Video Transcript
The Ratio Test
In this video, we’ll learn how to
determine if a series is convergent or divergent using the ratio test. We will see a variety of examples
of how the ratio test works. Let’s start by considering the
following series.
Now, this series here could
represent any series. We can define the limit of the
ratio of consecutive terms in this series as follows. We can say that this limit is equal
to 𝐿 and 𝐿 is equal to the limit as 𝑛 tends to ∞ of the absolute value of 𝑎 𝑛
plus one over 𝑎 𝑛. Now, this limit is crucial for us
when we’re using the ratio test. The ratio test tells us that if we
have some series, which is the sum from 𝑛 is equal to one to ∞ of 𝑎 𝑛, and if we
let 𝐿 be equal to the limit as 𝑛 tends to ∞ of the absolute value of 𝑎 𝑛 plus
one over 𝑎 𝑛. Firstly, if 𝐿 is less than one,
then the series converges absolutely. Secondly, if 𝐿 is greater than
one, then the series diverges. And thirdly, if 𝐿 is equal to one,
then the ratio test is inconclusive.
Let’s note that if number one here
is satisfied and 𝐿 is less than one, then what we mean by the series converging
absolutely is that both the sum from 𝑛 is equal to one to ∞ of 𝑎 𝑛 and the sum
from 𝑛 equals one to ∞ of the absolute value of 𝑎 𝑛 converge. This is different from conditional
convergence, which implies that only the sum from 𝑛 is equal to one to ∞ of 𝑎 𝑛
converges. So the type of convergence will be
showing here, which is the absolute convergence, is a stronger form of
convergence. Let’s also note that if we have
number three, so that’s when 𝐿 is equal to one, and we say that the ratio test is
inconclusive, which means the series could be absolutely convergent, conditionally
convergent, or divergent. And we’d need to try another
convergence test to find out.
Let’s now look at an example of how
the ratio test can be used.
Consider the series the sum from 𝑛
is equal to one to ∞ of 𝑛 cubed over three 𝑛 factorial. Determine whether the series
converges or diverges.
We can try using the ratio test to
determine whether the series converges or diverges. The ratio test tells us that for
some series, which is the sum from 𝑛 equals one to ∞ of 𝑎 𝑛 with, 𝐿 being equal
to the limit as 𝑛 tends to ∞ of the absolute value of 𝑎 𝑛 plus one over 𝑎 𝑛,
that firstly, if 𝐿 is less than one, then our series converges absolutely. Secondly, if 𝐿 is greater than
one, then our series diverges. And finally, if 𝐿 is equal to one,
then the test is inconclusive. Now, in our case, we can see that
𝑎 𝑛 is equal to 𝑛 cubed over three 𝑛 factorial. Let’s quickly note that this three
𝑛 factorial in the denominator is equal to three multiplied by 𝑛 factorial. And so, the factorial symbol
affects only the 𝑛 and not the three.
Next, we can write down what 𝑎 𝑛
plus one is. We simply change each 𝑛 in 𝑎 𝑛
to 𝑛 plus one. And we see that 𝑎 𝑛 plus one is
equal to 𝑛 plus one cubed over three 𝑛 plus one factorial. Now, we’re ready to find 𝐿. We have that 𝐿 is equal to the
limit as 𝑛 tends to ∞ of the absolute value of 𝑎 𝑛 plus one over 𝑎 𝑛. We substitute in 𝑎 𝑛 plus one and
one over 𝑎 𝑛. Now, we’re ready to start
simplifying.
First, we can cross cancel the
factor of three. Next, we know that we can rewrite
𝑛 plus one factorial as 𝑛 plus one multiplied by 𝑛 factorial. This will allow us to write our
limit as the limit as 𝑛 tends to ∞ of the absolute value of 𝑛 plus one cubed
multiplied by 𝑛 factorial over 𝑛 plus one multiplied by 𝑛 factorial multiplied by
𝑛 cubed. And so, we can cross out the factor
of 𝑛 factorial from the numerator and denominator. We can also cancel a factor of 𝑛
plus one from the denominator with one of the factors of 𝑛 plus one from the
numerator. What we’re left with here is the
limit as 𝑛 tends to ∞ of the absolute value of 𝑛 plus one squared over 𝑛
cubed.
