Hi everyone. Welcome back to integralcalc.com.
We're going to be talking about intervals
of convergence today. The problem that we’re
going to be using is the sum x minus 4 to
the n divided by the cube root of n.
So the first thing that we need to do when
we're talking about intervals of convergence
is the following. We substitute n plus 1 everywhere
we have n in this function. So in this case
that would be here in the exponent on the
numerator and what’s under the cube root
in the denominator.
So you can see we substituted for n plus 1
for both of those here and we take that whole
thing and we divide by the original function,
so everything down here is just the original
function.
And we find the absolute value of that whole
thing.
That’s what these brackets are here for,
absolute value. So, we find the limit as n
approches infinity of the absolute value of
this whole thing. We write that out and then
we start simplifying. So what we're going
to do is instead of dividing this big fraction
by this big fraction, we're going to multiply
the top fraction by the inverse of the bottom
fraction.
So what you can see that I did here is flip
this bottom fraction. This cube root of n
went to the numerator and the x minus 4 to
the n went to the denominator. We just flipped
it upside down and then we're multiplying
instead of dividing. That’s the first process
of simplification.
So now what we're going to do is simplify.
We can cancel out some of these x minus 4
terms because as you can see, we have x minus
4 in the numerator and x minus 4 in the denominator.
The one in the numerator is raised to the
n plus 1, the one in the denominator is raised
to the n. Remember that if you have something
in the numerator of a fraction and in the
denominator of a fraction, and the bases are
the same but the exponents are different,
you can add and subtract those things from
each other and simplify. So what we do is
we take the expoent in the numerator, the
n plus 1 and we subtract the exponent from
the denominator n and n plus 1 minus n is
just 1, which means we're just left with x
minus 4 to the power of 1 in the numerator.
So this x minus 4 is going to go completely
away in the denominator and we'll be left
with this x minus 4 to the 1 power in the
numerator.
So that's our next step of simplification.
Then we're going to go ahead and take the
x minus 4 and put it by itself and rearrange
these fractions so that we have all of our
n terms in fraction. Since everything here
is multiplied together, we can do that. We
can shift out the x minus 4 so I can kind
of get my xs and my ns separate, which is
a good tactic for simplifying in these kinds
of problems.
Now what I'm going to do is remove the x minus
4 completely from this absolute value brackets.
The reason that I can do that is because we're
taking the limit as n goes to infinity. So
in terms of this limit, all we care about
are ns so we leave the ns inside and we can
move the x minus 4 outside. When we move out
the x minus 4, because it's in absolute value
brackets, we have to leave the absolute value
brackets here on the x minus 4.
Then we can just take the limit as n approaches
infinity of this fraction here with the ns.
So now we're going to go ahead and simplify
further. Since both our numerator and our
denominator inside here are cube roots, we
can go ahead and take the cube root of the
numerator divided by the denominator.
Now that we've done that, we can go ahead
and plug in infinity for n and try and get
an idea of what this whole limit function
is going to simplify to.
As you can imagine, we have infinity divided
by infinity plus 1. Infinity is a hundred
million, gazillion. Adding 1 there isn't going
to make any difference. What we're going to
end up with is infinity over infinity.
And if we take the cube root of infinity,
we're still going to get infinity. I know
it's sometimes a difficult concept to grasp
but we have a little bit abstract. When we
think about an infinite number, even taking
the cube root of it, you're still going to
have an infinite number. So this whole thing
here will end up with infinity over infinity
which is going to simplify to 1, because infinity
divided by infinity just gives you 1.
So we end up with the absolute value of x
minus 4. This limit as n approaches infinity
of this whole thing is going to be one and
we’re going to end up with only the absolute
value of x minus 4.
That’s what this simplifies to. So now that
we have this absolute value of x minus 4,
we need to find out where in this range our
function is going to converge. And the way
that we do that is to find out where it's
less than 1. So we set it to less than 1 and
then to get rid of this absolute value brackets,
we can transform this inequality here to be
the quantity x minus 4 greater than negative
1 and less than 1.
Whenever you have this absolute value and
an inequality, you can take away the absolute
value brackets as long as you put the opposite
sign of this quantity here on the other side.
So since we have 1 here, we put negative 1
on the other side and we make this x minus
4 greater than.
So now that we've done that, we just want
to get x by itself in the middle so we add
4 to each of these three parts of the inequality,
the negative 1, the x minus 4 and the 1 because
adding 4 will get this x by itself.
So when we do that, when we add 4 to the left
hand side here, we get 3, we get x by itself
and we get 5 over here on the right.
