### Video Transcript

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.