Section 10.2 : More on Sequences
In the previous section we introduced the concept of a sequence and talked about limits of sequences and the idea of convergence and divergence for a sequence. In this section we want to take a quick look at some ideas involving sequences.
Let’s start off with some terminology and definitions.
Given any sequence \(\left\{ {{a_n}} \right\}\) we have the following.
- We call the sequence increasing if \({a_n} < {a_{n + 1}}\) for every \(n\).
- We call the sequence decreasing if \({a_n} > {a_{n + 1}}\) for every \(n\).
- If \(\left\{ {{a_n}} \right\}\) is an increasing sequence or \(\left\{ {{a_n}} \right\}\) is a decreasing sequence we call it monotonic.
- If there exists a number \(m\) such that \(m \le {a_n}\) for every \(n\) we say the sequence is bounded below. The number \(m\) is sometimes called a lower bound for the sequence.
- If there exists a number \(M\) such that \({a_n} \le M\) for every \(n\) we say the sequence is bounded above. The number \(M\) is sometimes called an upper bound for the sequence.
- If the sequence is both bounded below and bounded above we call the sequence bounded.
Note that in order for a sequence to be increasing or decreasing it must be increasing/decreasing for every \(n\). In other words, a sequence that increases for three terms and then decreases for the rest of the terms is NOT a decreasing sequence! Also note that a monotonic sequence must always increase or it must always decrease.
Before moving on we should make a quick point about the bounds for a sequence that is bounded above and/or below. We’ll make the point about lower bounds, but we could just as easily make it about upper bounds.
A sequence is bounded below if we can find any number \(m\) such that \(m \le {a_n}\) for every \(n\). Note however that if we find one number \(m\) to use for a lower bound then any number smaller than \(m\) will also be a lower bound. Also, just because we find one lower bound that doesn’t mean there won’t be a “better” lower bound for the sequence than the one we found. In other words, there are an infinite number of lower bounds for a sequence that is bounded below, some will be better than others. In my class all that I’m after will be a lower bound. I don’t necessarily need the best lower bound, just a number that will be a lower bound for the sequence.
Let’s take a look at a couple of examples.
- \(\left\{ { - {n^2}} \right\}_{n = 0}^\infty \)
- \(\left\{ {{{\left( { - 1} \right)}^{n + 1}}} \right\}_{n = 1}^\infty \)
- \(\left\{ {\displaystyle \frac{2}{{{n^2}}}} \right\}_{n = 5}^\infty \)
This sequence is a decreasing sequence (and hence monotonic) because,
\[ - {n^2} > - {\left( {n + 1} \right)^2}\]for every \(n\).
Also, since the sequence terms will be either zero or negative this sequence is bounded above. We can use any positive number or zero as the bound, \(M\), however, it’s standard to choose the smallest possible bound if we can and it’s a nice number. So, we’ll choose \(M = 0\) since,
\[ - {n^2} \le 0\hspace{0.25in}{\mbox{for every }}n\]This sequence is not bounded below however since we can always get below any potential bound by taking \(n\) large enough. Therefore, while the sequence is bounded above it is not bounded.
As a side note we can also note that this sequence diverges (to \( - \infty \) if we want to be specific).
b \(\left\{ {{{\left( { - 1} \right)}^{n + 1}}} \right\}_{n = 1}^\infty \) Show Solution
The sequence terms in this sequence alternate between 1 and -1 and so the sequence is neither an increasing sequence or a decreasing sequence. Since the sequence is neither an increasing nor decreasing sequence it is not a monotonic sequence.
The sequence is bounded however since it is bounded above by 1 and bounded below by -1.
Again, we can note that this sequence is also divergent.
c \(\left\{ {\displaystyle \frac{2}{{{n^2}}}} \right\}_{n = 5}^\infty \) Show Solution
This sequence is a decreasing sequence (and hence monotonic) since,
\[\frac{2}{{{n^2}}} > \frac{2}{{{{\left( {n + 1} \right)}^2}}}\]The terms in this sequence are all positive and so it is bounded below by zero. Also, since the sequence is a decreasing sequence the first sequence term will be the largest and so we can see that the sequence will also be bounded above by \(\frac{2}{{25}}\). Therefore, this sequence is bounded.
We can also take a quick limit and note that this sequence converges and its limit is zero.
Now, let’s work a couple more examples that are designed to make sure that we don’t get too used to relying on our intuition with these problems. As we noted in the previous section our intuition can often lead us astray with some of the concepts we’ll be looking at in this chapter.
