Okay, in this video I want to talk about Euler's constant, and Euler's constant, what is it?
Well, the idea is if we have, if we let t sub n = 1 + 1/2 + 1/3 + ... + 1/n - the natural logarithm of n,
If we look at the limit as n goes to infinity of t sub n, this sequence of numbers,
it approaches a number, and we use the little lowercase symbol gamma, that's what we're gonna use.
But the idea is that sequence t sub n approaches a finite number
and that's known as Euler's constant.
Now, one thing that's interesting to me right off the bat is if you, you know,
if you leave off the minus natural logarithm of n, you're looking at the harmonic series.
And we've seen that 1 + 1/2 + 1/3 + ... + 1/n, as you keep including more and more terms,
That harmonic series diverges, okay, so we're subtracting just a little bit, the natural logarithm of n,
and that, we're gonna show that that does converge.
And it's gonna take a little bit of work for sure, but nothing too bad.
The basic strategy, what we're gonna do is we're gonna show that the sequence t sub n is bounded,
So that means that there's- If you start looking at all the values t sub n there's a number above that it doesn't-
it never gets above a certain number and it never goes below a certain number,
so that's the idea of being bounded.
We're gonna show that t sub n is bounded and that t sub n is monotonic.
Monotonic.
And we're actually gonna show that it's monotonic decreasing.
So if a sequence is either increasing or decreasing we say it's monotonic, so,
we're gonna show it's monotonic by showing that it's decreasing, and then,
there's a theorem that says if you have a bounded monotonic sequence that it converges, okay.
So we're gonna show that we have a bounded, monotonic sequence and that we'll use a theorem that therefore it converges.
So, okay. Here we go. Let's get down to the nitty-gritty.
Alright so we're gonna use some geometric arguments here, so the first thing I'm gonna do is
I wanna show that t sub n is always greater than 0. That's the first thing I'm gonna do.
I'm gonna show that it's bounded below, whoops sorry about that, bounded below by zero.
And what we're gonna do is we're gonna use just a little geometric argument here,
so this is gonna be the curve,
so this is gonna be the curve 1/x,
and I'm just gonna make again a geometric argument, so here's gonna be, here's 1, 2, 3, etc.,
we'l make this generically n, and then we'll have n + 1 out here
so, I'm gonna make some rectangles here, and I'm gonna make my rectangles extend to the right.
So let's see, let's make one more...
Okay, so I've got some shaded rectangles here, and notice that the shaded rectangles
I should say the area of the shaded rectangles, well what is the area of the shaded rectangles equal?
Well, let's see, the first one has a height of 1 and a width of 1, so that's gonna equal 1.
Notice that the next one, since that over here at x = 2, since it's on the curve 1/x,
this is gonna have a height of 1/2, and again it has a width of 1, so, hey that rectangle has an area of 1/2,
the next one is gonna have an area of 1/3, etc.,
the very last rectangle I have here, again since it's sitting on the curve 1/x, it's gonna have a height of 1/n,
so the area of the shaded rectangles, hey, that's giving us that, you know,
basically looks like, right, the harmonic series.
So, alright, no big deal there, so okay.
So let's make another observation here, if I integrate from 1 up to n + 1 of 1/x dx,
so now I'm finding the area underneath the curve, okay, so I'm just gonna compare again,
clearly the area of the rectangles is gonna be larger than the area underneath the curve.
Well if we find the antiderivative of 1/x, that gives just the ln(x)
okay, normally it's an absolute value but here we're putting in positive values so it doesn't matter.
Okay, so in this case we're gonna have ln(n + 1) - ln(1), well ln(1) is 0,
so the area underneath the curve is gonna be equal to ln(n + 1).
Okay so again what we've just said, right, the area underneath the curve, which is ln(n + 1),
that's gonna be definitely less than the area of those shaded rectangles.
So ln(n + 1), that's gonna be smaller than 1 + 1/2 + 1/3 + ... + 1/n.
Okay, so again, the area underneath the curve is less than the area of those shaded rectangles.
But also another observation, well ln(n) is gonna be less than ln(n + 1),
because the natural logarithm is an increasing function, right?
ln(10) is less than ln(11), or ln(12), or a hundred or a million,
so definitely ln(n) is gonna be smaller than ln(n + 1).
So okay, I'm even gonna put that there.
So ln(n) is less than all this stuff, well that means if I subtract,
since this quantity is bigger than this quantity, well that means that 1 + 1/2 + 1/3 + ... 1/n,
if a subtract away the ln(n), that's going to be greater than 0, right?
Because, again, this quantity is bigger than this quantity. So if I subtract, hey, it's bigger than 0.
So, ta-da, boom, we've got that it's bounded below. Okay, so that's the first thing that we wanted to do.
Showing that it's bounded above, we'll talk about that in a second.
So the next thing I want to show, I'm going to show that this sequence, t sub n, is decreasing.
Okay, so that's my next goal, I'm going to show that it's decreasing, and to show it's decreasing, that means that
I would want to show, okay, what do I want to show? I want to show that t sub n - t sub n+1
Well I want to show that that's greater than 0, right, because I look at some term and then I look at,
you know, I get some value for t sub n. I get some value for t sub n+1. If t sub n is larger than t sub n+1,
well, when I subtract them, I'm going to get something greater than 0, and if t sub n is larger then t sub n+1,
hey, that means it's decreasing. Okay, so that's my next goal, is I want to show that this happens,
and that's again going to imply that our sequence is decreasing. Okay, we're having fun. I'm having fun.
