### Video Transcript

Using the limit comparison test,
determine whether the series the sum from π is one to β of one over four to the
power π plus six is convergent or divergent.

Weβve been given a series the sum
from π is one to β of one over four to the π plus six. The terms of our series are
therefore one over four to the π plus six. And weβre asked to determine the
convergence of this series using the limit comparison test. This test says that if we have two
series with terms π π and π π, which are positive, and the limit as π tends to
β of π π over π π is a constant πΆ, which is greater than zero and less than
β. Then either both series converge or
both series diverge.

We have a series with terms one
over four to the π plus six. In order to use the limit
comparison test, we need another series to compare in our ratio of terms. Weβd like to find another series
where the terms will make it easier to show that one term divided by the other as π
tends to β is constant. We also need to know whether this
second series converges or diverges. To find the terms of this second
series, what we can do is look at the behaviour of our terms as π tends to β. As π tends to β now in term one
over four to the π plus six, the six becomes superfluous, and our term turns to one
over four to the power π.

This could help us since this is a
term of a geometric series. So if we let the terms of our first
series equal to π π, thatβs one over four to the π plus six is equal to π π,
and the terms of our second series, one over four to the π equal π π. We can then find the limit as π
tends to β of π π over π π within our limit comparison test. Note that it doesnβt matter which
one is π π and which one is π π, whichever makes the calculation easiest. So weβre going to find the limit as
π tends to β of π π over π π, where π π is one over four to the π and π π
is one over four to the π plus six.

So we have the limit as π tends to
β of the one over four to the π divided by one over four to the π plus six. And thatβs equal to the limit as π
tends to β of the one over four to the π times four to the π plus six divided by
one. Thatβs the limit as π tends to β
of four to the π plus six divided by four to the π. Which is the limit as π tends to β
of one plus six, divided by four to the π. Since one is a constant, thatβs one
plus the limit as π tends to β of six divided by four to the power π. And six is a constant. We could take this outside, so that
gives us one plus six times the limit as π tends to β of one over four to the
π. And as π tends to β, one over four
to the power π tends to zero.

Our limit is therefore one which is
a constant greater than zero. So we found a second series where
the ratio of the terms of the two series as the π tends to β is a constant. Now, we need to find the
convergence of that second series. Because by the limit comparison
test, the convergence of the second series determines the convergence or divergence
of the first series. Our second series is the sum from
π is one to β of one over four to the power π. And we know that this is a
geometric series. A geometric series is a series with
the sum from π is zero to β of π times π to the power π. And thatβs π times π to the zero
plus π times π to the one plus π times π squared, et cetera. If the absolute value of π is less
than one, then the series sums to π over one minus π. If the absolute value of π is
greater than or equal to one, then the series diverges.

In our series, π is equal to one,
and π is equal to one over four. There is a difference, however. For the defined geometric series,
π begins with the value of zero. Whereas in our series, π starts at
one. Given that π, in our case, is less
than one, when we come to work out the sum of our series, weβll need to subtract
this first term. So letβs work out the sum of our
series, which is π over one minus π. The first term in the defined
series π times π to the power zero, in our case, will be one times one over four
to the power zero, And thatβs equal to one. So weβll need to subtract this from
the sum of our series.

That means we subtracted from π
over one minus π where, in our case, π is one and π is one over four. The sum of our series is therefore
the sum from π is one to β of one over four to the power π, which is equal to one
over one minus one over four minus one. Where this negative one corresponds
to π times π to the power zero. The sum of our second series is
therefore four over three minus one which is one over three. Our second series, the sum from π
is one to β of one over four to the power π, converges to the sum one over
three. By the limit comparison test, if
one of our series converges, then so does the other. This means that our original
series, the sum from π is one to β of one over four to the power π plus six, also
converges.