- So last time, where are we at? Well, we talked about sums, infinite sums, we call 'em series. And we were able to look
at a couple special cases, namely we had geometric
series and telescoping series and we can actually find
the answers for those cases. But a lot of time we can't
find an answer explicitly. We'll just be satisfied to
know that there is an answer. So that's where we're going next. We want to develop our tools
to decide is there an answer. So let's start with a
really simple observation. So let's suppose that the sum converges, in other words we can find an answer. The claim is that if I
look at an individual term, it has to go to 0. Now why is that? Well, if you think about
what's happening here, so remember this is our partial
sums and this we're writing in a slightly different way. Now, normally the way we
write this before is it says, look, if you want to
find the nth partial sum, what you do is you start with
the n minus 1 partial sum and then you add one more term. 'Cause to get to the next partial sum, you just add the last term. So if you move this across,
you get that the last term is the difference of two partial sums. Now, we say that the sum is converging, which means that this
partial sum converges to L and that partial sum is converging to L. So it's converging to
L minus L, which is 0. Alright? So if the sum converges
the terms have to go to 0. Now what you can do is
you can flip that around, there's a fancy word for this, it's called the contrapositive,
but just think of says, look, flip the order but do the opposite. So what's the opposite of
having the limit go to 0? Well the limit doesn't go to 0. What's the opposite of the sum converges? The sum diverges. So the conclusion is if
the limit does not go to 0, then the sum diverges. Okay, that's a pretty
straightforward easy idea. Now a word of warning
because a lot of people will hear this statement and they'll start to implicitly say, "Aha, if so, if the limit goes, sorry, if the limit doesn't go to 0 sum diverges, maybe if the limit does go to 0, does that imply converges?" And the answer is no, it does not. And we'll do an explicit example of that in just a moment here. So be careful. It's important to
understand what things say but also what they don't say. And we wanna make sure we
understand what they say. That's our goal. Alright, so here we have some examples. So what can we say about
the following series using the nth of term test? So the sum n equals 1 to infinity of n. Okay, well that's 1 plus 2 plus 3 plus 4. Okay. All right. What's going on? Well, the nth term, see
this, that is our ace of n. So the question we're
asking is what's happening to ace of n? So the limit as n goes to infinity of n. Well, we don't even even know,
need to know what it goes to. The question is, does it go to 0? The answer is no, no, no. It definitely does not go to 0. It gets huge. It's actually infinity. Good. We don't even need
to know that it's infinity. We just need it to know, not 0. All right, well the next one, sum, n equals 1 to infinity
of minus 1 to the n. And oh dear, this is wrong. Hmm, let's fix that. Okay, now we're okay. Now you might say, what was wrong? Well notice if n equals
1 should have minus 1 to the 1. So the first one should be
a minus 1, not a plus 1. Oops. Alright, well easy fix. In fact, I think we'll fix this one too. Alright, so what's happening? Well we have 1 minus 1 plus 1 minus 1 plus 1 minus 1. So, the question is, what happens here? So this is our ace event. So what's the limit, as n
goes to infinity of minus 1 to the n. And we should have, by the way here said, our conclusion is this
implies that the sum diverges, we can even be more precise. It diverges to infinity, it's infinite. Now, notice n goes
infinity minus 1 to the n. Well it's not 0, in fact it's not anything 'cause it alternates. Plus 1 minus 1 plus 1
minus 1 plus 1 minus 1. So since it doesn't go to 0, then that implies that this sum diverges. Now this is a little
bit different behavior. See up here we said it's
gonna head off to infinity, here it doesn't go to
anything in particular, it just oscillates. In fact, you can sort of see what happens. See you start with just
1, then 0 and back to 1 and back to 0, then 1, then 0. And it's just gonna keep doing that. All right, last one. Sum n equals 0 to infinity
of minus a 1/2 to the n. So that's 1 minus a 1/2 'cause that's minus 1/2 to the 1 plus a quarter because
that's minus 1/2 squared minus an 8th, plus 16th, so forth. So, alright, so that is our nth term. That's our ace event. So what's the limit as n goes to infinity of minus 1 1/2 to the n. Well the answer is 0, right? 'Cause it's a a small number. When you raise a small number
to n, that's gonna get very, very small and the limit goes to 0. Now what's the conclusion? 'Cause remember what are we using? We're using the nth term test. So when it doesn't go to 0,
our conclusion is diverges. When it does go to 0, our
conclusion is unknown. That's what we have. By the nth term test. Now does that mean we give up? No, it means we start
looking at other tools. Now, right now we don't
have very many tools, but in this case we say wait a second, this isn't just any sum,
