### Video Transcript

In this video, weβll learn how to
find the πth partial sum of a series and determine the convergence or divergence of
the series from the limit of its partial sum. We will primarily consider
something called telescoping series, but we will also look at geometric series and
how we can establish convergence or divergence of these series.

A series is not a list of terms in
a sequence, but the sum of the terms in that sequence. If a series has a finite number of
terms, then itβs quite straightforward to add each term and evaluate the series. But when there are an infinite
number of terms, itβs a little less straightforward. In fact, the series may not even
have a finite sum. For example, the series one plus
two plus three plus four plus all the way up to π and so on wonβt have a finite
sum. Whereas the series given by a half
plus a quarter plus an eighth plus a sixteenth and so on does because the πth term
of this series gets ever smaller and smaller, eventually approaching zero as π
approaches β.

We define an infinite series as the
sum of the terms π one, π two, π three all the way up to π π, and so on. And we denote it in short by the
sum from π equals one to β of π sub π. The thing is it doesnβt always make
sense to think about the sum of infinitely many terms. And so, we define something called
a partial sum. A partial sum of an infinite series
is the sum of a finite number of consecutive terms beginning with the first. It could be helpful to examine the
behaviour of partial sums of infinite series.

Formally, we say that given a
series the sum from π equals one to β of π sub π, π sub π is the πth partial
sum. And thatβs given by the sum from π
equals one to π of π sub π. This also leads us onto the
definition of convergence of these series. We say that if the limit as π
approaches β of the πth partial sum is some real number π, then the series the sum
of π π is called convergent. This real number π is called the
sum of the series. Of course, if this is not true, we
say that the series is called divergent. Now, often, partials sums can be
computed with the sum function. Other times, weβll need to spot
patterns or apply formulae to look for evidence of convergence of these series. Letβs Look at an example.

Find the partial sum from the
series the sum from π equals one to β of π to the power of one over π minus π to
the power of one over π plus one. Is the series convergent or
divergent?

Remember, the partial sum of our
series is the sum of the first π terms. In general, the πth partial sum of
the series, the sum from π equals one to β of π sub π, is the sum from π equals
one to π of π sub π. And thatβs π sub one plus π sub
two plus π sub three all the way up to π sub π. We generally define this as π sub
π. So, letβs find the πth partial sum
for our series.

We know itβs going to be π sub one
plus π sub two plus π sub three all the way up to π sub π. But here, π sub π is π to the
power of one over π minus π to the power of one over π plus one. π sub one is found by replacing π
with the number one. So, we get π to the power of one
over one minus π to the power of one plus one, which is simply π minus π to the
power of one-half. Then, π sub two is π to the power
of one-half minus π to the power of one over two plus one, or π to the power of
one-third.

In the same way, we can define the
remaining terms as shown. This means π sub π, the πth
partial sum, is given by π minus π to the power of one-half. Thatβs π sub one. Plus π to the power of one-half
minus π to the power of one-third. Remember, thatβs π sub two. Plus π to the power of one-third
minus π to the power of a quarter. That was π sub three. And we continue adding these terms
until we get to π sub π, which is π to the power of one over π minus π to the
power of one over π plus one.

But if we look really carefully at
our πth partial sum, we should see that some of the terms will cancel. We have negative π to the power of
one-half plus π to the power of one-half. Well, thatβs zero. We then have negative π to the
power of one-third plus π to the power of one-third, which is also zero. But letβs add in π sub π minus
one. And when we do, we see that this
process repeats all the way up to negative π to the power of one over π plus π to
the power of one over π. And so, what this means is all
weβre left with to describe our πth partial sum is π minus π to the power of one
over π plus one. And so, the partial sum for our
series is π minus π to the power of one over π plus one.

The next part of this question asks
us whether the series is convergent or divergent. Well, remember, we say that if the
limit as π approaches β of the πth partial sum is equal to some real number π,
then that means the series, the sum of π π, is convergent. So, letβs evaluate the limit as π
approaches β of π sub π.

In that case, thatβs the limit as
π approaches β of π minus π to the power of one over π plus one. Now, actually, π itself is
completely independent of π. And we should notice that as π
grows larger, one over π plus one grows smaller. As π approaches β, one over π
plus one approaches zero. So, the limit as π approaches β of
the πth partial sum is π minus π to the power of zero. But of course, π to the power of
zero β in fact, anything to the power of zero β is equal to one. So, our limit is equal to π minus
one. This is indeed a real number. And so, we can say that our series
is convergent.

Now, notice how the partial sum of
our series eventually had just a few terms after cancelling. This is called the method of
differences. And the series is called a
telescoping series. This is a series whose partial sums
eventually only have a fixed number of terms after cancelling. Letβs have another look at a series
of this form.

