Let's integrate.
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Series and integrals are not entirely unrelated.
Well, here I got a graph and I've drawn a bunch of boxes in red on my graph.
Each of these boxes has width one and the heights
of the boxes, given by the terms in a sequence.
So the first box has height a sub 1, the next box has height a sub 2, the
next box has height a sub 3, the next box has height a sub 4 and so on.
Since the boxes each have width 1, it's easy to calculate their areas.
The area of the first box is 1 times it's height, which is a sub 1.
So the area of the first is a sub 1.
The area of the second box is it's width 1 times it's height
a sub 2, so the area of the next box is a sub 2.
And so on.
The area of the next box is a sub 3, the area of the next box is a sub 4.
What if I want to calculate the area of all
the red boxes, I imagine this red boxes go on forever.
Well, that's really an infinite series, right?
Its a sub 1 plus a sub 2 plus a sub 3 plus a sub 4 and
so on, so I could write the total area
in all the red boxes as these infinite series.
There is more going on in this graph, I don't
just have these boxes, I've also got this blue curve.
That's the curve y equals f of x and
one special feature about this function f is that, I've rigged it so
that a stubborn is f of n and we can see that here.
Because this blue curve passes through the corners of these boxes.
Now what if I wanted to calculate the area under the
blue curve and to the right of the line x equals 1?
Well that would be this interval.
The interval from one to infinity of f of x, dx.
How does this blue region compare to this red region?
Well this blue region sits inside all of the red boxes.
What is that mean?
That means that this is the red region, this is this blue
region, but this blue region sits inside all of the red boxes.
And that means that the area of the red region
must be larger than the area of the blue region.
Why does that matter, though?
Well the takeaway message here is that
this sum is even larger than this interval.
And that means, if I know that this integral, say, is infinity, if
the area under the blue curve and to the right of the line x equals 1 is infinite.
Then what do I know about this series?
Well that series then must diverge.
The picture includes some as of yet unspecified assumptions.
So more precisely, I need to assume
that f is a decreasing and positive function.
And I should explicitly mention that a sub n is equal to f of n and with those
conditions, than if this integral diverges, then the sum also diverges.
We're in a position to prove this.
This is what I'd like to show. Suppose that little f is a decreasing but
positive function and building a sequence a sub n out of the function f.
And I'm supposing that the integral f of x dx goes from 1
to infinity diverges and I want to show that the series diverges as well.
Well to get a handle on a series, I'm going to look at this.
The sum little n goes from 1 to big N the a
sub n, and somehow, I want to relate this sum to an interval.
I won't get there all at once but I'll do
it in steps.
First thing I can say is this, that the sum of
the a sub n's is equal to the sum, little n
goes from 1 to big N of the integral x goes from n to n plus 1 of f of n dx.
And although this looks somewhat fearsome It's true for a not very deep reason.
the upside here is what?
F of n is equal to a sub n,
and consequently, this is just the integral of a constant
on an interval of length 1.
So if you integrate a sub n, x goes from n
to n plus 1, well, you're just getting a sub n.
So this is the sum of a sub n which is exactly what I've got here.
But when I write it like this, I'm
giving myself some more freedom to mess around with
this integral, instead of saying f of x I could say f of the floor of x.
Well why's that
helpful well what's floor of x.
The floor of x is the greatest integer less than or equal to x.
So if x is between n and n plus 1, n's an integer, then the floor of
x is just equal to n and that means, except at the endpoints, f of the floor
of x is equal to f of n on the interval between n and n plus 1.
And that means this integral is just,
again, a sub n.
So the sum of the a sub n's is the sum of, well that's just a sub n.
Now I'm writing a sum of a bunch of integrals, all the
integrands look the same but the bounds of integration are changing, right.
The first integral goes from 1 to 2, the next integral goes from
2 to 3, the next integral goes from 3 to 4 and so on.
But if you add up a bunch of integrals,
all of which lie end-to-end like that, you can just write that as a single integral.
So, this sum of integrals is equal to just the integral, x goes
from 1 to big N plus 1, of f of the floor of x
dx. All right.
So I've replaced, this sum with this integral.
Now what can I do?
Oh, I know that f is a decreasing function, so if
I plug in a bigger input, the functions output is smaller.
That tells me that the integral of f of floor of x, x goes from 1 to big N plus 1.