it's a geometric sum. In fact it's a geometric
sum and here the r in absolute values less than 1 so because it's geometric, we can actually make a
conclusion, namely we can conclude that this does converge and
not only does it converge, we can say what it converges to and it converges to the first term is 1, so 1/1 minus and the thing we multiply at every stage is minus a 1/2. So that's minus, minus makes a plus, one plus a 1/2 is 3 halves. So one over three halves,
which becomes two thirds. So that's what that's converged to. But if we only are using the
nth term test, no conclusion, we don't know, not enough information. Alright, well let's start
with a very famous sum. And it even has a name. So it must be famous. This is called the harmonic sum. And what is it? Well, you start from 1 to infinity of 1/n. So you get 1/1 plus 1/2 plus
1/3 plus 1/4 and so forth. All these things. Now let's use our nth term test. So the nth term test, what
does that allow us to say? Well what we do is we look at the limit as n goes to infinity of
the nth term which is 1/n. Well what does that do? It gets very small, so it equals 0. Alright, conclusion. Well, that implies we don't know. Okay? So at this point we
can't make a conclusion just from the nth term test. Well, but maybe there's
something else we can use to help us make a decision. So you'll notice that we've
done some grouping here. So the 1 is by itself,
1/2, well, hmm, all right, we can say certainly that's
a 1/2 because a 1/2 is a 1/2. Now notice the next terms, 1/3 plus a 1/4. Which one is the smaller term? Is it the 1/3 or is it the 1/4? The 1/3, well that has a
smaller number downstairs. Does that make it the smaller value? No, no, no, it's the 1/4. 1/4 is the smaller one. Now, so let's suppose I
replace the 1/3 here by 1/4. So I could certainly say that
this part is greater than a 1/4 plus a 1/4, which is a 1/2. Okay, let's go to the next grouping here. A 1/5, a 1/6, a 1/7, an 1/8. Which one's the smallest? It's the 1/8. So if I replace each one of
these by the smallest one, then I'll certainly make the
whole expression smaller. So this is bigger than an 1/8 plus an 1/8 plus an 1/8 plus an 1/8. Well how many 1/8s do we have? That's 4/8s, which is a 1/2. Now you probably seen
where we're going here. See this next big block. Well what can we do? Well, let's replace all the
terms by the smallest term, that's the 1/16. So, there's a 1/16 from the 1/9, the 1/16 from the 1/10 from the 1/11, the 1/12, the 1/13, the 1/14, the 1/15. And the 1/16 just stays as 1/16. And if you count there's
going to be 8 of them. So that's a 1/2. Now can we keep it going? Yeah, we can. So 1/17 up to where, where do
we make the next cutoff at? Can you see it? Where are cutoffs happening? 2, 4, 8, 16. The next one, 32. And here, if you now replace
everything in this block by 1/32, what you'll get, this is bigger than
1/32 added up 16 times, which is a 1/2. And we can keep this going forever. So what does that tell us about our sum? This says that the sum
from n equals 1 to infinity of 1/n is bigger than 1 plus
a 1/2 from the first block, plus a 1/2 from the second block, plus a 1/2 from the third block. And in fact, each one of these blocks is at least as big as a 1/2. Of course the length of the
blocks gets bigger and bigger. But the thing is, there's
a lot of room in infinity. So even though the the blocks
themselves are getting bigger and bigger, there's still
plenty of blocks left. So we're gonna have enough space. So in fact it's bigger than 1 plus a 1/2 added up infinitely often. But what happens if you add
up a 1/2 infinitely often? Well the answer is you
get big, it's infinity. So the sum is bigger than infinity. Well there aren't that many
things which are bigger than infinity. In fact, there's really
nothing above infinity. So we come to the conclusion
that the original sum, the sum n equals 1 to
infinity of 1/n is infinite. Alright, good. Now this is a very famous series, in fact we're gonna put a box around that. Now if you remember, whenever you see a box around something, that's a sign that we should
pay attention to this formula because it's probably gonna
show up multiple times in the future. And the more that we see it, then the better we're
gonna get at using it. Now you have some space
left and some time, I'll just mention a few things here. This it illustrates by the way why you have to be careful
with the nth term test. Just because something is
going to 0, the nth terms are going to 0 does not mean
when you add everything up, it's gonna be finite. It might be that things are going to 0, but they're going to 0 too slowly. See here we have the 1/n that's going to 0 in some sense in a very slow way. So it's getting there,
but it's taking its time. If you can compare that to
something we had a moment ago, see minus a 1/2 to the
n, that also goes to 0. But in some sense it's
going to 0 a lot faster. If you put in n equals 100, the number you get is super, super small. So small that your calculator
probably will say it's 0. If you put in n equals 100 here, the number you're gonna
get is not that small. So somehow there's always