Next, we can distribute the square
in the numerator. Here, it is clear that our highest
power of 𝑛 in the numerator is 𝑛 squared and our highest power of 𝑛 in the
denominator is 𝑛 cubed. Since we’re taking the limit as 𝑛
tends to ∞ and the highest power of 𝑛 in the denominator is greater than the
highest power of 𝑛 in the numerator, this tells us that this limit will be equal to
zero. What we have found here is that our
value of 𝐿 for this series is zero and zero is less than one. So when we look at our rule for the
ratio test, we see that we’ve satisfied condition number one. And this tells us our solution to
the question, which is that the sum from 𝑛 equals one to ∞ of 𝑛 cubed over three
𝑛 factorial converges absolutely.
Let’s now move on to our second
example.
True or false: The series, which is
the sum from 𝑛 equals one to ∞ of one over 𝑛 squared plus one, is convergent by
the ratio test.
We’ll need to apply the ratio test
to this series. The ratio test tells us that for
the series, which is the sum from 𝑛 equals one to ∞ of 𝑎 𝑛, where 𝐿 is equal to
the limit as 𝑛 tends to ∞ of the absolute value of 𝑎 𝑛 plus one over 𝑎 𝑛. Then, firstly, if 𝐿 is less than
one, then the series converges absolutely. Secondly, if 𝐿 is greater than
one, then the series diverges. And thirdly, if 𝐿 is equal to one,
then the test is inconclusive.
From the series given in the
question, we have that 𝑎 𝑛 is equal to one over 𝑛 squared plus one. Therefore, 𝑎 𝑛 plus one is equal
to one over 𝑛 plus one squared plus one, which after we distribute the parentheses,
we can see is equal to one over 𝑛 squared plus two 𝑛 plus two. Now, we’re ready to find our value
of 𝐿. We have that 𝐿 is equal to the
limit as 𝑛 tends to ∞ of 𝑎 𝑛 plus one over 𝑎 𝑛. Substituting in our values for 𝑎
𝑛 and 𝑎 𝑛 plus one, we can see that this limit is equal to the limit as 𝑛 tends
to ∞ of the absolute value of 𝑛 squared plus one over 𝑛 squared plus two 𝑛 plus
two.
Now, we can divide both the
numerator and denominator of the fraction here by 𝑛 squared. And we’ll do this because we’re
trying to find the infinite limit of a rational function. And 𝑛 squared is the highest power
which occurs in our fraction. So we have the limit as 𝑛 tends to
∞ of the absolute value of one plus one over 𝑛 squared all over one plus two over
𝑛 plus two over 𝑛 squared. Next, we used the fact that the
limit as 𝑛 tends to ∞ of one over 𝑛 is equal to zero. Therefore, any term within our
limit, which uses one over 𝑛 or one over 𝑛 squared, will tend to zero as 𝑛 tends
to ∞.
And so our limit will become the
absolute value of one plus zero over one plus zero plus zero. And this is simply equal to
one. Therefore, we found that 𝐿 is
equal to one. Looking at the ratio test, we can
see that this satisfies condition number three. Therefore, we can say that the
ratio test is inconclusive. It doesn’t tell us whether this
series is absolutely convergent, conditionally convergent, or divergent. Therefore, the answer to this
question is false.
In the next example, we’ll see how
we can use this test to check convergence of a series containing unknown
constant.
Consider the series, which is the
sum from 𝑛 equals zero to ∞ of 𝑎 𝑛, where 𝑎 𝑛 is equal to 𝑏 to the 𝑛 over 𝑛
plus 𝑐 factorial for some integers 𝑏 and 𝑐, which are both greater than one. Part i, calculate the limit as 𝑛
tends to ∞ of the absolute value of 𝑎 𝑛 plus one over 𝑎 𝑛. And part ii, hence decide whether
the series converges or diverges.
