Now we'll look at three examples
of determining if an infinite geometric series
converges or diverges.
An infinite geometric series fits the form
of this series here, where we have this sum of a
times r raised to the power of n
from n equals zero to infinity.
If we look at the first several terms
of a geometric series, defined successive terms,
the pattern is we multiply by r,
which is called the common ratio.
So notice how from the first term to the second term
we multiply by a factor of r.
From the second term to the third term
we multiply by a factor of r and so on.
R is called a common ratio, and if we cannot
determine r by looking at the formula or the terms,
we can find r by taking any term in the series a sub n
and dividing by the previous term
which would be a sub n minus one.
A geometric series is probably the easiest series
to determine whether it converges or diverges.
If the absolute value of r is less than one,
the series converges and it converges to a,
the first term, divided by the quantity one minus r.
And if the absolute value of r is greater than
or equal to one, the series diverges.
Let's look at our examples.
Here we have the sum of 1/3 raised to the power of n
from n equals zero to infinity.
So right away we probably recognize that r
is equal to 1/3, and therefore the geometric series
will converge, but let's take the time
and generate the first several terms of our series.
So we'd have 1/3 to the zero power
plus 1/3 to the first plus 1/3 to the second
plus 1/3 to the third and so on.
So the series would be one plus 1/3
plus, 1/3 squared is 1/9, plus,
1/3 cubed is 1/27, and so on.
So notice how for our geometric series
the first term a equals one and the common ratio r
we can tell is equal to 1/3.
Notice how if we take any term in the series
and divide by the previous term we also get 1/3.
1/3 divided by one is 1/3, 1/9 divided by 1/3
is also 1/3 and so on.
So the geometric series with the absolute value
of r, in this case equals the absolute value
1/3, which equals 1/3, is less than one
and therefore the series converges.
And we can actually find the sum
or determine what the series converges to
and taking a and dividing by the quantity one minus r.
So a, the first term, is one and r is 1/3,
so the infinite sum is equal to one
divided by one minus 1/3, which should be
one divided by 2/3, and dividing by 2/3
is equivalent to multiplying by the reciprocal,
so this would be one times 3/2, or just 3/2.
So we can say, the series converges to 3/2, or 1.5.
Which would be the sum of the infinite geometric series.
Next we have the summation of 5/4, raised to the power of n
from n equals zero to infinity.
Once again, notice how we probably recognize
that r is equal to 5/4 and therefore the series
will diverge, but again, let's take the time
to generate the first few terms of our series.
So we'd have 5/4 raised to the zero
plus 5/4 raised to the first plus 5/4 squared and so on.
So 5/4 to the zero is one, plus 5/4 to the first
which is 5/4, plus 5/4 squared, 25/16 and so on.
So for this geometric series the first term a,
again, is equal to one and r, the common ratio,
we can probably tell from the formula would be 5/4.
We could also take the second term
and divide by the first term to get r of 5/4.
But notice how here the series is going to diverge,
so we'll say, the geometric series with absolute value
of r equals, in this case, absolute value of 5/4,
which equals 5/4, is greater than or equal to one
and therefore the series diverges.
Let's take a look at one more.
Here the formula's a little bit different.
We have the summation of four times 2/5
raised to the power of n, when n equals zero to infinity.
Notice in this case r would be 2/5
and therefore this series will converge,
but again, let's find the first few terms of our series.
The first term would be four times 2/5 raised to the zero,
plus we'd have four times 2/5 to the first,
plus four times 2/5 to the second and so on.
So the first term would be four times 2/5 to the zero,
that's one, so four times one is four,
plus here we'd have four times 2/5 to the first,
if we want 4/1, so that would be 8/5,
plus then we'd have four times 2/5 squared
which would be four times 4/25,
so we have 16/25 and so on.
So notice for this geometric series a, the first term,
equals four, the common ratio r, looking at our formula,
would be 2/5.
Of course, if we took 8/5 and divided by four
we would also get 2/5, and let's go ahead and show that.
8/5 divided by four is equivalent
to multiplying by the reciprocal of 1/4,
if we simplify before multiplying, common factor of four,
and we do get 2/5.
And so looking at our notes, because the absolute value
of r is less than one the series converges.
So we say, the geometric series
with absolute value of r equals
absolute value of 2/5, which equals 2/5,
which is less than one, converges and therefore
the sum of the series where the series converges
to a divided by the quantity one minus r.
So the infinite sum is equal to a, which is four,
divided by one minus r, or one minus 2/5.
So we'd have four divided by 3/5,
which is equivalent to four times the reciprocal of 5/3,
so we should put this 4/1, and so the infinite sum,
or the series converges to 20/3.
So we can say the series converges to 20/3.
Which means the infinite sum is 20/3.
I hope you found this helpful.