a question about, okay, how quickly do you go to 0? And do you go to 0 fast enough. Now there's another thing, and when I was a student back in the day, a long, long time ago, and well our college had
this for Valentine's Day, you could print a little tiny
message to your sweetheart and it could be sort of a,
they call it a love line. And I thought, you know what, it would be fun to do
something kind of mathy, right? 'Cause you know I'm a math person, I was a math person back then
too, still I'm a math person. And so I wrote the following
and it said my love for you. See I'm romantic. I used the word love. My love for you is like the harmonic sum. And, why is that? Well, growing little by little but infinite in the total. Oh my gosh. See if you need to, you
can borrow this line. It's okay, I don't mind. You can use this to woo a
very math oriented person. 'Cause they'll say, "Ah, that
is like the harmonic sum." See, it does grow slowly 'cause
the terms are going to 0. But when you add 'em
all up, it's infinite. Ah, what a beautiful line. Alright, good. So now we're gonna move
into a another test. So we have the nth term test,
we're gonna add another test. And this is the integral test, which means all those fun things
we learned about integrals are still useful to know. So assess the following. Suppose f of x is continuous,
positive and decreasing. So in other words, f is a nice function. So if we have our sum of f of n and we look at that and
we have the integral of f of x dx and we look at that, they both converge or they both diverge. So in other words, we can
connect the behaviors together. Now why do we care about this? Well, finding sums is tough,
very, very challenging. But finding integrals is tough. But we have tools. See the integration, it's
in the continuous world and there's better tools for that. Now why is this true? Well, when we think of this sum, we really can think of this like, okay, we're adding a sort of discreet thing. So one way we can think of
it is we can think of it as we call it a Riemann sum. And so that is, we're gonna think of it as a bunch of rectangles. So we're gonna have two areas. One area is a bunch of rectangles
that represents the sum. And the other area is what
comes from the function. And we're gonna say,
okay, look what happens. So one way I can do it
is I can think about suppose this is k and then
I, you know, keep going. Now one thing I can do is
for k I can have it go up and to the left. I can just do it at every stage. And if I do that, I get
a bunch of rectangles and you can see the areas, add 'em up and the rectangles, where are they? They're all underneath the curve. Now the other thing I
could do is I could go to k and I could go not to
the left, go to the right and I could do that at every single one. And again, a bunch of
rectangles and the function. Now, what's the story? Well the claim is if 1
converges the other converges. So let's suppose that we
know that this sum converges. So that has finite area. Well now let's come to this picture. Look at the area. The area underneath the
curve which is the integral is smaller than the area that
comes from adding that up. So smaller area, if this
is finite, that is finite. Now if the integral is finite,
then is the other picture it says look the integral
that says there's finite area underneath this curve and
here we can add up these, that's also gonna be a finite area. What about diverging? Well let's suppose the integral diverges, well now come to this picture. If the integral diverges, that means that the area
underneath this curve is infinite. And by the way, it's the
tail that you worry about. You don't worry about
the start, it's the tail. But now look at the rectangles. The rectangles have more
area than the curve. So if the curve has infinite area, the rectangles also have
infinite area, so it diverges, similar when you can go over here. So that's what the integral test says. It says, "Hey, I can think of
these as being tied together." But actually the integral test, there's a little side note here and I think it's worth mentioning. So even though we know
things converge or diverge, there's always the question of, well what do I do if I want
to get a good estimate? And so a reasonable test
should help us understand how do we get a good estimate. So, that happens here
with the integral test. So there's a way to get a good estimate. And here's what you do. So you may wanna say,
how much is my error? 'Cause I'm not gonna add
up in too many things but maybe I'll say, do I
need to add up 100 things, do I need to add up 1000 things? How many things do I need to add up before I get a good estimate? And here's the the intuition. So if I have the sum from
say n equals k to infinity of my f of n, well that's approximately the sum from n equals k up
to some capital n of f of n plus an error. In fact, we can replace that approximately that's equal, right? Now the question of
course is what can we say about the error? And here's the beautiful thing. The error, is less than
or equal to the integral from capital n to infinity of f of x dx. Now, why is that? Well, here's sort of the intuition. So you have this function going down and you've added up a bunch of rectangles and you've added up to some point here. So what's happening?