For part 1, we simply need to find
the limit as 𝑛 tends to ∞ if the absolute value of 𝑎 𝑛 plus one over 𝑎 𝑛. We’ve been given 𝑎 𝑛 in the
question and it’s equal to 𝑏 to the 𝑛 over 𝑛 plus 𝑐 factorial. Therefore, we can say that 𝑎 𝑛
plus one is equal to 𝑏 to the 𝑛 plus one over 𝑛 plus one plus 𝑐 factorial. So we can substitute in 𝑎 𝑛 and
𝑎 𝑛 plus one into our limit. So we have that our limit is equal
to the limit as 𝑛 tends to ∞ of 𝑎 𝑛 plus one. So that’s 𝑏 to the 𝑛 plus one
over 𝑛 plus one plus 𝑐 factorial multiplied by one over 𝑎 𝑛. So that’s 𝑛 plus 𝑐 factorial over
𝑏 to the 𝑛.
Immediately, we can see that we can
do some cross cancelling here. We have 𝑏 to the 𝑛 plus one in
the numerator of the fraction on the left. And we can cross cancel this with
the 𝑏 to the 𝑛 in the denominator of the fraction on the right. In doing this, we’ll simply be left
with 𝑏 in the numerator of the fraction on the left. Next, we can rewrite 𝑛 plus one
plus 𝑐 factorial. 𝑛 plus one plus 𝑐 factorial is
equal to 𝑛 plus 𝑐 plus one factorial. And this is also equal to 𝑛 plus
𝑐 plus one multiplied by 𝑛 plus 𝑐 factorial. And we can substitute this in to
our fraction in our limit.
We’re left with the limit as 𝑛
tends to ∞ of the absolute value of 𝑏 multiplied by 𝑛 plus 𝑐 factorial over 𝑛
plus 𝑐 plus one multiplied by 𝑛 plus 𝑐 factorial. Therefore, this factor of 𝑛 plus
𝑐 factorial in the numerator and denominator can be cancelled. We’re left with the limit as 𝑛
tends to ∞ of the absolute value of 𝑏 over 𝑛 plus 𝑐 plus one. Here, we’re taking the limit as 𝑛
tends to ∞. All the other terms in our fraction
are constant. Since 𝑛 appears in the denominator
of the fraction and nowhere else, this means that the limit as 𝑛 tends to ∞ of this
fraction will be equal to zero. This is because as 𝑛 gets larger
and larger and larger and closer to ∞, the fraction with 𝑛 in the denominator will
get closer and closer to zero. And so, we can say that this limit
is equal to zero. Since we’ve found the value of our
limit, this concludes the first part of the question. Let’s now move on to part 2.
In part 2, we’re required to use
our answer from part 1 to decide whether the series converges or diverges. Since in part 1, we calculated the
limit as 𝑛 tends to ∞ of the absolute value of 𝑎 𝑛 plus one over 𝑎 𝑛, this
pushes us towards using the ratio test. Now, the ratio test tells us that
for some series, which is the sum from 𝑛 equals one to ∞ of 𝑎 𝑛 with 𝐿 being
equal to the limit as 𝑛 tends to ∞ of the absolute value of 𝑎 𝑛 plus one over 𝑎
𝑛, that if 𝐿 is less than one, then our series converges absolutely. If 𝐿 is greater than one, then our
series diverges. And if 𝐿 is equal to one, then the
test is inconclusive.
Now, the first thing we may notice
is that the ratio test applies for series, which go from 𝑛 equals one to ∞. However, our series goes from 𝑛
equals zero to ∞. However, this does not matter. We’re still able to apply the ratio
test. This is because the sum from 𝑛
equals zero to ∞ of 𝑎 𝑛 is equal to 𝑎 nought plus the sum from 𝑛 equals one to ∞
of 𝑎 𝑛. So we can apply the ratio test to
the sum from 𝑛 equals one to ∞ of 𝑎 𝑛. And in our case, 𝑎 nought is equal
to one over 𝑐 factorial, where 𝑐 is an integer which is greater than one. So we can see that 𝑎 nought will
be a constant. And so, the result of the ratio
test on our series from 𝑛 is equal to one will also apply to our series from 𝑛 is
equal to zero.
Let’s now apply the ratio test. In part 1, we found that the limit
as 𝑛 tends to ∞ of the absolute value of 𝑎 𝑛 plus one over 𝑎 𝑛 is equal to
zero. And so, we have that 𝐿 is equal to
zero. Zero is less than one. And so, we have satisfied part
number 1 of the ratio test. This tells us that the sum from 𝑛
is equal to one to ∞ of 𝑎 𝑛 converges absolutely. We also have that 𝑎 nought is
finite. Hence, we can come to the
conclusion that the sum from 𝑛 is equal to zero to ∞ of 𝑎 𝑛 converges
absolutely.
