would diverge so this means, let’s go
ahead and continue on. we can divide the numerator
and denominator by n squared, we can apply
L’Hopital’s rule because this is infinity
over infinity as the condition check but at
any rate, we do have a series that may converge
so let’s go on to that integral test. We
are going to go over here and we are going
to let f(x), don't forget to say 'Let f(x)’
equal x over x squared plus 3. now I am going
to try and not remember if I have said something
of that or not in the intro because I have
shot this entire video but we are setting
up that function, that f(n) that agrees with
a sub n any time that n is a positive integer
and this needs to be positive. Okay well if
you are letting x represent basically n, well
then this is a continuous function, really
really interested about when x is 1,2,3,4,5
and so on. So this is always going to give
us positive answers, it’s continuous, the
denominator is never going to be equal zero
and even if it was possible for this denominator
to equal zero, that wouldn’t be a problem
or there won’t be a problem as long as it’s
not somewhere along our domain that we care
about as far as series. For example if it
was undefined at x=-2. Not a problem! So we
are going to let, we are going to find the
derivative of f(x). It’s going to be low
derivative of high. so we are looking at x^2 +3 times
1 minus high derivative low the numerator is x, the
denominator, the derivative of the denominator
is 2x over low low, x squared plus 3 squared.
Okay, so we are looking at x squared plus
3 minus 2 x squared over x squared plus 3
square. 1 minus 2 is -1. Now, is the first
derivative negative? Is this function monotonic
decreasing or non-increasing?
The denominator is always going to be positive,
all we have to do is see where the numerator
is going to be less than zero. Subtract both
sides by 3, divide both sides by -1, square
root. And x has to be greater than the square
root of 3. Which is one in change so we have
a derivative which is negative for when x
is greater than or equal to 2. Now this original
series we are trying to determine if this
diverges where n starts at one. But the first
few terms in the series does not determine
or influence whether or not the series converges
or diverges. So it’s not an issue whether
we have the first derivative that is decreasing
or negative once you get to 2 and beyond.
That's not an issue. Convergence and divergence
is about what happens as n approaches infinity.
so that is not going to be an issue which
means that we have f(x) is positive, its continuous
and its decreasing as long as x is greater
than or equal to two. That means we can apply
the integral test to see whether this thing
converges or diverges. All right let me get
this out of the way and make sure I have plenty
of room. And we have the definite integral
from 1 to infinity of, don’t know really
why I am using parenthesis x over x squared
plus 3 dx. Because the fraction bar's a grouping
symbol so really no reason for another grouping
symbol. The denominator has a degree of two,
the numerator has a degree of one, so probably
can make a u prime over u situation show up.
So if we let u equal x^2 +3 then du is going
to be 2x dx. And I don’t want to go through
a whole u substitution. I just want to like
a scratch note over the side cause I don’t
want to change the lower and upper limits
if I don’t need to cause this is a very
easy chain rule to finish. I just need a 2
in the numerator. So let’s just go ahead
and introduce that. If I am going to introduce
a multiplication of two, of course we need
to balance that with a division of 2 and here
we go, we have u prime over u. well the integral
of u prime over u is the natural log of the
absolute value of u. Excellent. Now don’t
really need the absolute value symbols there
because again x is, well we have x squared
plus 3, that’s always going to be positive
anyway and this x is really representing,
we care about working with the sequence when
n has got a domain of all positive integers.
So they are redundant. We have 1/2 the natural
log of infinity squared plus 3 minus the natural
log of one plus 3. Well whatever. That’s
infinity! So when we take the definite integral
of f(x), we get infinity. The definite integral
is increasing without bounds. So thus the
definite integral diverges and therefore the
given series diverges.
BAM! And that is the end of our first example.
Example number two coming up right now. And
for Example number two, we have the series
of natural log of n over n to the fourth power.
And over here I have got the convergence,
it is just a summary of what I said with the
issue of when we are doing the checks. The
convergence or divergence of a series is not
affected by deleting the first n terms. If
all conditions are met for all x which is
greater than or equal to which is greater
than 1, you can use the integral from n to
infinity of f(x) dx to test for convergence.
So having to delete or having an issue with
your first couple of terms there like we had
with the first derivative not being negative
until x was greater than or equal to 2 is
not an issue. So let’s make sure... little
extra coffee to keep me going here! Let me
make sure that it is not a waste of time to
apply the integral test with our problem here.