You see the error really that's the sum from n equals
capital n plus 1 to infinity of f of n. That's our error. Okay? That's not very surprising. See if you add those
up, that's what you get. So what do we say? Well here's capital n. So you go to n plus 1, n plus 2, so forth and so on and
you make your rectangles. And the way you do it is you go up and to the left and you're gonna do that all the way down, so forth and so on. Now if you add up from
n plus 1 to infinity, f of n, that's these rectangles here and that's smaller than
the integral from capital n to infinity of f of f x dx. So you can always figure
out how far do I have to go? You just say look how far do I have to go to make this part of the integral small? So that's a way that we can
be confident that we say, look, if we stop after 1000 terms, the rest of the value is at most this. And so that gives us a way
to get an error because look, an estimate is only as good as your error. If you have a huge error,
it's a terrible estimate. If you have a small error,
it's a good estimate. So it's good to know things. Alright, well that's a little aside. Let's ask a question. Does the sum from 0 to
infinity of 1/n squared plus 1 converge? Now if you use the nth term test, we can see that it goes to 0 'cause it's 1 over a big number. Okay? So by the nth term test we don't know. Alright? But we have a better test. So what we can think
of this here as being, as we can think of this
like well this is f of n where f of x is what? Well think of this as where
you see an n put an x. So it's 1 over x squared plus 1. And now we're gonna ask the question. Okay, well I can say look,
this sum from n equals 0 to infinity of 1/n squared plus 1 is somehow tied to the
integral 0 to infinity of 1 over x squared plus 1. So if I'm having a hard
time with that sum, I ask, okay, can I have a
hard time with this integral? I'm gonna just pause for a second and say, if you're ever worried about the start, because notice there is
some technical things you need to be positive and decreasing and it might not be positive
and decreasing at the start. You don't care about the start. Remember when we're talking
about converge diverge, we always wonder about
what happens at the end, the infinity, that's the important part. So it's the tail, the
tail that tells the story. So now let's work on this integral. So integral 0 to infinity of
1 over x squared plus 1 dx. This sounds familiar, like we've heard of this
integral somewhere before. Hmm, huh? Hmm, what do you think? Does it? Well, hmm. Well I think we have, that's arctangent. Ah, arctangent, you beautiful function. It's been too long. All right, so it's arctangent
of x from 0 to infinity. Now of course you can't plug in infinity. So when we talk about arctangent infinity, what do we mean? What happens to arctangent
of x as x goes to infinity. Well what does happen to arctangent of x is x goes to infinity. The answer is it goes to pi/2. Subtract arctangent is 0. Now do we care about the value? No. What do we care about? The question is, does this converge, in other words, is it finite? Well it does turn out that
pi/2 is in fact finite. It's less than infinity. Hard to believe, but true. Or maybe it's not so hard to believe. So what does that say? That says that this
integral we know converges, we just checked it. So that says because they're tied together that the sum converges. Now, does the sum converge to pi/2? No, not even close. Well, why not? Well you can actually
see it if you think about the pictures here because if you look because
of the way the pictures work, you can see that there's
little bits of error. See it's these little tiny pieces and these little tiny
pieces, they're small, so I mean pi/2 is not a bad estimate, but still the moral is the
difference between these is contained in how much area
there is between the curve and these little pieces of rectangle. So we are not claiming
that the 1/n squared plus 1 goes to pi/2. That is not true. But we can claim it goes to
something, it does converge and you can always find it
numerically pretty well. It turns out that with
more sophisticated tools, it is actually possible to get
an explicit value for this. But that's outside the scope of our class. Alright, well now that we
have the integral test, we're gonna look at a very important test and that is p-Series. So we're gonna look at
the sum of 1/n to the p. This is a very famous series, in fact, this actually
has a name in mathematics. This is known as the zeta function. And I apologize, my zetas