Find the partial sum of the series
the sum from π equals one to β of one over two π plus one times two π minus
one. Is the series convergent or
divergent?

It might not instantly be obvious
how weβre going to find the πth partial sum of this series, but π sub π here is a
fraction. Weβre going to manipulate the
expression for the πth term of our sequence by writing it in partial fraction
form. Remember, this is a way of
simplifying the fraction to the sum of two or more less complicated rational
functions.

In this case, we write one over two
π plus one times two π minus one as π΄ over two π plus one plus π΅ over two π
minus one. We want to make the expression on
the right-hand side look like that on the left. And so, we create a common
denominator. And we achieve this by multiplying
the numerator and denominator of our first fraction by two π minus one and of our
second fraction by two π plus one. Once weβve done this, we can simply
add the numerators. And so, on the right-hand side, we
get π΄ times two π minus one plus π΅ times two π plus one all over two π plus one
times two π minus one.

Notice now that the denominators on
the right- and left-hand side of our equation are equal. This, in turn, means that for these
two fractions to be equal, their numerators must themselves be equal. That is to say, one equals π΄ times
two π minus one plus π΅ times two π plus one. And then, we need to work out the
values of π΄ and π΅. And we have a couple of ways that
we can do this. We could distribute the parentheses
and equate coefficients on the left- and right-hand side of our equation. Alternatively, we can substitute
the zeros of two π plus one times two π minus one into the entire equation. That is, let π equal one-half, see
what happens, and let π equal negative one-half and see what happens.

By letting π be equal to one-half,
we get one equals π΄ times one minus one plus π΅ times one plus one. But of course, π΄ times one minus
one is π΄ times zero, which is zero. And thatβs the whole purpose of
substituting these zeros in; it leaves us with an equation purely in terms of one of
our unknowns. In this case, this simplifies to
one is equal to two π΅. And if we divide through by two, we
find π΅ is equal to one-half.

Letβs repeat this process for π is
equal to negative one-half. Our equation becomes one equals π΄
times negative one minus one plus π΅ times negative one plus one. π΅ times negative one plus one is
π΅ times zero, which is zero. And so, this time, we have an
equation purely in terms of π΄. Its simplifies to one equals
negative two π΄. And if we divide through by
negative two, we find π΄ is equal to negative one-half. And so, we replace π΄ with negative
one-half and π΅ with one-half. And we see that our πth term, one
over two π plus one times two π minus one, can be written as negative one-half
over two π plus one plus a half over two π minus one.

Letβs clear some space and simplify
this a little. We can write a half over two π
minus one as one over two times two π minus one. Similarly, we can write negative a
half over two π plus one as negative one over two times two π plus one. To find the πth partial sum, weβre
going to list out the first few terms of our series. π sub one is found by replacing π
with one. This is the first term in our
series. We get one over two times two minus
one minus one over two times two plus one. This simplifies to a half minus
one-sixth.

Then, π sub two is found by
replacing π with two. And we get one over two times four
minus one minus one over two times four plus one. Thatβs a sixth minus a tenth. In the same way, we find π sub
three to be equal to a tenth minus fourteenth, π sub four to be equal to a
fourteenth minus an eighteenth, and so on. And so, we find the π partial sum
to be the sum of all of these all the way up to π sub π. So, thatβs a half minus a sixth
plus a sixth minus a tenth plus a tenth minus fourteenth plus a fourteenth minus an
eighteenth all the way up to one over two times two π minus one minus one over two
times two π plus one.

But now look what happens. Negative a sixth plus one-sixth is
zero. Negative a tenth plus one-tenth is
zero. Negative one fourteenth plus one
fourteenth is zero. And this continues all the way up
to one over two times two π minus one. And so, weβre actually only left
with two terms in our πth partial sum. They are a half and negative one
over two times two π plus one. We are almost done. But weβre going to add these
fractions once again by creating a common denominator.

This time, we achieve this simply
by multiplying the numerator and denominator of our first fraction by two π plus
one. Which gives us two π plus one
minus one over two times two π plus one. And one minus one is zero. Then, we see that these twos
cancel. And we have our expression for the
πth partial sum of our series. Itβs π over two π plus one.

The second part of this question
asks us to establish whether the series is convergent or divergent. And so, we recall that if the limit
as π approaches β of π sub π exists as some real number π, then the sum of π
π, this series, is convergent. Well, in our case, π sub π is π
over two π plus one. So, we need to find the limit as π
approaches β of π over two π plus one. We canβt use direct substitution
because if we were substitute in π equals β, we get β over β, which we know to be
undefined.

