Sequences and series (part 2)

Finding the sum of an infinite geometric series.

Welcome back.

So where we left off in the last video, I'd shown you

this thing called the geometric series.

And, you know, we could
have some base a.

It could be any number.

It could be 1/2,
it could be 10.

But that's just--
but some number.

And we keep taking it to
increasing exponents, and we

sum them up, and this is
called a geometric series.

And so I want to figure out the
sum of a geometric series of,

you know, when I have some base
a, and I go up to some

number a to the n.

What-- is this a to
the-- why did I write

a to n minus 2 there?

That should be a to the big N.

My brain must have been
malfunctioning in

the previous video.

That always happens when I
start running out of time.

But anyway.

Let's go back to this.

So I defined s as
this geometric sum.

Now I'm going to
define another sum.

And that sum I'm going
to define as a times s.

And that equals-- well, that's
just going to be a times

this exact sum, right?

And that's the same
a as this a, right?

That a is the same as this a.

So what's a times
this whole thing?

Well, it's the a times a
to the zero is-- let me

write it down for you.

So this'll be a because I just
distribute the a, right? a

times a to the zero, plus a
times a to the 1, plus a times

a squared, plus all the way a
times a to the n minus one,

plus a times a to the n.

I just took an a and I
distributed it along

this whole sum.

But what is this equal to?

Well, this is equal to
a times a to the zero.

That's a one-- a to the first
power-- plus a squared, plus a

cubed, plus a to the n, right?

Because you just add the
exponents, a to the n.

Plus a to the n plus 1.

So this is as.

And we saw before that s
is just our original sum.

That is just a to the zero,
plus a to the 1, plus a

squared, plus up, up, up, up.

All the way to plus
a to the n, right?

So let me ask you a question.

What happens if I
subtract this from that?

What happens?

If I say, as minus s.

Well, I subtracted this from
here, on the left hand side.

What happens on the
right hand side?

Well, all of these
become negative, right?

Let me do it in a bold color.

This becomes-- because I'm
subtracting-- negative,

negative, these are
all negatives.

Negative.

Negative.

Well, a to the first,
minus a to the first.

That crosses out. a squared
minus a squared crosses

out. a to the third,
it'll all cross out.

All the way up to a
to the n, right?

So what are we left with?

We're just left with minus
a to the zero, right?

We're just left with that term.

And we're just left
with that term.

Plus a to the n plus 1.

And of course, what's
a to the zero?

That's just 1.

So we have a times s
minus s is equal to a to

the n plus 1 minus 1.

And now let's
distribute the s out.

So we get s times a minus
1 is equal to a to the n

plus 1 minus 1, right?

And then what do we get?

Well, we can just divide
both sides by a minus 1.

Let me erase some of
this stuff on top.

I think I can safely erase
all of this, really.

Well, I don't want
to erase that much.

I want to erase this stuff.

That's good enough.

OK.

So I have just-- dividing both
sides of this equation by a

minus 1, I get s is equal to a
to the n plus 1 minus

1 over a minus 1.

So where did that get us?

We defined the geometric
series as equal to the sum.

From k is equal to 0,
to n of a to the k.

And now we've just derived
a formula for what that

sum ends up being.

Equals a to the n plus 1
minus 1 over a minus 1.

And why is this useful?

We now know, if I were to say,
well, what is-- let me clean

up all of this, as well.

Let me clean up all of
this and we can-- OK.

So if I said, you figure out
the sum of, I don't know, the

powers of 3 up to 3 to the, I
don't know, 3 to

the tenth power.

So, you know, 3.

So 3 to the zero, plus 3 to the
one, plus 3 squared, plus all

the way to 3 to the tenth.

So this is the same thing as
the sum of k equals zero

to 10, of 3 to the k.

Right?

So this formula we just figured
out, a is 3 and n is 10.

So this sum is just going to be
equal to 3 to the eleventh

power minus 1 over 3 minus 1.

Which equals-- well, I
don't know what 3 to

the eleventh power is.

Minus 1 over 2.

So that's kind of useful.

That is a number.

Although you'd have to memorize
your exponent tables to the

eleventh power to do that.

But I think you get the idea.

This is especially useful if we
were dealing with-- well, if

the base was a power of ten,
it would be very, very easy.

But what I actually want to do
now is I want to take this and

say, well, what happens
if n goes to infinity?

Let me show you.

So what happens?

So there's two types of series
that we can take-- that's

not what I wanted to do.

There are two types of series
that we can take that we

can find the sums of.

There's finite series,
and infinite series.

And in order for an infinite
series to come up to a sum

that's not infinity, they
need to-- what we say--

they need to converge.

And if you think about what has
to happen for them to converge,

every next digit has to
essentially get smaller and

smaller and smaller, as
we go towards infinity.

So let's say that
a is a fraction.

a is 1/2.

So how does a geometric series
look like if we have 1/2 there?

So let's say that we're taking
the geometric series from k

is equal to 0 to infinity.

So this is neat.

We're going to take an infinite
sum, an infinite number of

terms, and let's see if we can
actually get an actual number.

You know, we take an infinite
thing, add it up, and it

actually adds up to
a finite thing.

This has always amazed me.

And the base now is
going to be 1/2.

It's 1/2 and it's going to
be 1/2 to the k power.

So this is going to be what?

1/2 to the zero, plus 1/2,
plus-- what's 1/2 squared?

Plus 1/4, plus 1/8, plus 1/16.

So as you see, each term is
getting a lot, lot smaller.

It's getting half of
the previous term.

Well, let's say, what happens
if this wasn't infinity?

What happens if this was n?

Well, then we'd get plus 1
over 2 to the n, right?

1/2 to the n is the same
thing as 1 over 2 to the n.

And if we look at the formula
we figured out, we would say,

well, that is just equal to
1/2 to the n plus 1, minus

1, over 1/2 minus one.

And that would be our answer.

We'd have to know what n is.

But now we want to know what
happens if we go to infinity.

So this is essentially
a limit problem.

What happens-- what's the
limit, as n goes to infinity,

of 1/2 to the n plus one
minus 1 over 1/2 minus 1?

Well, all of these are constant
terms, so nothing happens.

So what happens as this term,
right here, goes to infinity?

What's 1/2 to the
infinity power?

Well, that's zero.

That's an unbelievably
small number.

Take 1/2 to arbitrarily large
exponents, this just goes to 0.

And so what are we left with?

We're just left with this
equals minus 1 over 1/2 minus

1, or we could multiply the top
and the bottom by negative 1.

And we get 1 over 1 minus 1/2.

Which equals 1 over 1/2,
which is equal to 2.

I find that amazing.

If I add 0 plus 1/2 plus 1/4
plus 1/8 plus 1/16 and I never

stop-- I go to infinity-- and
not infinity, but I go to 1

over essentially 2 to the
infinity-- I end up with

this neat and clean number.

2.

And this might be a little
project for you, to actually

draw it out into like maybe a
pie and see what happens as

you keep adding smaller and
smaller pieces to the pie.

But it never ceases to amaze
me, that I added an infinite

number of terms, right?

This was infinity.

And I got a finite number.

I got a finite number.

Anyway, we ran out of time.

See you soon.