tend to look like C sometimes. This is called zeta of p. And so famous, there's actually a million dollar question tied to understanding this series. So, alright, does this converge? Does it not converge? All right, so what can we do? Well we ought to figure out for which p does the sum 1/p converge. So what do we do is we say look the sum, we're going to connect it to the integral. So we need to ask ourselves a question, when does the integral 1 to infinity of what's our function? Well look, n, that's our index. So n is becoming our x. So it's 1/x to the power p dx. Alright, so if I wanna know
when does this sum converge? I ask when does this integral converge? All right, now 1/x to P,
you gotta be a little bit careful here 'cause there's two cases. One case is p equals 1. Now when p equals 1, you get
the integral 1 to infinity of 1/x. Now once you got 1/x, it is natural log of x and
we go from 1 to infinity. Now what happens to natural
log as x goes to infinity? The answer is it gets huge. It does it very slowly, you know, log it drags its heels. I don't wanna get big, I wanna stay small and it doesn't want to
go arbitrarily large, but it does, it does. It takes its time but it gets there. Now you might say, hey, you
forgot to subtract that. Yeah, it doesn't matter. Natural log of 1 is 0, but it doesn't matter once you're infinity it's hard to recover. So that tells us it
definitely is not gonna work when p equals 1. Well that doesn't surprise us, right? Because remember what
happens when p equals 1, that's the sum of 1/n. That's the harmonic sum and we
already know that it's going to diverge when it's p equals 1. Okay? So that agrees with
what we've seen before. Alright, well what if p is not 1? Well then the way we should
think about the integral is we should think of it as
x to the power minus p dx. 'Cause that's sort of the
better way to write it. 'Cause our other rule is we're
thinking like x to the power. Now how do we integrate that? Well the rule is you
add 1 to the exponent, so that's x to the power of minus p plus 1 and then you divide by that new exponent. So 1/minus p plus 1 and
you go from 1 to infinity. Now the 1 doesn't matter,
it's the infinity, the infinity, the tail,
what's happening at the tail. That's going to be the
what dictates the story. So let's think about it. So there's a couple of cases and here the 1/ minus p plus
1 is when that's the constant. So you're getting a huge
number raised to the power of minus p plus 1. Now if minus p plus 1
is a positive number, then you're getting a huge number raised to a positive exponent. That's gonna be big, right? 'Cause a big number positive
exponent like you know, n to the power a 1/2, n to the power of 3. It's big, big, big, big. Now what happens if minus
p plus 1 is less than 0, that means it's a negative exponent, which really means it's downstairs. So it's a huge number downstairs. So that's going to be a finite number. Now technically I'm not gonna,
that's just the one end. We can even say what
it's gonna go to 0 minus and then when you plug
in 1, 1/1 minus p, okay? So that's what we mean. But again, we don't care
about what the value is. We care about whether it's finite or not. So, hmm, let's rearrange this. So here I could say
instead of minus p plus 1 is greater than 0, I could
say, well if p is less than 1, 'cause that's what that would say. Because here this would
say if p is greater than 1. So putting this together,
p equals 1 diverges, p less than 1 diverges, p greater than 1 converges. So the integral, this converges when p is greater than 1. Now this is important, you know it's important
'cause it even has a name. So we're gonna keep
referring back to this over and over again. P-Series and it's kind
of an interesting thing because what does this say? Well this says look, if
you have the sum n equals 1 to infinity of 1/n, well that's diverging. But what if you have the
sum n equals 1 to infinity of 1/n to the 1.000000001. What happens? It converges. So, somehow it's like this
1/n is like just barely short because these two,
they're very, very close, especially for the first, you
know, couple million terms. You wouldn't be able to
really distinguish anything but somehow it in the long term,
see the long-term behavior, that's where the
interesting things happen. So we're gonna keep seeing
p-Series over and over again. So the main things to
remember is that if you have 1/n to the p, you look at that p. If p is greater than 1 converges, p less than equal to 1
diverges and you're done. And it's good to know that. On a side note, I mentioned
this was a famous sum and it has been studied. So I'll give you an example. If you add up the sum
n equals 1 to infinity of 1/n squared, it's known
exactly what that is. That's equal to pi squared/6, right? Well I don't know, it's
equal to something, right? It's kind of like pi, really?