So we're looking at x greater than 3 and x
less than 5. So, on the range 3 to 5.
Now we know that the series is going to converge
between 3 and 5, we just need to check our
end-points 3 and 5 to determine whether the
series will still converge at the exact point
3 and/or the exact point 5. And the way that
we do tha is by plugging in 3 and 5 separately
to our original function.
We plug in 3 for x here and we start simplifying.
We're going to get negative 1 here, this 3
minus 4 is negative 1, we leave the ns alone.
And now it comes down to different tests to
determine whether or not this term converges.
In this case, we're going to be looking at
the alternate series test because whenever
you have a negative 1 raised to the power
of n, you know that it's going to be alternating
series. This is your dead give away right
here.
And you can see alternating series. You have
this negative 1.
It doesn't particularly matter. In the example
shown in the table, we have n minus 1 and
in ours we only have n. But this negative
1 raised to the n is a giveaway that you're
talking about an alternating series. So in
that case when you have an alternating series,
we need to test two things. One, that the
limit as n approaches infinity of the series
is equal to 0, and also that the limit as
n approaches infinty is going to decrease.
So you may have noticed that these two functions
here look slightly different than these above
in the table. What we do to test is just take
the negative 1 and replace it with 1. We have
to do that inorder to test.
So if you replaced this negative 1 with one,
we'll get 1 to the n and 1 to the n is always
going to be 1.
So that's why we end up with 1 in the numerator
here. We leave the cube root of n in the denominator
and now we're testing these two functions.
First to make sure that it equals 0, and secondly
to make sure that it decreases as n increases.
So let's go ahead and test this first condition
here.
The way that we do that is by plugging in
infinity for n and when we're going to take
the cube root of infinity, we're still going
to have an infinite number. So we'll end up
with 1 over infinity.
And if you can imagine, 1 divided by the biggest
number ever is going to give you 0.
1 divided by 1 is 1.
1 divided by 1 is point 10.
That number gets lower and lower and lower,
so the limit is going to 0 because this eventually
is going to reduce all the way to 0. Not quite
but the limit is going to be 0.
So that coindition is met. The limit as n
approaches infinity of this function does
infact equal 0.
If both of these conditions are met, the series
converges at 3.
So we need to check this other one now.
We need to check whether or not, as n gets
larger, this whole function here decreases.
So let's go ahead and plug in two test conditions
to make sure. I'm going to plug in 8 for n
just to make this work out nicely. If i plug
in 8 for n, I end up with 1/2, if I plug in
27 for n, I'm going to end up with 1/3.
So you can see, as n goes from 8 to 27, as
n increases, the over all value of the function
is going to decrease. We’re going to get
1/2 and then 1/3. So that's decreasing, which
means this whole function is decreasing and
that condition is met which means the series
does converge at 3.
So we’ve checked that end-point. Now we're
just going to put the information that it
converges at 3 on hold. We’re putting that
on hold for a second and we have to go check
5, because remember x was between 3 and 5.
So we plug in 5 for x and when we simplify,
we end up with 5 minus 4, we get 1 to the
n in the numerator. 1 to the n, no matter
what n is, it's always going to be 1 so we
just end up with 1 on the numerator here.
Then to simplify this further, I can change
this cube root of n in the denominator to
n to the 1/3.
The square root of n is the same as n to the
1/2 while the cube root of n is the same as
n to the 1/3.
Whenever you end up with 1 divide by n to
an exponent, you know that you're talking
about a p-series test, because that's actually
what the p-series looks like here.
We end up with 1 divided by n to p.
So in our case, 1/3 is going to be equal to
p.
You're always looking for the exponent, so
whatever the exponent is, you set p equal
to. In our case, p equals 1/3 and the p-series
test claims that if p is less than or equal
to 1, then the series diverges at this point.
If, it's greater than 1 then the series converges.
In our case, 1/3 is less than 1 which means
that our series diverges at the point 5.
So everything between 3 and 5 in our series
is converging. We tested before to make sure
that the series at the point 3 converged and
in fact it did converge. But in this point,
5 it diverges, which means that our final
answer looks like this.
This is the way that we describe convergence
when we're talking about the interval of convergence.
Everything between 3 and 5 converges. The
series also converges at the point 3 so we
do this hard bracket { [ }. Since the series
diverges at the point 5, we have the soft
curve bracket { ) } here around 5. So this
{ [ } means converges at 3, all the way from
3 to 5 but diverges { ) }at the point 5 itself.
So that's going to be our final answer and
I hope that was helpful to you guys and I'll
see you in the next video. Bye.
