Professor Dave again, let’s check out some tests.
We just learned about convergent series, which are ones where the sum of all the terms in
the series is some finite number.
But it can often be difficult to calculate the precise value of the sum, and there are
only certain special cases where we can easily get the sum using simple computation.
Luckily there are special tricks we can use to help us find sums, or at least estimate
them, and one such trick is called the integral test.
Take something like the sum of one over N squared, from one to infinity, giving us one
over one squared, plus one over two squared, plus one over three squared, and so forth,
and we can easily plot these points graphically.
If we also show the function Y equals one over X squared, essentially connecting the
dots, we recognize that this is the precise example we looked at when we learned about
improper integrals, and we know the area under the curve over this infinite interval is actually finite.
Therefore, we shouldn’t be surprised to learn that this similar looking series is
convergent, especially since the sequence one over N squared does go to zero as N approaches infinity.
In fact, we can actually use this function to estimate the sum of the series.
Let’s make some rectangles under the curve, each with a base of one and a height equal
to the value of the function at the right endpoint of the interval.
The area of each rectangle is equal to one of the terms in the series, because this rectangle
has an area of one over two squared, and the second term in the series is one over two squared.
This rectangle has an area of one over three squared, and the third term in the series
is one over three squared.
So every rectangle all the way to infinity has a corresponding term in the series, and
we are leaving out only the first term, one.
The sum of the areas of these rectangles is pretty close to being the area under the curve,
which we know is just the integral of the function.
So the sum of the series will be approximately equal to one, to account for the first term
in the series that has no corresponding rectangle, plus this improper integral, which approximates
the sum of all the rectangle areas and therefore accounts for the rest of the terms in the series.
We know how to evaluate the integral, we just find the antiderivative, which is negative
one over X, and we evaluate at infinity and then at one, which gives us zero minus negative
one, or positive one.
The one from before plus the one from the integral equals two, so that’s our estimate
for this sum.
Now we know that the rectangles don’t quite fill up the total area, as there are some
empty spaces, so the integral is bigger than what we really wanted, and therefore the actual
value for the sum of the series must be a little less than two, but we’ve still done
a pretty good job at getting some information about the series.
How can we summarize this technique in a general way?
We can say that if we have some function, F, that is positive, continuous, and decreasing
from one to infinity, and we also have a sequence that is derived from this function, in otherwords,
A sub N equals F of N, then the series that can be formed from this sequence, from one
to infinity, will be convergent only if the corresponding improper integral is convergent.
If the integral is divergent, the series will be divergent.
If the series is indeed convergent, then we can also estimate its sum by calculating the
improper integral in some way that is similar to the example we just performed.
And that is the essence of the integral test.
Let’s just mention a few important truths that can be found by applying the integral test.
Any series in the form of one over N raised to some exponent is called a P-series.
This includes our first example, one over N squared.
As it happens, any P-series will be convergent if P is greater than one, and it will be divergent
if P is less than or equal to one.
So if we have one over N cubed, it will converge.
If we have one over the cube root of N, or one over N to the one third, it will diverge.
There is one more test we can learn that will be an important technique, and it is called
the comparison test.
With this technique, we will take some series and compare it with a similar series that
we know to be convergent or divergent.
For example, say we are looking at the series one over the quantity two to the N, plus one,
from one to infinity.
This looks like it could be tricky to assess, but what if we compare it to a simpler series,
one over two to the N.
This is a simple geometric series, and as we have already seen, the fraction gets smaller
and smaller as the exponent gets larger and larger, so the sequence converges to zero,
and as we saw in the previous tutorial, this is indeed a convergent series with a sum that
is equal to one.
If we know that this series converges, what can we say about the original one?
Well, this plus one term in the denominator means that each term in this series must be
smaller than each corresponding term in the other series, and if every term is smaller,
then this series must also converge, and the sum of this series must also be less than one.
So we aren’t calculating the precise sum with the comparison test, but it is a useful
tool for determining whether a series is convergent or divergent.
If every term in a series is smaller than each corresponding term in another series
that is known to be convergent, then it too, must be convergent, as we just saw.
On the other hand, if every term in a series is larger than each corresponding term in
another series that is known to be divergent, then it too, must be divergent.
Apart from understanding the comparison test logically, in this manner, we can also represent
it a little more formally, by comparing two series, A sub N and B sub N, and setting up
these relationships between them, whereby if B converges or diverges, and these conditions
are met for A and B, A will also converge or diverge accordingly.
Let’s just do one example to make sure we can apply this.
Take the series five over the quantity (two N squared plus four N plus three).
We can simplify this to five over two N squared to use the comparison test.
What do we know about this simpler series?
Well we can manipulate series just like we do integrals, and five halves is a constant
that can be pulled outside of the sum, as doing this is essentially like factoring five
halves out of all the infinite terms in the series.
Then we are left with something very familiar, and we know that this series is convergent.
Our original series has terms that must be smaller than the terms in this series, because
there is more in the denominator, and N is always positive, so the fraction is always
smaller than if those terms weren’t there.
Therefore this series is also convergent, and its sum must be less than that of the
one we are comparing it to.
Now that we have the basics down for assessing whether a series is convergent or divergent,
let’s check comprehension.