I think this is interesting. How are we going to do this, Patrick? Well, we're going to make a couple observations
here, and again, we're going to just look at- Make another little geometric argument.
So the first thing I'm going to do is I'm actually going to come up with a formula for t sub n - t sub n+1.
Okay, so t sub n, we said that's equal to 1 + 1/2 + 1/3 + ... + 1/n - ln(n).
Well, t sub n+1, that's going to be 1 + 1/2 + 1/3 + ..., we would have 1/n. Well, this one goes out to 1/(n+1)
though, so we'll have 1/n + 1/(n+1) - ln(n+1). Okay, so that's our t sub n and our t sub n+1.
I hope- I don't think I have yet. Hopefully I don't misspeak too much through this. Lots of little subscripts and n's.
So I'm just going to subtract, okay? So if we subtract, what's going to happen? Well,
1-1 cancels, the halves cancel, the thirds cancel, etc. The 1/n and the 1/n, those are going to cancel.
Well, I've got this negative ln(n), that's still going to be remaining. When I subtract, I'm going to have-
I'm going to end up with a negative 1/(n + 1), and then when I subtract, I'm going to end up with a positive
ln(n+1). So I'm just going to rewrite this as the ln(n+1) - ln(n) - 1/(n+1). So, again, this is basically what you
would get- Well, not basically. This is what you would get if you take t sub n and subtract away t sub n+1.
You would get that number, okay, after you plug it in and do the arithmetic.
Okay, so why on Earth do we care about this? Because we're going to actually make this appear in our-
When we do our little geometric argument. So, alright, let me get another piece of paper. So, we'll come back
to this in just a second. Alright, so again, let's go back to our curve. So, here's 1/x.
There's 1/x, and I'm going to look at this between the values of n and n+1. Okay, so the first thing to notice
is if we integrate from n to n+1 of 1/x dx, we can make a couple arguments here. So, one thing we can say
for sure, by- So again, this is going to be the area underneath the curve. If I make my little, you know,
box at this height. So the area of the box here, well, what is the area of that box? The area of that box is going to
be 1/(n+1), because again, I'm putting it into the function 1/x. Well, the area underneath the curve 1/x dx,
that's going to be all this stuff. Well, certainly this is larger than 1/(n+1). Again, just making that geometric
argument. But let's actually compute this.
So just like before, the antiderivative 1/x, that's the natural logarithm of the absolute value of x.
Again, we're assuming these that are, you know, positive values. So again, I'm just going to drop the absolute
value. So, we'll have the ln(x) from n up to n+1. Well, if we substitute that, we'll have ln(n+1) - ln(n).
Okay, so maybe that's starting to look like what we had just a second ago. And again, we're saying that this
quantity is going to be larger than 1/(n+1).
Well, same argument that we used, kind of, you know, previously. Since this quantity is greater than
this quantity, you know, I can just subtract from both sides.
That means that ln(n+1) - ln(n) - 1/(n+1), that's going to be greater than 0, okay? That's greater than 0.
But, oh yes, you know, this is- This now ties back into what we had a second ago.
Okay, because we said that t sub n - t sub n+1, that's equal to ln(n+1) - ln(n) - 1/(n+1).
We've now shown that this quantity is greater than zero, and what we said at the very beginning, was
we said if we can justify that that's true, that shows that our sequence is in fact decreasing. Okay.
We're getting close. Okay, so we've got that it's bounded below. We got that it's decreasing.
It's clearly bounded above. Notice that t sub 1, if we compute t sub 1, okay? So this is now a different little
thing. You know, t sub 1, that's just- If we go back to our formula, t sub 1 would say, basically, you just go up to
one, and you subtract away ln(1), which is 0, so t sub 1 is equal to 1. Since our sequence t sub n is decreasing,
we know that this is the largest value that you could ever get out of our sequence. So we've got that it's bounded
below. We've got that it's bounded above. We've shown that it's monotonic. So, QED, the proof is complete, right
We've shown every thing that we needed to show. We've got that it's bounded above and below.
We've shown that it's bounded, check. We've shown that it's monotonic, check. And by the bounded, monotonic
theorem, convergence theorem, it says since this happens, it says that our sequence t sub does in fact
converge.
What does it converge to? That's a different question. Whoops, I'm doing two things.
So t sub n does, in fact, converge. Now we haven't shown what value it converges to, but that's okay.
We've shown it's convergent and that's what I wanted to do in this video.
Oh, by the way, before I click off here, Euler's- This Euler constant, sometimes people use e for Euler's constant.
And I've always heard this- This is Euler's number, okay? e is Euler's number.
The Euler constant is the number that we're discussing in this video. It has tons of applications.
I remember reading in, you know, Laplace transforms and harmonics, and all kinds of things.
You know, Google it if you're interested, for sure. Check it out, it's definitely, you know, not one of these
just contrived little numbers just- Hey, look at this, it converges.
It definitively has practical applications, so check it out, and alright, I hope this video was interesting to you.
And hopefully if you're studying Euler's constant, it gave you a little insight on showing how to actually
show that this sequence does in fact converge.