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Section 8.1 Introduction to Infinite Sequences

Subsection Sequences as Functions

Before we move on to our actual goal of analyzing infinite series, we will construct infinite sequences. The big thing to remember here is that, when we build and analyze these sequences, we are are really building and analyzing functions. We want to keep this idea of sequences as functions in the forefront, since it will help us as we think about accumulating these function values into infinite series.

Activity 8.1.1. Building our First Sequences.

We might already have some familiarity with sequences. Here, we’ll focus less on some of the detailed mechanics and just think about these sequences as functions.
(a)
Describe a sequence of numbers where you use a consistent rule/function to build each term (each number) based only on the previous term in the sequence. You will need to decide on some first term to start your sequence.
(b)
Describe a different sequence of numbers using the same rule to generate new terms/numbers from the previous one. What do you need to do to make these two sequences different from each other?
(c)
Describe a new sequence of numbers where you use a consistent rule/function to build each term based on its position in the sequence (i.e. the first term will be some rule/function based on the input 1, the second will be based on 2, you’ll use 3 to get the third term, etc.). We will call the position of each term in the sequence the index.
(d)
Describe another, new, sequence of numbers where you use a consistent rule/function to build each term based on its index. This time, make the terms get smaller in size as the index increases.

Definition 8.1.1. Explicit Formula.

An infinite sequence defined using an explicit formula is one where the \(k\)th term of the sequence is defined as a function output of \(k\text{,}\) the term’s index.
Using notation, we might say that \(a_k = f(k)\) where:
  • \(a\) is the ``name’’ of the sequence (similar to how \(f\) and \(g\) are common names of functions).
  • \(k\) is the index of the term, typically a non-negative integer.
  • \(f(k)\) is the function that we use to generate the terms.

Definition 8.1.2. Recursion Relation.

A sequence is defined using a recursion relation is one where the \(k\)th term of the sequence is defined as a function output of the previous term, the \((k-1)\)st term. The sequence also needs some initial term to base the subsequent terms from.
Using notation, we might say that \(a_{k} = f(a_{k-1})\text{.}\)
These definitions are relatively limited. You might, for instance, know of a very famous sequence that is defined recursively by having each term being the sum of the two previous terms. Our study of sequences will be brief and all pointing towards infinite series, so there are a lot of nuances about sequences that we will skip.

Activity 8.1.2. Returning to our First Sequences.

Let’s return back to the four sequences we created in Activity 8.1.1.
(a)
For each of the sequences, how are we going to define them? Explicit formulas? Recursion relations? How do you know?
(b)
Now, for each sequence, define the sequence formally using either an explicit formula or recursion relation, whichever matches with how you described the sequence in Activity 8.1.1.

Example 8.1.3. Practice Writing some Terms.

For each of the following sequences, write out the first handful of terms. There isn’t a set amount, but you should write out enough to get a feel for the sequence structure and how the different ways of defining the sequences work. In each, you can start the index \(k\) at 1 and count upwards (\(k=1,2,3,...\)).
(a)
\(a_1 = \frac{1}{3}\) and \(a_k = 2(a_{k-1})^2\)
(b)
\(b_k = \frac{\sin(k)}{k^2}\)
(d)
\(d_k = \frac{k + e^k}{e^k}\)

Activity 8.1.3. Describing These Sequences.

Let’s look at the sequences from Example 8.1.3. Go through the following tasks for each sequence.
(a)
What do you think each sequence is “counting towards” (if anything)?
Hint.
If you’re not sure, maybe you need to write out a few more terms! You can also change how you write the numbers themselves: in some cases, fractions might be helpful, but in others it might be useful to write the numbers in decimal form. Maybe you’ll approximate values of the sine or exponential functions, or maybe you’ll leave them as \(\sin(2)\) or \(e^3\text{.}\)
(b)
Can you show that the sequence is counting towards what you think it is with a limit (or show that it’s not counting towards anything)?
Hint.
Some of these limits, as \(k\to \infty\text{,}\) will be tricky to work with! When might you want to use The Squeeze Theorem? When might you want to use L’Hôpital’s Rule?

Activity 8.1.4. Write the Sequence Rules.

