[Tutorial Instructor]
Welcome to a lesson on
absolute and conditional
convergence of an infinite series.
If a given series converges
and the absolute value
of the given series converges,
then the given series is
absolutely convergent.
However, if the given series converges
but the absolute value of
the given series diverges,
then the series is
conditionally convergent.
So to determine absolute
versus conditional convergence,
we're going to take a
look at the given series
as well as the absolute
value of the given series.
Let's go ahead and give it a try.
The given series is an alternating series
so we'll start by applying
the alternating series test.
So looking at this infinite
series we know that a sub n
would be the nonalternating
part of this formula,
so a sub n would be one over n plus two.
So using the alternating
series test we'll start by
taking the limit of this to make sure
that it's equal to zero.
Our numerator is fixed,
our denominator is
increasing without bound,
so this limit does equal zero,
and now we need to make
sure that a sub n plus one
is always less than or equal to a sub n.
To determine a sub n
plus one, we'll replace n
with n plus one.
So we have one over n plus one plus two,
which would be n plus three,
less than or equal to a sub n,
which is one over n plus two.
Well these denominators
will always be larger
than these denominators here,
and therefore this fraction will always be
less than or equal to this fraction here.
So both conditions of the
alternating series test
have been met.
So this given series is convergent.
But now I want to determine if this series
is absolutely convergent or
conditionally convergent.
So now we'll take a look at the summation
of the absolute value of
this, so we'll just have
the summation of one over n plus two.
Let's go ahead and do
that on the next screen.
If this series converges,
then the original series
is absolutely convergent,
and if this series diverges,
then the original series is
conditionally convergent.
To determine if this series
is convergent or divergent,
let's use the integral test.
So we'll let f of x equal
one over x plus two.
So if we apply the integral test,
then, if we take the
integral from one to infinity
of one over x plus two, with respect to x.
So if this improper integral
converges, so does the series.
If it diverges, so does the series.
So this'll be equal to the
limit as b approaches infinity
of the integral from one
to b of one over x plus two
with respect to x.
If we let u equal x plus two,
then the u would just equal the x.
So if the antidigit of
one over u would just be
natural log u, so in this case we'd have
natural log x plus two, but
this is a definite integral,
so we're going to have the
limit as b approaches infinity
of natural log b plus
two minus natural log
of one plus two, or natural log three.
Well as b approaches infinity
so does the natural log of b plus two,
so we can state now that
since the summation of one
over n plus two diverges,
by the integral test, the
original alternating series
is conditionally convergent.
Let's go back to the previous
screen just for a moment.
We have the original series, which,
by the alternating series test converged.
Then we considered the absolute value
of this infinite series as we see here,
and since it diverged, that tells us
that the original series is
conditionally convergent,
not absolutely convergent.
Let's take a look at another one.
Notice the given series is alternating,
so we'll apply the
alternating series test.
So a sub n is going to equal one over n
times the cube root of
n, which is the same as
n to the one third.
So let's go ahead and
rewrite this as one over n
to the four thirds power.
Now we'll take the limit of a sub n,
and this limit does equal zero,
so now we just need to check to make sure
that a sub n plus one
is less than or equal to
a sub n.
Well a sub n plus one would be n plus one
raised to the four thirds power,
and a sub n is just one
over n to the four thirds.
Notice these fractions here would have
a larger denominator than
these fractions here,
and therefore these
fractions will always be
less than or equal to
these fractions here.
So we have met both conditions
from the alternating series test,
so this given alternating
series does converge.
But we want to know if it's
conditionally convergent
or absolutely convergent.
So now we'll consider the summation
of one over n times the cubit of n,
which as we said before is the same as
one over n to the four thirds power.
To determine if this series converges,
we can apply the p series test.
By the p series test with
p equal to four thirds,
which is greater than one,
this series converges.
So since the original
alternating series converged,
and the absolute value
of the alternating series
also converges, we can state that
the original alternating series
is not only convergent but
it's absolutely convergent.
And I think we have time for one more.
Let's start by applying the
alternating series test,
where a sub n is going to
equal five n divided by
nine n minus two.
If we take the limit of a sub
n as n approaches infinity,
because the degree of the
numerator and denominator
are the same, this limit
is equal to the ratio
of the leading coefficients,
or five ninths.
So this alternating series fails
the n'th term divergent test
and therefore is divergent.
And I think we're out of time.
Thank you for watching.
