### Video Transcript

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.