Yeah, it really is a pi. Wow. Weird. But it gets even stranger. Now if you add up 1/n to the fourth, that's also known and I don't
know off the top of my head, but it is known, well what if you add up 1/n cubed? Well, what is that? Well, the answer we're not exactly sure. Now when I say we're not exactly sure, we can approximate that to
you know, a million digits, that's not a problem. The problem is, does this
equal something fancy like pi squared/6? It's not known. We don't know. There's a lot of things we don't know. And you might think that, wow, I thought math had all the answers. No, not yet. Isn't it great? Isn't it wonderful? There's still things we don't know. Ah, what a great time to be alive. Alright, let's do another problem. Show that for 0 less than r, less than 1 that this sum nr to the n converges. Alright, cool. Well let's think about that. So what do we have, nr to the n? Now r is small. So there's a couple of things
that are nice about this. First off, that means
that these are positive. Now you might say, okay, is it decreasing? Well, gotta be careful there 'cause what are we gonna do? We're gonna really think about
this as the integral test. So the function, we're
gonna think of this as xr to the x power. Is this function decreasing? The answer is eventually yes. What happens though is
at the start it goes up, but eventually it goes down. Now, because it's increasing
at the start, you might say, well the integral test can't be used. But remember the integral test says, I only care about the end. So don't worry about the
start, worry about the end. So we'll say look, because in the end it
does decrease down to 0 we can use the integral test. So let's use the integral
test and what do we get? Well, we have, we wanna
look at the integral from 0 to infinity of xr to the x dx because what we're doing is
we're replacing our n's by x's. So we're thinking as this as our function. Now how do we integrate this? Well it might help to recall things. First off, let's recall
the integral of r to the x. What is that integral? Now you might say e to the
x is really easy, right? E to the x is e to the x. Well what's the integral of r to the x? Well if you don't remember,
here's how you can rederive. You want to have something
that looks like e to something. So can we replace r by
something that looks like e to something and we can, it's e to the natural log of r to the x. Or if you like, that's e
to the log of r times x dx. So it's like e to a constant x. So the way you integrate e to a constant x is you have e to that constant times x divided by your constant and then you throw in the plus e of course 'cause you always wanna make
sure you get all the points. E to the natural log of r
times x, well work backwards. That's r to the x. So you get 1/natural log
of r times r to the x. Okay? So that's the
integral of r to the x. How do you integrate xr to the x? It seems that this is something that has multiple parts, right? There's x, there's r to the x, how would we do integration
in that setting? Well, we can do integration by parts. All right, so if we do integration by
parts, we say, all right, we have two pieces, which
part do we integrate and which part do we differentiate? Well the r to the x, essentially it's gonna
look like r to the x. Now that's whether you
integrate or differentiate. The x, you wanna make that simpler. If you integrate, that
becomes a century x squared, more complicated, differentiate
1 that's beautiful. So we'll make this part be
the part we differentiate, this part, the part we integrate. So that du is equal to
1 dx and then the r dx, we just did that. That's 1/natural log
of r times r to the x. So we get uv 1/natural
log of r xr to the x, evaluate that from 0 to infinity, minus the integral of du. So that 1/ log r's a constant
if you can pull it out. So 1/log r integral r to the x dx. Now here when you plug things in, in infinity, what wins between polynomial and exponential and the
exponential by the way, exponential decay. Well the answer is the
exponential decay beats out. So polynomial's grow
slower than exponentials. So when you go to
infinity, that goes to 0. Well what happens when you plug in 0, well you get 0 times something. So this part is 0. Okay, that's good. What's happening over here? Well, whoops. Forgot the bounce 0 to infinity. Alright, well so we integrate that again. So minus 1/natural log of r, then integral r to the x there's
another 1/natural log of r times r to the x from 0 to infinity. And, what do we have? Well, when you go to infinity, these terms at the start of constant, you get r raised to huge
power, that's gonna go to 0. So you're gonna get 0
subtract when you plug in 0, r to 0 is 1. So subtract minus 1/log r times 1/log r. Now this turns out to be
1/natural log of r squared. Do we care about the value? No, not really. What do we care about? We
care about is it finite? And yeah, this is definitely finite. What does that tell us? It tells us the integral converges. But remember the integral
test says these things have the same behavior and
therefore the conclusion is so you say, so by integral test since the integral converges
then the sum converges. And that's the conclusion. Alright, so there we go. Now we know it converges, but we don't know what it converges to. We don't know that because that's, remember the integral test doesn't say what things converge to. See like this 1/n squared,