Let’s now move on to our final
example.
Consider the series, which is the
sum from 𝑛 is equal to one to ∞ of negative one to the 𝑛 multiplied by two 𝑛
factorial over three to the power of three 𝑛. Determine whether the series
converges or diverges.
So we can try and work out whether
this series converges or diverges by applying the ratio test. The ratio test tells us that for
the series, which is the sum from 𝑛 is equal to one to ∞ of 𝑎 𝑛 with 𝐿 is equal
to the limit as 𝑛 tends to ∞ of the absolute value of 𝑎 𝑛 plus one over 𝑎 𝑛
that if 𝐿 is less than one, then the series converges absolutely. If 𝐿 is greater than one, then the
series diverges. And if 𝐿 is equal to one, then the
ratio test is inconclusive.
So looking at the series given to
us in the question, we can see that 𝑎 𝑛 is equal to negative one to the 𝑛
multiplied by two 𝑛 factorial over three to the power of three 𝑛. And so, we get that 𝑎 𝑛 plus one
is equal to negative one to the 𝑛 plus one multiplied by two multiplied by 𝑛 plus
one factorial over three to the power of three times 𝑛 plus one. This can be rewritten as negative
one to the power of 𝑛 plus one multiplied by two 𝑛 plus two factorial over three
to the power of three 𝑛 plus three.
We can find the value of 𝐿 by
taking the limit as 𝑛 tends to ∞ of the absolute value of 𝑎 𝑛 plus one over 𝑎
𝑛. And so substituting in our values
for 𝑎 𝑛 plus one and 𝑎 𝑛, we obtain that 𝐿 is equal to this. Here, we can do some cross
cancelling. We can cross cancel a factor of
negative one to the power of 𝑛 to leave us with negative one in the numerator. And we can cross cancel a factor of
three to the power of three 𝑛 to leave us with three cubed in the denominator. Now, we have that 𝐿 is equal to
the limit as 𝑛 tends to ∞ of the absolute value of negative one multiplied by two
𝑛 plus two factorial over three cubed multiplied by two 𝑛 factorial.
Firstly, since this is an absolute
value, we can simply get rid of the negative one. Next, we can take two of the terms
out of two 𝑛 plus two factorial. And we can rewrite two 𝑛 plus two
factorial as two 𝑛 plus two multiplied by two 𝑛 plus one multiplied by two 𝑛
factorial. Substituting this into the
numerator of our limit, we notice that we can cancel a factor of two 𝑛 factorial
from the top and bottom. And now, we have obtained that 𝐿
is equal to the limit as 𝑛 tends to ∞ of the absolute value of two 𝑛 plus two
multiplied by two 𝑛 plus one over three cubed.
Here, the 𝑛 terms are only in the
numerator of our limit. Therefore, as 𝑛 gets larger and
larger and larger, the numerator of our fraction will also get larger and larger and
larger. And so, we can say that this limit
must be equal to ∞. So we found that 𝐿 is equal to ∞,
which is clearly greater than one. And so, this gives us the second
part of the ratio test, which tells us our solution to the question. Which is that the sum from 𝑛 is
equal to one to ∞ over negative one to the power of 𝑛 multiplied by two 𝑛
factorial over three to the power of three 𝑛 diverges.
We have now covered a variety of
examples of how we can use the ratio test to show that a series converges absolutely
or diverges. We’ve also seen how sometimes the
ratio test is not useful in finding the convergence or divergence of a series. Let’s now recap some key points of
the video.
Key Points
The ratio test: suppose we have the
series, which is the sum from 𝑛 is equal to one to ∞ of 𝑎 𝑛. We can let 𝐿 be equal to the limit
as 𝑛 tends to ∞ of the absolute value of 𝑎 𝑛 plus one over 𝑎 𝑛. Then, firstly, if 𝐿 is less than
one, the series converges absolutely. Secondly, if 𝐿 is greater than
one, the series diverges. And thirdly, if 𝐿 is equal to one,
then the ratio test is inconclusive. If the ratio test is inconclusive,
we would need to try a different convergence test to work out if our series is
absolutely convergent, conditionally convergent, or divergent.