We’ll look at some sequences by writing out the first handful of terms. From there, our goal is to write out more terms and eventually define each sequence fully.
For each sequence, write an explicit formula and a recursion relation to define the sequence. You can choose whether to start your index at \(k=0\) or \(k=1\text{.}\)
(a)
\(\{a_k\} = \left\{4, \frac{2}{3}, \frac{1}{9}, \frac{1}{54}, ...\right\}\)
Hint 1.
It might be helpful to write these numbers using a common denominator! Or at least some of the numbers. Alternatively, you can try a common numerator (which is very fun to do, since we normally don’t do that).
Hint 2.
If you are recursively multiplying by a number each time, what will that look like in the explicit formula? How do we represent repeated multiplication?
(b)
\(\{b_k\} = \left\{\frac{3}{5}, \frac{2}{5}, \frac{5}{17}, \frac{3}{13}, \frac{7}{37},...,\right\}\)
Hint 1.
You can rewrite these fractions! Have any of them been “reduced?”
Hint 2.
Rewrite \(\frac{2}{5}\) and \(\frac{3}{13}\) by scaling the numerator and denominator by 2. Can you find a formula for the numerator and denominator separately? This one is very difficult to find a recursion relation for, so feel free to only write it explicitly!
(c)
\(\{c_k\} = \left\{\frac{1}{5}, \frac{3}{5}, 1, \frac{7}{5},...\right\}\)
Hint 1.
This one will definitely be helpful to rewrite so that all fractions have a common denominator.
Hint 2.
If you are recursively adding something, how does that show up in the explicit formula? How do we repeatedly add?
(d)
What kinds of connections do you notice between the explicit formulas and the recursion relations for these sequences?
Before moving on, we should think about a couple of notes:
  • When we add something recursively (where we add the same thing repetitively to get from the \(k\)th term to the \((k+1)\)st term), this is the same thing as multiplication in an explicit formula!
    \begin{equation*} \underbrace{3 + 3 + ... + 3}_{k \text{ times}} = 3k \end{equation*}
  • Similarly, when we multiply something recursively, we can think about this as an exponential in the explicit formula!
    \begin{equation*} \underbrace{3\cdot 3 \cdot 3 \cdots 3}_{k \text{ times}} = 3^k \end{equation*}
  • In general, it can be pretty difficult to find either of these formulas for a given sequence of numbers. In fact, in any sequence where only the first few terms are given, we can find an infinite number of formulas that provide those first few terms and then deviate on to any other numbers. We cannot easily extrapolate onto only one “pattern” or formula. Because of this, we’ll try to limit our work as much as we can to situations where we have the formula defining the sequences.

Subsection Graphing Sequences

We have tried introducing and talking about sequences as special types of functions, mapping natural number inputs to real number outputs. If we are committed to thinking about sequences as functions, with maybe some special context, then we should really investigate how one of our primary representations of functions (graphs) manifests itself in this new context.
There really is not too much to think about here! We can focus on the domain of these functions. If we define a sequence \(\{a_k\}\) explicitly, then we have some function \(a_k = f(k)\text{,}\) and we can plot this sequence function in the same way that we normally would any other function \(g(x)\text{.}\) We will use the horizontal axis for the inputs and the vertical axis to represent the outputs, and try to visualize the graph as the set of all of the pairs of inputs with their (single) corresponding output.
The only new feature, then, is that these functions have only non-negative integer inputs. So when we plot the points, we do not get some nice curve acting as a visual representation of the function: we get discrete points floating on the 2-dimensional plane, each with some consistent horizontal spacing between them.
Consider, for example the following function and sequence:
\begin{equation*} f(x)= \frac{x}{x^2+1} \text{ for } x\geq 0 \end{equation*}
and
\begin{equation*} a_k = \frac{k}{k^2+1} \text{ for } k = 0, 1, 2, ... \end{equation*}
We can graph the curve \(y=f(x)\) in the normal way, as a smooth curve starting at the point at \(x=0\text{.}\)
A green curve, starting at (0,0) and increasing until a point at y=1 and then decreasing towards the x-axis asymptotically.
Figure 8.1.4. The function \(f(x)=\dfrac{x}{x^2+1}\) plotted on the interval \([0,\infty)\text{.}\)
When we plot the sequence \(\left\{a_k\right\}\text{,}\) though, we will visualize just the points on \(f(x)\) at non-negative integer inputs.
A barely visible plot of the same green curve, but with points at x=0, x=1, ..., x=10 highlighted.
Figure 8.1.5. The sequence \(\{a_k\}_{k=0}^{\infty}\) for \(a_k = \dfrac{k}{k^2+1}\)
We will typically not plot these with the smooth curve of the corresponding continuous function plotted, but in this first example it is useful to highlight how we think about this sequence as a function.
Let’s continue to think about these sequences as just functions in a special kind of context. How does this discrete context change how we talk about functions and what kinds of terminology we use?