integral 1/x squared you wouldn't ever get a
pi squared over six out. So there's some sort
of, there's a gap there. So we don't know what it converges to. That'd be a very different
type of a question. So let's go to the next question. Where does it converge to? Oh, okay. All right, well, hmm hmm. Now, like I said, different
type of a question. We can't use the integral test to find what things converge to. We can't use the nth term test. When we talk about convergence test, they're only useful to
give us yes no answers. Does it converge? Does it not converge? Yes or no? So when it says find something,
you have to use other tools. So let's think about
what we're looking for. So we're gonna wanna find
the sum n times r to the n for absolute value of r, less than 1. We wanna figure out what that equals. So let's think about
what the sum looks like. So the sum n equals 0 to
infinity of nr to the n. I can think of it in the following way. Well, that's 0 times r to the 0. All right, well that's not too surprising. Plus 1 times r to the 1, plus 2 times r to the 2, plus 3 times r to the 3, plus 4r to the 4, plus 5r to the 5, plus 6r to the 6, and so forth and so on. Now, all right, what do we do? So we wanna find these sums. So the problem is that this
is not just a geometric, if there wasn't the n at the start, if it's just r to the n, life is good, geometric, we're done. But it's not. That extra n throws a a
little bit of a wrench into our process. But here's sort of an idea. Let's think of it as the following. Look, I've got a single r. How many r squares do I
have? Well, I have two. So I'm gonna think of it as r squared and then I'm gonna start
another sum and r squared. Alright, how many rq'd do I have? Well, I have 3 of them. So, I'm gonna have an r cubed. I'm gonna put an r cubed
here, and then, well, I still have one to go. I'll put another r cubed there. How about r to the fourths? Well, I've got four of them. So r to the fourth and r to the fourth and r to the fourth and then I need to start a
new one, r to the fourth. How about r to the fifth? Well, all right, I'll do that. I'll break these up as r to the fifth. Now why am I doing this? Well, I'm doing this because
I want to reduce to something I already know about. What do I already know about? I already know about geometric sums. So what I'm trying to
do is I'm trying to make things look like geometric sums, and I'm trying to find
ways to break things up into nice little pieces. And you'll see what's happening here. Well, one thing that's happening is we're getting a lot of terms, but one of the things we can
think about this as being here is we can really think about, you know, what do we really have
here is we can think of each line here as being
its own geometric sum. 'Cause that's really, if you
look at every single line, that's what's happening, right? 'Cause you have r, r squared,
r cubed, r to the fourth, and so forth. Here, it's a geometric sum. The first term is r and every time you multiply by r. Here there's a geometric sum. First term is r squared
every time you multiply by r. Here, geometric sum, first
term is rq, multiply by r. Now let's add these up.
So, what are we doing? Well, we're saying, look, instead of trying to find one sum, we're gonna add up infinitely
many infinite sums. So, wow, okay, let's see what we get. Now remember the rule, if
you have a geometric sum, you start with the first term. So for the first 1, we're gonna get r and then you divide it by 1
minus what you multiply it by. Every time here, we're
multiplying by 1 minus r, right? So every time we multiply by r, so it's divided by 1 minus r. If we go to the next line, our first term is r squared. What do we divide by? Well, we divide by, we
look at what we multiply, see every time we multiply by r. So again, 1 minus r. The third line, it's geometric,
first term is r cubed. So we're gonna get r
cubed, divide by 1 minus r, fourth line, r to the fourth. So r to the fourth, divided by 1 minus r and that continues. So, if we look at these individual sums, that's what we get. Now remember, that's not our answer. We now need to add all these up. So we need to add up r/1 minus r, plus r squared/1 minus r,
plus r cubed/1 minus r, plus r to the fourth/1 minus r, plus r to the fifth/1 minus r. Now, what kind of sum do we have here? It's geometric, right? Because our first term is r/1 minus r. To get to our next term, multiplied by r. Get to the next term
after that, multiply by r. And every time we multiply by r. So we take our first term, divide that by 1 minus
what we multiply by, and therefore we get r over
we bring that upstairs, 1 minus r squared. So that's our conclusion. We have that our sum is r,
divide that by 1 minus r, quantity squared. Now that's, you know, not so obvious what to do or how to begin. But I wanna say that within
a couple of weeks from now, we're gonna get to the stage
where we feel comfortable with this. This is actually gonna be
a fun problem for us to do in a few weeks. So expect to see something like this again in the near future, but not today because
we're done for today. And we'll see you again next time. Have a good day. Bye.