### Video Transcript

Determine whether the series which
is the sum from π equals one to β of negative one to the π plus one multiplied by
five over π plus one minus five over π plus two converges or diverges.

Now, looking at our series, we can
see that it looks like an alternating series. This is because of the negative one
to the π plus one term. Now, an alternating series can be
of the form as shown here. However, weβre required to that π
π must be greater than zero. In our case, π π is equal to five
over π plus one minus five over π plus two. We can check whether this is
greater than zero by combining the two fractions. We multiply the fraction on the
left by π plus two over π plus two and multiply the fraction on the right by π
plus one over π plus one. After doing this, weβll have found
a common denominator of π plus one multiplied by π plus two. Then, we can combine the two
fractions. This will give five lots of π plus
two minus five lots of π plus one over π plus one multiplied by π plus two.

Now, we can multiply through the
numerator. And for the final step, we can
simplify this numerator. What weβre left with is that π π
is equal to five over π plus one multiplied by π plus two. Now, π can be any integer between
one and β. Therefore, π is always
positive. And if π is always positive, this
means that our π π must also always be positive. Now, we have confirmed that our
series is in fact an alternating Series. We can try and use the alternating
series test, which tells us that an alternating series converges if the limit as π
tends to β of π π is equal to zero and π π is a decreasing sequence.

Letβs start by finding the limit as
π tends to β of π π. We have that the limit as π tends
to β of π π is equal to the limit as π tends to β of five over π plus one
multiplied by π plus two. Now, all of the πs in our limit
are on the denominator of our fraction. And all of these πs are
positive. Therefore, as π gets larger and
larger and larger, the denominator of this fraction will also get larger. This means that the whole fraction
itself will tend towards zero. Hence, we can say that this limit
must be equal to zero. Therefore, weβve satisfied the
first condition of the alternating series test.

Next, we need to check whether the
sequence π π is decreasing. If the sequence is decreasing, then
π π must be greater than π π plus one, since each term will be smaller than the
previous one. We can rearrange this
inequality. And here, what we have to check is
that π π minus π π plus one is greater than zero. For π π, we have five over π
plus one times π plus two. Then, for π π plus one, we have
five over π plus two times π plus three. We can find the difference here by
multiplying the fraction on the left by π plus three over π plus three and the
fraction on the right by π plus one over π plus one.

After combining the two resulting
fractions, what weβre left with is five lots of π plus three minus five lots of π
plus one over π plus one times π plus two times π plus three. Simplifying this fraction, what
weβre left with is 10 over π plus one times π plus two times π plus three. Since π must be between one and β,
this tells us that π itself must be greater than or equal to one. For any π value, which is greater
than or equal to one, our fraction here, which is π π minus π π plus one, must
be greater than zero. This is because all of the terms in
this fraction will be positive. So we have satisfied the condition
required for π π to be a decreasing sequence. And therefore, the second condition
for the alternating series test is also satisfied.

From this, we can come to the
conclusion that the series which is the sum from π equals one to β of negative one
to the π plus one multiplied by five over π plus one minus five over π plus two
converges.