In other words, let’s do the nth term test.
The limit as n approaches infinity of the
natural log of n over n to the 4th. We are
going to have a very similar expression later
on in this example, I am going to show all
this work once. We look at this and we let
n approach infinity well that’s going to
be infinity over infinity which is a form
of indeterminates, a form which allows us
to apply L'Hopital's Rule. Which means can
we take the derivative of the numerator and
denominator separately and get this into a
format that will easily allow us to evaluate
the limit. So we have the limit as n approaches
infinity, I hope you are really comfortable
with that by now, since we are talking about
series and sequences. But the derivative of
the natural log of n is 1/n, derivative of
n to the fourth is 4n^3 and we can clean that
up and all we care about is finding the limit
as n approaches infinity so 1 over infinity
is zero. Over infinity the cube power is going
to be infinity to the fourth which is 0 over
infinity which is zero. So, then again this
doesn’t determine if this converges but
since it is. Is equal to...whoo! Is equal
to zero, then we have the possibility of that
series converging? If that were 2 or 3 or
something like that or it wasn’t equal to
zero, we'd be guaranteed that this diverges
and continuing on would be a waste of time
and effort. So yes, we might be converging!
Awesome! Alright so we are going to let f(x)
equal to the natural log of x over x to the
4th. F(x) needs to be positive, needs to be
continuous and needs to be decreasing. well
when n is starting at 2 and going up, we are
guaranteed that this is going to give us a
positive answer, this is undefined at zero
but I don’t really care about continuity
at zero, I care about it from two and beyond
so we have a positive continuous function,
all we have to do is make sure that it is
decreasing or at least non increasing. And
we are going to do that by looking at the
derivative. So we have f prime of x is equal
to low to high minus high to low, all over
low low, which is going to be x to the fourth
squared. That means we are going to have a
common factor of, well lets clean that up
a little bit, x to the fourth over x is x
to the third minus 4 x to the third, x to
the eighth. Now the numerator, both terms
have a constant... common factor of x^3. Taking
that out... the benefit of cleaning this up
a little bit is not so much, I don’t need
to find out when the numerator is going to
equal zero, or do a lot of tremendous amount
of fancy calculus or algebra but if it seems
like there’s a lot going in the numerator
so much so that I can’t look at it and automatically
tell that it’s always positive or always
negative, the denominator is x^8 or now its
x^5, that x is really going to need to be
greater than 2 so that's always going to be
positive. We need to see when or if the numerator's
always negative. So I just want a simpler
numerator to work with. So when is f prime
going to be negative, when is this function
decreasing. Well we have 1-4lnx needs to be
less than zero. So we have -4lnx is less than
-1. We have lnx<1/4. Taking both sides to
be exponent of e, we have... whoops, Divide
by a negative number yeah Switch that sign.
so we have x needs to be greater than e^1/4
which means x needs to be greater than 1 point
something, right off the top of my head..
! Wrong page! Or is it? Ahh there it is..
X needs to be greater than 1.284 which means
x needs to greater than or equal to 2 because
really we are trying to emulate that idea
that domain for integer or all positive integers.
So really we care about x being greater than
or equal to 2. So this means that, one more
time... f is continuous, it is always positive,
and it is decreasing as long as x is greater
than or equal to 2 which means that we can
apply our integral test. Okay which means
that we need to look at, just put it over
here I guess, that the definite integral from
2 to infinity of the natural log of x over
x to the 4 dx. Now we are looking at the definite
integral from 2 to infinity of x to the -4
natural log of x dx. That is going to require
integration by parts 
and when setting up your integration by parts,
you are looking for a part of the integrand
that is most complex that you can still integrate
using the most basic integration rule and
not too sure about easily integrating the
natural log of x, but we can integrate the
x to the -4 dx and we can take the derivative
and get a simpler term when we are taking
the derivative of the natural log of x. so
we are going to let u equal to the natural
log of x, then our du is 1/x dx, our dv x
to the -4 dx. Integrating both sides to find
our v and we have v is equal to, raise that
power by 1 and divide by it, yeah and that’s
it. Okay, now again I don’t want to change
the lower and upper limits if I don’t need
to, I am just going to pull this off to the
sides and start from scratch and say the indefinite
integral of x to the -4 natural log of x dx
is equal to well uv, so u v, keep the negative
upfront on top or whatever, negative natural
log of x over 3 x^3 so there's my uv minus
indefinite integral of v du, okay cleaning
that up a little bit, I am going to bring
the -1/3 out front, x^-1 times x^-3 is x^-4.
