### Video Transcript

Consider the series, which is the
sum from π equals zero to β of π π, where π π is equal to π to the π over π
plus π factorial for some integers π and π, which are both greater than one. Part i, calculate the limit as π
tends to β of the absolute value of π π plus one over π π. And part ii, hence decide whether
the series converges or diverges.

For part 1, we simply need to find
the limit as π tends to β if the absolute value of π π plus one over π π. Weβve been given π π in the
question and itβs equal to π to the π over π plus π factorial. Therefore, we can say that π π
plus one is equal to π to the π plus one over π plus one plus π factorial. So we can substitute in π π and
π π plus one into our limit. So we have that our limit is equal
to the limit as π tends to β of π π plus one. So thatβs π to the π plus one
over π plus one plus π factorial multiplied by one over π π. So thatβs π plus π factorial over
π to the π.

Immediately, we can see that we can
do some cross cancelling here. We have π to the π plus one in
the numerator of the fraction on the left. And we can cross cancel this with
the π to the π in the denominator of the fraction on the right. In doing this, weβll simply be left
with π in the numerator of the fraction on the left. Next, we can rewrite π plus one
plus π factorial. π plus one plus π factorial is
equal to π plus π plus one factorial. And this is also equal to π plus
π plus one multiplied by π plus π factorial. And we can substitute this in to
our fraction in our limit.

Weβre left with the limit as π
tends to β of the absolute value of π multiplied by π plus π factorial over π
plus π plus one multiplied by π plus π factorial. Therefore, this factor of π plus
π factorial in the numerator and denominator can be cancelled. Weβre left with the limit as π
tends to β of the absolute value of π over π plus π plus one. Here, weβre taking the limit as π
tends to β. All the other terms in our fraction
are constant. Since π appears in the denominator
of the fraction and nowhere else, this means that the limit as π tends to β of this
fraction will be equal to zero. This is because as π gets larger
and larger and larger and closer to β, the fraction with π in the denominator will
get closer and closer to zero. And so, we can say that this limit
is equal to zero. Since weβve found the value of our
limit, this concludes the first part of the question. Letβs now move on to part 2.

In part 2, weβre required to use
our answer from part 1 to decide whether the series converges or diverges. Since in part 1, we calculated the
limit as π tends to β of the absolute value of π π plus one over π π, this
pushes us towards using the ratio test. Now, the ratio test tells us that
for some series, which is the sum from π equals one to β of π π with πΏ being
equal to the limit as π tends to β of the absolute value of π π plus one over π
π, that if πΏ is less than one, then our series converges absolutely. If πΏ is greater than one, then our
series diverges. And if πΏ is equal to one, then the
test is inconclusive.

Now, the first thing we may notice
is that the ratio test applies for series, which go from π equals one to β. However, our series goes from π
equals zero to β. However, this does not matter. Weβre still able to apply the ratio
test. This is because the sum from π
equals zero to β of π π is equal to π nought plus the sum from π equals one to β
of π π. So we can apply the ratio test to
the sum from π equals one to β of π π. And in our case, π nought is equal
to one over π factorial, where π is an integer which is greater than one. So we can see that π nought will
be a constant. And so, the result of the ratio
test on our series from π is equal to one will also apply to our series from π is
equal to zero.

Letβs now apply the ratio test. In part 1, we found that the limit
as π tends to β of the absolute value of π π plus one over π π is equal to
zero. And so, we have that πΏ is equal to
zero. Zero is less than one. And so, we have satisfied part
number 1 of the ratio test. This tells us that the sum from π
is equal to one to β of π π converges absolutely. We also have that π nought is
finite. Hence, we can come to the
conclusion that the sum from π is equal to zero to β of π π converges
absolutely.