Example 2 Determine
if the following sequences are monotonic and/or bounded.
We’ll start with the bounded part of this example first
and then come back and deal with the increasing/decreasing question since
that is where students often make mistakes with this type of sequence.
First, n is
positive and so the sequence terms are all positive. The sequence is therefore bounded below by
zero. Likewise each sequence term is
the quotient of a number divided by a larger number and so is guaranteed to
be less than one. The sequence is then
bounded above by one. So, this
sequence is bounded.
Now let’s think about the monotonic question. First, students will often make the mistake
of assuming that because the denominator is larger the quotient must be
decreasing. This will not always be
the case and in this case we would be wrong.
This sequence is increasing as we’ll see.
To determine the increasing/decreasing nature of this
sequence we will need to resort to Calculus I techniques. First consider the following function and
We can see that the first derivative is always positive
and so from Calculus I we know that the function must then be an increasing
function. So, how does this help
us? Notice that,
Therefore because and is increasing we can also say that,
In other words, the sequence must be increasing.
Note that now that we know the sequence is an increasing
sequence we can get a better lower bound for the sequence. Since the sequence is increasing the first
term in the sequence must be the smallest term and so since we are starting
at we could also use a lower bound of for this sequence. It is important to remember that any number
that is always less than or equal to all the sequence terms can be a lower
bound. Some are better than others
A quick limit will also tell us that this sequence
converges with a limit of 1.
Before moving on to the next part there is a natural
question that many students will have at this point. Why did we use Calculus to determine the
increasing/decreasing nature of the sequence when we could have just plugged
in a couple of n’s and quickly
determined the same thing?
The answer to this question is the next part of this
[Return to Problems]
This is a messy looking sequence, but it needs to be in
order to make the point of this part.
First, notice that, as with the previous part, the
sequence terms are all positive and will all be less than one (since the
numerator is guaranteed to be less than the denominator) and so the sequence
Now, let’s move on to the increasing/decreasing
question. As with the last problem,
many students will look at the exponents in the numerator and denominator and
determine based on that that sequence terms must decrease.
This however, isn’t a decreasing sequence. Let’s take a look at the first few terms to
The first 10 terms of this sequence are all increasing and
so clearly the sequence can’t be a decreasing sequence. Recall that a sequence can only be
decreasing if ALL the terms are decreasing.
Now, we can’t make another common mistake and assume that
because the first few terms increase then whole sequence must also
increase. If we did that we would also
be mistaken as this is also not an increasing sequence.
This sequence is neither decreasing or increasing. The only sure way to see this is to do the
Calculus I approach to increasing/decreasing functions.
In this case we’ll need the following function and its
This function will have the following three critical points,
Why critical points?
Remember these are the only places where the function may change sign! Our sequence starts at and so we can ignore the third one since it
lies outside the values of n that
we’re considering. By plugging in some
test values of x we can quickly
determine that the derivative is positive for and so the function is increasing in this
range. Likewise, we can see that the
derivative is negative for and so the function will be decreasing in
So, our sequence will be increasing for and decreasing for . Therefore the function is not monotonic.
Finally, note that this sequence will also converge and
has a limit of zero.
[Return to Problems]