Alright... integrating again. Now I can see
that we are going to have a denominator of
9 and I am going to bring this x^3 down so
I am going to go ahead and multiply this first
fraction by 3 in the numerator and denominator
and get -3lnx, it's going to be -1, over 3..
Well it's going to be 9 now, right? See I
don’t just show this work for you guys,
I need it too! 3 times 3 is 9, x^3. Okay...
let me just make sure I haven’t written
something incorrectly! Beautiful! Alright
so that means we have done this indefinite
integration so this is equal to, okay so now
plugging in our infinity, we have got.. Okay
now, couple of things here, let’s go ahead
and sort of factor out the -1 out of the two
terms in the numerator here and negative times
negative is positive and looking over here
probably should use some limit notations since
this is infinity over infinity, but like I
said in the previous page or that I erased,
when I wanted to find out the limit as n approaches
infinity of this and I was doing the whole
L'Hopital's thing. That's really kind of what's
going on here. Basically what we are doing
here is
the limit as n, not n but x approaches infinity
of -3 times ln x -1 over 9 x to the third
and we kind of need to look at this term separately
and see if it approaches infinity or some
kind of constant or zero and again I don’t
want to show that L'Hopital's rule again just
because of time but we are going to have this
kind of very similar situations we had up
here natural log of x over x to the fourth,
we have a natural log of basically x only
now it's over x^3, that denominator is still
going to go faster than the numerator and
L'Hopital's rule will show us again that that
is going to be zero so really our final answer
here is 3 ln2 +1 over 2 cubed is 8and 8 times
9 is 72. That is a real solution not like
an infinity or negative infinity. so my definite
integral converges and the integral test said
that either both the series and the integral
diverge or both the series and the integral
converge as long as you've met the requirements
of the integral test and we did so therefore
since this is a real solution, 'the given'
is not one word! And there, that's not the
given series, there you go. That converges
and that is the end of our second example.
It's really the end of the major example I
am going to do just two more up here showing
you that it's going to be very beneficial
to do the nth term test and make sure that
you do meet all three requirements of the
integral test before you actually apply it.
BAM!
And for our last two examples, and an update
for what’s going on with the camera equipment,
I do believe the Nikon D7000, at least audio,
some issues are going on. Sound quality so
far is great with the camcorder playing with
the lighting try and make it match up with
this camera a little better. So, again any
feedback would be good. I think I am going
to have to keep using my camcorder for a while
but at any rate, you just care about the math,
right?! We have... I mean that's what we are
here for! So we have this series and, did
we, can we just immediately almost, can we
immediately determine if it diverges? Well,
is the limit as n approaches infinity for
n+5 over n-1 is that equal to zero or is it
not equal to zero? Dividing the numerator
and denominator by n, we get the limit, we
can probably see the answer by now, all of
our awesome calculus skills, we have 1+5/n
over 1-1/n and that certainly is going to
approach 1. Which means that it is not equal
to 0. So the given, by the nth term test,
the given series diverges! Now over here if
we let a sub n, find the limit as n approaches
infinity of a sub n, this is going to have
a numerator where the numerator stays bound
between -1 and 1 and the denominator is going
to grow without bounds so the limit as n approaches
infinity for a sub n here, that is going to
be zero. So this sequence, or this series
excuse me, may converge. But, remembering
that when we say let f(x)= cos x over 2 to
the x, f needs to be positive needs to be
positive, continuous and decreasing. Our denominator's
always going to be positive with a positive
base of 2 being raised to any number but the
numerator, y= cos x and there’s nothing
else going on up here to make it more complicated,
the cosine of zero is 1, cosine of pi/2 is
0, the cosine of pi is -1 and so on.. You
know what the cosine function looks like.
It's going to alternate between being positive
and negative so this function is not going
to be always, f(x) is not going to be positive
for all n greater than or equal to 1 and that
is our last two examples. I am Mr. Tarrou!
Thank you for sticking with me through the
technical issues. Yeah... that’s your integral
test. I am Mr. Tarrou. BAM! Go do your homework.
I said ‘I am Mr. Tarrou’ twice... Now
three times!
