This video is going to be about the remainder estimate for the integral test and what the remainder
[with] this basically this estimate remainder estimate says is that suppose you have a convergent series
and
It says remember s sub n is the partial sum of the first n terms
s
Represents the sum if you add up all infinitely many terms so basically it says the difference between your estimate and the true value
we're going to call that R sub N, and
It says that remainder
if so if you sum up the first [n] terms
the remainder is going to be less than [or] equal to the integral from 0 to [infinity] of f of x where a
sub [N] equals f sub n f of N and
It says [it'll] be bounded above by N. Plus 1 to infinity
All right
So in this problem
I want to look at the question. It says if you use the first [ten] terms of this
From 2 to infinity of 1 over N. Natural log of N. Squared how accurate would your estimate be?
So first off really we should look at the the integral from 2 to infinity
1 over so we'll turn our function. I like to use x's you can use ends if you want
I like to just use x's because it makes my natural logarithms not have so many ends in there
Okay, so to integrate this remember we replace our upper limit of integration
With a [t] and in this case, we'll let t go to infinity, [okay]
[and] now we have to integrate this so I just kind of integrate the improper integral
1 over x natural logarithm of x quantity squared
Dx and I think we can integrate this just by using a good [old] U substitution
will let U equal the natural logarithm of x
so d U will be 1 over x
Dx
[ok] so we're going to replace our natural logarithm of x
Here, so we're going to call that [ru] [Ru] [part], so instead of having Ln of x squared. We're going to have U squared
But then what are we left with we're left with 1 over x Dx which?
Conveniently we have so we have d U equals 1 over x DX
So we can replace the 1 over x DX with Simply 1
d you
Okay
So really we're integrating one over U squared
well [1] over U squared remember we can rewrite [that] as
U to the negative 2nd
D u and this is now the integral that we're calculating
[ok], so if you integrate [u] to the [negative]
[2nd] you add [1] so you'll get u to the negative first over negative 1 plus C
so we can rewrite this as negative 1 over [u]
Plus C. And again we know that this is I'm going to be negative 1 over the natural logarithm of x
Plus C
Ok so that's what we get when we integrate this thing so what that tells us is
What that tells us is when we integrate this thing then we're going to get the limit from t. Goes to infinity
so if you integrate 1 over x natural log of x quantity squared
we said you get negative Ln of x remember for definite integrals, you can drop the
Limit Via this arbitrary constant
[and] then we'll be calculating this from 2 to t
Ok so now we need to evaluate this part
So it says we'll have the limit as t goes to infinity from negative 1 over the natural logarithm of t
minus negative 1 over the Natural logarithm of 2
When we plug that in and as t goes to infinity remember the natural logarithm of t goes to infinity?
[so] the denominator of the first parts getting arbitrarily large the Ln of t
So what that tells us?
Is that in fact?
Well think about it 1 over a big number is a number close to zero so that first limits going to [zero] we have two
Negatives that makes a positive 1 over the natural logarithm of two
ok the integral test says if you get a
You'll get a positive finite number finite being the key part
if the improper integral converges so does the series [ok] so we do have a convergent series by the
by the integral test
All right, so it says now. It says if we use the first 10 terms how accurate what's your estimate [b]. So remember?
So really we're going to only use the right-hand side of that inequality
remember, it's said if you use in this case 10 terms [it] set our integral our remainder will be less than or equal to the
integral [from] 10 to infinity
of the function
Okay, well
What does that mean it well my function in this case again?
We're going to use the one over x natural log of x quantity squared
instead of using a two
Well now I'm going to use a ten we still have the limit as t [goes] to infinity
[I] think maybe I left that off a second ago
So what's going to change? Well? The only thing that's going to change is our lower limit of integration would turn into a ten
Okay, well that means over here when we plug it in
Instead of having negative 1 over the natural logarithm of 10 or excuse [me] of x we'll get the natural logarithm of 10
So what does that mean it means our last calculation again will still give this first limit going to zero?
but it says [that'll] be less than
Well, the integral will be 1 over the natural logarithm of 10
So it says if we sum up the [first] 10 terms
Our remainder will be correct with error at most
[oh], I can't write it trying to write and talk our error will be at most 1 over the natural logarithm of
10
[ok], so that's the basic idea on at least one use of the this integral test this estimation
So basically in the process of showing that it's convergent
You know really the only thing that's going to change is your lower limit of integration
Will change if you're just trying to find an upper estimate?
For your for your error
And I think equivalently if you use the other formula the only other thing that would change would be the lower bound
[I] think you could also show that you would get 1 [over] the natural logarithm of 11
Being your lower estimate
Okay, so that's supposed to be 11 natural logarithm of 11, all right?
I hope this video helps it makes a little [bit] of sense
like I said to be able to use it you have to make sure your integration techniques are okay, because
You know any of the series that you see?
You know it's [anymore]. You may have a variety of integration techniques
When you have to integrate these so anyways [alright] good luck out there. I hope this helps