Now, Part B well we have a Maclaurin series
now on that general term so whoops now
we do we're going to determine the
interval of convergence using that
general term which means that hopefully
as you know maybe watching this as
you're learning this concept I'm not
just reviewing we're going to do the
ratio test which means we're going to
take the limit as n approaches infinity
of a sub n plus 1 over a sub n because
this is basically sort of a well power
series and we're going to make sure then
the bet ratio is less than 1 so that
power series converges and then we're
going to test the endpoints of this
interval of convergence because this is
just going to give us what kind of
initially the radius of convergence and
then we'll we'll know what the open
interval is for the interval of
convergence so we have a sub n plus 1
which means that we're going to just
look at this general term and everywhere
we see an N replace it with an N so n
plus 1 so we have negative 1 to the n
plus 1 plus 1 which is n plus 2 we have
times X to the n plus 1 plus 1 which is
n plus 2 over n plus 1 as a factor times
3 to the n plus 1 now when I go ahead
and and we have that a sub n plus 1 term
now divided by a sub n I'm going to flip
that up and change that to a
multiplication of the reciprocal just to
save some space on the chalkboard so
we're going to x naught divided by a sub
n term but multiplied by the reciprocal
of the a sub n term so that means we're
gonna multiply by n times 3 to the N
power over negative 1 to the n plus 1
times X to the n plus 1 and we're going
to make sure that that
limit stays less than one okay so is
there anything you know we can cancel
out well here we have n plus 1 basis of
negative 1 and in the numerator over
here we just simply have one more well
I'm going to show this cancellation but
ultimately we have absolute value
symbols so that negative once-in-a
cancel out anyway in the numerator of
our first factor we have n plus 2 bases
of X and here we have n plus 1 basis of
X so here we have one more base of X
then we have in this denominator so
that's going to cancel out we have n
bases of 3 in the numerator and we have
n plus 1 basis of 3 in this denominator
so these cancel out we're going to have
one base of 3 one factor of 3 left and
that denominator and nothing's going to
cancel with this n over n plus 1 so
we're looking at the limit as n
approaches infinity they did tell us
ashore work this might be a slight
overkill for some of us but I think it's
good
negative well negative what n times X
over 1 plus 1 times 3 again for the
ratio test we need to make sure the
limit as n approaches infinity of this
is equal to or less than 1 now the
absolute values are going to basically
just take care of this negative force
and remember we're letting n approach
infinity not X so you're looking at just
this n over N plus 1 and we don't need
to show like l'hopital's rule or some
kind of process to show that the limit
of basically n over n plus 1 is equal to
1 they're not expecting that much work
to be shown we're just looking now at
and the absolute value of x over 3 is
less than 1 which means that if we
multiply both sides by 3 just as a
review we could look at this expression
and we could see the center of
convergence which we don't really need
to because we're told that it's a
Maclaurin
series which is basically just a special
type of Taylor series where the center
of convergence is equal to zero but we
could see that this is the absolute
value of X minus zero which means that
the center of convergence is 0 and then
that absolute value of this binomial
that has a coefficient of on the X of 1
see I got rid of the coefficient of 1/3
minus 0 is less than 3 when you have it
in this X minus C absolute value that is
less than basically you're looking at
are your radius of convergence so R I
can't call it an interval of convergence
yet because we haven't tested the
endpoints but we have so far a open
interval of negative 3 to 3 now that's
not interval of convergence but it's
like we're almost there now all we have
to do is test the endpoints we have a
center of convergence of 0 and a radius
of convergence of 3 so now that we have
this open interval that is the beginning
of us developing the final answer of
what our interval of convergence is we
need to test the endpoints ok so we're
going to do that by using this general
term and right not this whole tale
Maclaurin series but just the series in
terms of Sigma notation so with the
endpoint of negative x equals negative 3
we have the series where and starts at 1
and goes to infinity which I should be
including that correct yes okay we have
use and again this general term we've
got negative 1 to the n plus 1 power x
now not X because we're replacing it
with negative 3 so negative 3 to the n
plus 1 power over n times 3 to the N and
all I have to do is show the work as you
start to determine whether the series
converges or diverges so that you know
whether to include the endpoint of
negative 3 or not and I'm going to show
a lot of work just to make sure that I
don't miss the question and
well you understand what I'm doing so
first thing I'm going to do is bring out
this negative one as its own factor so
we can see how it combines with this
negative one to the n plus one power
when you multiply like bases of course
we add the exponents so rewrite in that
negative 3 to the n plus 1 power we have
with the expansion now negative 1 to the
n plus 1 times negative 1 to the n plus
1 times 3 to the n plus 1 multiplying
these like bases the two bases that have
the value of negative 1 we have negative
1 n plus n is 2 n + 1 plus 1 is equal to
2 so we have this series where negative
we have negative 1 to the 2n plus 2
times 3 to the n plus 1 power over n
times 3 to the N power this negative 1
its exponent is 2 n if you take a number
multiply it by 2 it's even if you add
two it's to leave it a negative 1 to an
even power is positive so the negative
on goes away we have 3 to the n plus 1
in the numerator and we have 3 to the N
in the denominator so the numerator has
one more base or factor of 3 if you will
so this is going to the 3n and the
denominators and a cancel out with all
but one of the threes in the numerator
and now in our series notation all we
have left is 3 over N and I'm going to
go ahead and just to make sure it's
perfectly clear we're going to pull this
factor of 3 out in front of the Sigma
notation so that we can see that we have
3 times the summation where n starts at
1 and goes to infinity of 1 over N well
that series but basically the series of
1 over N is a harmonic series which
diverges so this is harmonic give a
reason please harmonic series diverge so
for the purpose of this problem finding
the interval of convergence at least in
terms of the interval
of convergence X cannot be equal to
negative three okay so this part the
season the left of negative three is
going to stay now let's test X is equal
to positive three and see if this is
going to be an open interval or closed
interval on our interval of convergence
one more time right out that series
notation that Sigma notation or the
series with Sigma notation for in series
with Sigma notation so we have summation
where n starts at one goes to infinity
one more time of negative one to the n
plus one power times 3 to the n plus one
over N times three to the N problems
like this is really why in Chapter nine
with all of your interval of convergence
tests is for this final step of finding
your Taylor and Maclaurin series testing
the endpoints so we have 3 to the n plus
1 over 3 to the N so in the denominator
all of those factors of 3 are going to
cancel out leaving you with just one
factor of 3 in the numerator and there
is nothing to take care of that negative
1 to the n plus 1 so this is again
moving that 3 out front just to make
this abundantly clear we have 3 times
the summation where again starts at 1
and goes to infinity of negative 1 to
the n plus 1 power over N and that is a
alternating harmonic series which
converges
thus 3 can be into our interval of
convergence and the interval of
convergence starts at with the open you
know it's open on the left side so from
negative 3 open to 3 closed or you can
say that negative 3 is less than X which
is less than or equal to 3 is our
interval
of convergence for our series negative
one to the n plus one power X to the
negative x to the n plus one over three
times 3 to the N
great that is Part B Part C coming up
right now
you