- \(\left\{ {\displaystyle \frac{n}{{n + 1}}} \right\}_{n = 1}^\infty \)
- \(\left\{ {\displaystyle \frac{{{n^3}}}{{{n^4} + 10000}}} \right\}_{n = 0}^\infty \)
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 its derivative.
\[f\left( x \right) = \frac{x}{{x + 1}}\hspace{0.75in}f'\left( x \right) = \frac{1}{{{{\left( {x + 1} \right)}^2}}}\]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,
\[f\left( n \right) = \frac{n}{{n + 1}} = {a_n}\]Therefore because \(n < n + 1\) and \(f\left( x \right)\) is increasing we can also say that,
\[{a_n} = \frac{n}{{n + 1}} = f\left( n \right) < f\left( {n + 1} \right) = \frac{{n + 1}}{{n + 1 + 1}} = {a_{n + 1}}\hspace{0.5in} \Rightarrow \hspace{0.5in}{a_n} < {a_{n + 1}}\]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 \(n = 1\) we could also use a lower bound of \(\frac{1}{2}\) 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 however.
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 example!
b \(\left\{ {\displaystyle \frac{{{n^3}}}{{{n^4} + 10000}}} \right\}_{n = 0}^\infty \) Show Solution
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 is bounded.
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 see this.
\[\begin{align*}{a_1} & = \frac{1}{{10001}} \approx 0.00009999 & \hspace{0.5in}{a_2} & = \frac{1}{{1252}} \approx 0.0007987\\ {a_3} & = \frac{{27}}{{10081}} \approx 0.005678 & \hspace{0.5in}{a_4} & = \frac{4}{{641}} \approx 0.006240\\ {a_5} & = \frac{1}{{85}} \approx 0.011756 & \hspace{0.5in}{a_6} & = \frac{{27}}{{1412}} \approx 0.019122\\ {a_7} & = \frac{{343}}{{12401}} \approx 0.02766 & \hspace{0.5in}{a_8} & = \frac{{32}}{{881}} \approx 0.03632\\ {a_9} & = \frac{{729}}{{16561}} \approx 0.04402& \hspace{0.5in}{a_{10}} & = \frac{1}{{20}} = 0.05\end{align*}\]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 derivative.
\[f\left( x \right) = \frac{{{x^3}}}{{{x^4} + 10000}}\hspace{0.75in}f'\left( x \right) = \frac{{ - {x^2}\left( {{x^4} - 30000} \right)}}{{{{\left( {{x^4} + 10000} \right)}^2}}}\]This function will have the following three critical points,
\[x = 0,\,\,\,x = \sqrt[4]{{30000}} \approx 13.1607,\hspace{0.25in}\,\,\,\,x = - \sqrt[4]{{30000}} \approx - 13.1607\]Why critical points? Remember these are the only places where the derivative may change sign! Our sequence starts at \(n = 0\) 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 \(0 < x < \sqrt[4]{{30000}} \approx 13.16\) and so the function is increasing in this range. Likewise, we can see that the derivative is negative for \(x > \sqrt[4]{{30000}} \approx 13.16\) and so the function will be decreasing in this range.
So, our sequence will be increasing for \(0 \le n \le 13\) and decreasing for \(n \ge 13\). Therefore, the function is not monotonic.
Finally, note that this sequence will also converge and has a limit of zero.
So, as the last example has shown we need to be careful in making assumptions about sequences. Our intuition will often not be sufficient to get the correct answer and we can NEVER make assumptions about a sequence based on the value of the first few terms. As the last part has shown there are sequences which will increase or decrease for a few terms and then change direction after that.
Note as well that we said “first few terms” here, but it is completely possible for a sequence to decrease for the first 10,000 terms and then start increasing for the remaining terms. In other words, there is no “magical” value of \(n\) for which all we have to do is check up to that point and then we’ll know what the whole sequence will do.
The only time that we’ll be able to avoid using Calculus I techniques to determine the increasing/decreasing nature of a sequence is in sequences like part (c) of Example 1. In this case increasing \(n\) only changed (in fact increased) the denominator and so we were able to determine the behavior of the sequence based on that.
In Example 2 however, increasing \(n\) increased both the denominator and the numerator. In cases like this there is no way to determine which increase will “win out” and cause the sequence terms to increase or decrease and so we need to resort to Calculus I techniques to answer the question.
We’ll close out this section with a nice theorem that we’ll use in some of the proofs later in this chapter.
Theorem
If \(\left\{ {{a_n}} \right\}\) is bounded and monotonic then \(\left\{ {{a_n}} \right\}\) is convergent.
Be careful to not misuse this theorem. It does not say that if a sequence is not bounded and/or not monotonic that it is divergent. Example 2b is a good case in point. The sequence in that example was not monotonic but it does converge.
Note as well that we can make several variants of this theorem. If \(\left\{ {{a_n}} \right\}\) is bounded above and increasing then it converges and likewise if \(\left\{ {{a_n}} \right\}\) is bounded below and decreasing then it converges.