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.