What we can do, though, is
manipulate the fraction a little. We divide both the numerator and
denominator by the highest power of π in our denominator. In this case, we divide by π. So, on the numerator, we get π
over π, which is equal to one. And on the denominator, the first
part is two π over π, which is two. And then, we add one over π.

And now, we only have one term
involving an π. Itβs this, one over π. Now, as π grows larger, one over
π grows smaller, eventually approaching zero. And so, we find the limit as π
approaches β of π sub π of our πth partial sum simplifies to one over two plus
zero, which is just one-half. Since this exists as a real number,
we can say that our series is convergent.

So far, weβve looked at telescoping
series, but there are a handful of series whose partial sums can be calculated using
formulae. One of the key ones of these is the
geometric series. Now, itβs outside of the scope of
the video to prove these results, but we can quote them.

In a geometric series, each turn is
obtained from the preceding one by multiplying it by the common ratio π. We say that the πth partial sum of
a geometric series is given by π sub π equals π times one minus π to the πth
power over one minus π. Importantly, we say that the series
is convergent if the absolute value of π is less than one and its sum is then given
by π over one minus π. We also say that if the absolute
value of the common ratio π is greater than or equal to one, the series is
divergent. Weβll now have a look at an example
of this form.

Find the partial sum for the series
the sum from π equals one to β of two times one-half to the power of π minus
one. Is the series convergent or
divergent?

We could begin by listing out the
first few terms of this series. The πth partial sum is the sum of
the first π terms. So, itβs π one plus π two plus π
three plus π four all the way up to π sub π, where here π sub π is two times a
half to the power of π minus one. And this means that π sub one is
found by replacing π with one. We get two times a half to the
power of one minus one, which is two times a half to the power of zero, or just
two.

The second term, π sub two, is two
times a half to the power of two minus one, which is two times one-half to the power
of one, or just two times a half. The third term is two times a half
to the power of three minus one, which is two times one-half squared. And we continue this process all
the way up to the term two times a half to the power of π minus one.

Now, at first glance, it might look
like this is a really complicated partial sum. But, in fact, we have a special
type of series. Itβs a geometric series. Each term is obtained from the
preceding one by multiplying it by the common ratio π. Now, there is a formula we can
quote to find the partial sum of a geometric series. Itβs given by π times one minus π
to the πth power over one minus π. So, if we can define π and π from
our series, weβll be able to quite easily find the πth partial sum.

Well, we know that the first term
in our series is two times a half to the power of zero. Well, a half to the power of zero
is one. So, thatβs two times one. And we can, therefore, say that π
is equal to two. And then, we can see that π is
equal to one-half. Each time, our term is being
multiplied by one-half. And so, substituting these values
of π and π into the formula for the πth partial sum, and we get two times one
minus a half to the πth power over one minus a half.

Now, in fact, the denominator of
this fraction, one minus a half, becomes one-half. And, of course, dividing by
one-half is actually the same as multiplying by two. So, we can multiply the numerator
of our expression by two. And weβre left with the πth
partial sum. Itβs four times one minus a half to
the πth power.

The second part of this question
asks us to decide whether this series is convergent or divergent. Well, in fact, we could find the
limit as π approaches β of π sub π. Alternatively, we can quote a
general result. We say that a geometric series with
a common ratio of π is convergent if the absolute value of π is less than one. And itβs divergent if the absolute
value of π is greater than or equal to one.

Well, here, π is equal to
one-half. And the absolute value of one-half
is one-half, which is less than one. And so, in this case, we can say
that this series is convergent. In fact, weβre also able to
generalise this and say that if the series is convergent, itβs sum is given by π
over one minus π. In this case, thatβs two over one
minus one-half, which is equal to four.

In this video, weβve learned that a
partial sum of an infinite series is a sum of a finite number of consecutive terms
beginning with the first. Given a series the sum from π
equals one to β of π sub π, its πth partial sum is defined at π sub π. Which is π one plus π two plus π
three all the way up to π sub π. We say that if the limit as π
approaches β of π sub π is equal to some real member π, then the series is called
convergent. This number π is called the sum of
the series. And if this is not true, the series
is called divergent.

We looked primarily at telescoping
series. And those are a series whose
partial sums eventually only have a fixed number of terms after cancelling. And this cancellation process is
often known as the method of differences. Finally, we looked at geometric
series. And in a geometric series, each
term is obtained from the preceding one by multiplying it by the common ratio
π.

The partial sum of a geometric
series is given by π times one minus π to the πth power over one minus π. We also saw that our geometric
series are convergent if the absolute value of π is less than one. And if the absolute value of π is
greater than or equal to one, the geometric series is divergent.