Is bigger than or equal to the integral cause f of x, x goes from
1 to big N plus 1, because f of the floor of x is at least as big as f of x.
How else do I know? I know that this integral diverges.
Which means what?
Well, it means that as long as n is big
enough, I can make this as big as you'd like.
All right, all you need to do is tell me how big you want this to
be and I'll tell you some big enough ends, so that this thing is really that large.
But, this thing is
smaller than this sum.
So what that means is that, the sum can be made as big
as you like, as long as you choose big N and big enough.
But if you can make the sum really, really large, that means
the sequence of partial sums isn't getting close to any finite value.
And what that means, is that this series diverges.
We can go the other way too.
We can integrate to prove convergence.
So again, I've got this integral and this series and I somehow want to relate them.
Well let's suppose that this integral is finite.
Then I'd like to be able to conclude that this series converges.
With all the same conditions as before, right, I
want f to be a positive decreasing function and such.
Let's prove it.
So here's what I've got. I've got that
this integral is finite, converges, that I
want to know that the series converges as well.
So here's the integral.
The integral from one to infinity of f of x dx.
How does that integral compare just the integral from one to big N of f of x dx.
Well f's a positive function.
So these things differ by the integral from
n to infinity, which is some positive number.
So this thing here is definitely an over estimate of this thing.
So the integral from
1 to infinity is bigger than the integral from 1 to big N of f of x, dx.
I'm going to relate that integral to this integral, the integral from
1 to big N of f of the ceiling of x.
Remember what the ceiling is, the ceiling is
the smallest integer bigger than equal to x.
So how does f of x relate to f the ceiling of x?
Well the ceiling of x is always at least just as big as x.
But because f is
decreasing, f sends bigger inputs to smaller outputs.
So f for the ceiling of x is less than or equal to f of x.
And that means this intergral is an overestimate of this intergral.
Now I can split this integral up and instead of integrating from 1
to big N all in one fell swoop, I can instead integrate like this.
I can integrate from little n to little n plus 1 and
add up all those integrals as n goes from 1 to big n minus 1.
But on each of these intervals, I know exactly what the ceiling of x is.
If x is between little n and little n plus 1, then the ceiling of x is just n plus 1.
So now I've got the sum of the integral of f of n plus 1.
But what is f of n plus 1?
Remember f of n plus 1 is just a sub n plus 1.
And I'm integrating this constant of an interval of length 1, so this
integral just ends up being a sub n plus 1 and that means these are equal.
Now I could re-index this, you know, the first
time that I'm adding here, the n equals 1 term,
is a sub 2, the next term is a sub 3, and eventually I plug in a little n equals
big N minus 1.
Which gives me a sub big N here, so I could instead write that
as just the sum little n equals 2 to big N of a sub n.
So all together, right, what I have shown is that this
integral, which has some finite value is an overestimate of this.
Which is almost the partial sum, it's just missing a sub 1,
but what that means is of a sequence of partial sums for this
series Is bounded and I already know that it's monotone
because all the things that I'm adding up are positive.
So I've got a sequence which is bounded
and monotone and therefore, by the monotone conversions theoreom,
this sequence of partial sums converges, which is exactly
what it means to say that this series converges.
We're doing more than just proving convergence.
What I mean is that I've got
this inequality here.
The big deal here is that I could add a sub 1 to both sides
and if I add a sub 1 to both sides, now I've got this inequality.
But this is just a sub 1 plus a sub 2 plus a sub
3 plus dot dot dot plus a sub big N, which I already know.
Is at least as large as the integral from 1 to big N plus 1 of f of x dx.
Let's now take a look at this. I could now take
a limit as big N approaches infinity and
here I would be getting what. Well this would just become the series,
the sum of all the a sub n's. And this is now the integral from 1 to
infinity of f of x dx. So I really know is that the value of this
series, is trapped between this integral and the same integral plus the first term.
Let me summarize what we have.
Same setup as always, I'm supposing I've got some function little f which
is decreasing but positive and I built my sequence out of that function.
Then if I've got that, this integral
converges, if that integral has a finite value,
that happens if and only if this series converges, right?
Because you've shown both directions, I've shown that
if this integral diverges, then the series diverges.
But I've also shown that if this interval has a finite value, then
the sequence of partial sums is
monitored and bounded therefore the series converges.
And more than that, I've also got these bounds.
I also
know that the value of this sequence is
trapped between these two values.
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