It’s Professor Dave, let’s look at alternating series.
We’ve examined arithmetic sequences and geometric series, so let’s look at another
type of series now.
An alternating series is one in which the signs of the terms in the series alternate
between positive and negative values.
How can we produce a series like this?
In order to produce an alternating series, there will usually be a term involved whereby
the number negative one is raised to some exponent, like N or some term with N in it.
For example, take the series negative one to the N minus one power, over N, from one
to infinity.
For the first term, we are raising negative one to the zero power, which is one, and then
we divide by one, so we just get one.
But then, for the second term, negative one is raised to the first power, so we leave
the negative one, divide by two, and we get minus one half.
For the third term, the negative one is squared, to give positive one, and we get one third.
If we continue in this manner, we see how the number in the denominator increases by
one each time, as expected, but because the term in the numerator alternates between positive
and negative one, the terms in the series alternate between positive and negative, all
the way to infinity.
So any time you see a series and there is a negative one in parentheses being raised
to the N power, or something similar, you know that you are looking at an alternating series.
Because of these alternating signs, it becomes trickier to assess whether the series will
converge or diverge, but we can use something called the alternating series test.
This test says the following.
Let’s say we have some series in the form of negative one to the N minus one, times
B sub N, which gives us B one, minus B two, plus B three, minus B four, and so forth,
where B sub N is greater than zero.
If B sub N plus one is less than or equal to B sub N, or in otherwords if each term
in the series is smaller than the last, disregarding sign, and also if the sequence B sub N goes
to zero in the limit of infinity, then the alternating series will be convergent.
So for example, with that first series we looked at, we can pull the term in the numerator
to the side, and B sub N is therefore represented by one over N. Each term in this series is
smaller than the previous, and in the limit of infinity, one over N goes to zero, so both
criteria are met, and this is indeed a convergent series.
Let’s apply this to a few series to make sure this sinks in.
Take this one that is the same as our previous example, except with root N in the denominator.
Does each term in this series get smaller?
As N increases, its square root increases, so the fraction gets smaller as we go.
And in the limit of infinity, this term does become zero.
Both criteria are met, and this sequence converges.
What about this one?
Negative one to the N times the quantity three N minus one over the quantity two N plus one.
Here does each term get smaller?
Well we have a larger multiple of N on top.
As N gets very large, the plus and minus one terms will be negligible, and we will just
have the numerator getting increasingly larger than the denominator.
The terms are therefore not getting smaller, and in this case that means that the series
is divergent.
How about this one, with negative one to the N plus one, times N squared over the quantity
N cubed plus four?
As we might have realized by now, the slight discrepancies in the exponent associated with
negative one are not really relevant, so let’s look at the rest.
With a higher power of N in the denominator, we can be sure that each term in the series
is getting smaller.
Does it go to zero in the limit of infinity?
Well just plugging in infinity, we get infinity over infinity, so let’s divide top and bottom
by N squared.
That gives us one over the quantity N plus four over N squared.
Now plugging in infinity, we get one over the quantity infinity plus zero, or one over
infinity, which is zero.
Both criteria are met, and this series converges.
Let’s do just one more, negative one to the N times N over the natural log of N. Do
the terms increase or decrease?
Well the natural log of N will always be smaller than N, and it will increase at a slower rate
than N does, as we can see by plugging in a few values for N, so the terms are increasing,
and this series is divergent.
So now we know how to look at alternating series and test for convergence, but we must
now describe two different types of convergence.
Let’s say we have some series, A sub N, from one to infinity.
Now let’s take the absolute value of that series, which means we are taking the absolute
value of every term in that series.
If A sub N is convergent, and the absolute value of A sub N is also convergent, then
that series is called absolutely convergent.
Taking our first example, negative one to the N minus one, over N squared, we said that
this series was convergent.
Now let’s take the absolute value of this series.
This means that whenever the negative one gives us a negative value for a particular
term in the series, we just make it positive instead, so rather than an alternating series,
we just get a regular p-series with a P value of two, as though the negative one term was
not present.
We know that this series is also convergent, for reasons discussed in the previous tutorial.
Since they are both convergent, then the original series is absolutely convergent.
On the other hand, let’s modify this slightly and take away the exponent in the denominator,
leaving simply N. That will give us a different alternating series, which we know is also convergent.
However, when we take the absolute value of this series, which turns all the terms positive,
we get another p-series, but this time with a P value of one.
Any p-series with a P value of one or less will be divergent, and so we say that the
original series is conditionally convergent.
This is different from being absolutely convergent, because taking its absolute value produces
a series that is divergent.
There is a test that is helpful in determining whether a series is absolutely convergent,
and it is called the ratio test.
We can place A sub (N plus one) over A sub N to form a ratio, and take the absolute value.
Then we take the limit of this ratio as N approaches infinity.
This limit will equal L. If L is less than one, then the series A sub N is absolutely
convergent.
If L is greater than one, the series is divergent.
And if the limit equals one, the ratio test is not useful, and it can’t tell us any
information about the series.
Let’s try one example that requires this ratio test.
How about the series N to the N power over N factorial.
We need to get the ratio of A sub (N plus one) over A sub N.
For A sub (N plus one), we just put N plus one wherever N was, so we get the quantity
N plus one raised to the N plus one power, over N plus one factorial.
Then we divide that by the original series, N to the N over N factorial, and we take the
absolute value of all that.
But everything is positive, so the brackets just go away.
Now, dividing by a fraction is the same as multiplying by its reciprocal, so let’s
flip this denominator and bring it up here.
Looking at all this, we should be able to cancel some things if we get clever.
For one thing, we have N plus one factorial, as well as N factorial.
Well we know that N plus one factorial will be N plus one, times N, times N minus one,
and so forth, so really, we can just say that this is equal to N plus one, times N factorial,
since the rest of this is just N factorial anyway.
Then on the top, raising this binomial to the N plus one power is the same as N plus
one to the N power times N plus one to the first power.
This is just using a property of exponents in reverse.
Now things look pretty good.
We can cancel out an N plus one from top and bottom, and we can do the same with N factorial,
leaving us with N plus one to the N power over N to the N. Since these have the same
exponent, let’s pull it out here, to get N plus one, over N, all to the N power.
We just have to be clever one more time and divide through by N to get one plus one over
N, and recognize that this is actually equivalent to E, as was described earlier in the series.
E, or Euler’s number, is around 2.72, and that is greater than one, so according to
the ratio test, this series must be divergent.
Now that we know about alternating series and how to test them for convergence, let’s check comprehension.