Subsection Sequence Terminology

If a sequence is a function (and we’re saying in this introductory section that it is), then we can think of all of the different terminology and adjectives that we use to describe functions. How many of them are relevant to sequences?
For now, we’ll talk about sequences in two ways: their direction and the size of their terms.

Definition 8.1.6. Direction of a Sequence.

We say that a sequence \(\{a_k\}_{k=1}^\infty\) is increasing if, for all \(k=1,2,3...\text{,}\) \(a_{k+1}\gt a_k\text{.}\) If \(a_{k+1}\geq a_k\) for all \(k=1,2,3,...\) then we say that \(\{a_k\}\) is non-decreasing.
We say that a sequence \(\{a_k\}_{k=1}^\infty\) is decreasing if, for all \(k=1,2,3...\text{,}\) \(a_{k+1}\lt a_k\text{.}\) If \(a_{k+1}\leq a_k\) for all \(k=1,2,3,...\) then we say that \(\{a_k\}\) is non-increasing.
We say that \(a_k\) is constant if \(a_{k+1}=a_k\text{,}\) but this is a very boring sequence and we will likely not think terribly hard about these kinds of sequences.
Sometimes we might say that a sequence is eventually non-increasing if there is some \(K\gt1\text{,}\) and the sequence is non-increasing for \(k=K, K+1, K+2, ...\text{,}\) and similarly for eventually non-decreasing.
A good example of a sequence that is eventually decreasing is the one we plotted in Figure 8.1.5. We can see that the sequence increases from \(k=0\) to \(k=1\) (since \(a_0\lt a_1\)), but then decreases after that.
We could think about the corresponding continuous function (the one plotted in Figure 8.1.5) and find the point at which our function starts decreasing: we just need to refer back to Theorem 4.2.6 First Derivative Test to find the interval(s) for which \(f(x)=\dfrac{x}{x^2+1}\) decreases.

Definition 8.1.7. Monotonic Sequences.

For the sequence \(\{a_k\}_{k=1}^\infty\text{,}\) we say that it is monotonic if the sequence is either non-increasing or non-decreasing.
We can, again, include a little modifier to talk about a sequence being eventually monotonic.

Definition 8.1.8. Bounded Sequences.

We say that a sequence \(\{a_k\}_{k=1}^\infty\) is bounded below if there is some real number \(M\) such that \(a_k\geq M\) for all \(k=1,2,3,...\text{.}\)
Similarly we say that a sequence \(\{a_k\}_{k=1}^\infty\) is bounded above if there is some real number \(N\) such that \(a_k\leq N\) for all \(k=1,2,3,...\text{.}\)
If a sequence has both an upper bound and a lower bound, then we often just say that the sequence is bounded.
Lastly, we’ll focus on the end-behavior of a sequence. We’ll think about convergence of a sequence in the same way that we did for Improper Integrals: does the limit exist?

Definition 8.1.9. Sequence Convergence.

For the sequence \(\{a_k\}\text{,}\) if \(L\) is some real number and \(\lim_{k\to\infty}a_k = L\text{,}\) the we say that the sequence \(\{a_k\}\) converges to \(L\text{.}\) If this limit does not exist, we say that the sequence \(\{a_k\}\) diverges.
This theorem seems to be a bit obvious to many students: why would we care about this, when we can just find a limit of the explicit formula for a sequence? We’ll see throughout the rest of this chapter that this theorem is one of the most important and most useful results in our study of infinite sequences and infinite series. For now, though, let’s use it to find the limits of some recursively defined sequences.

Subsection Some Cool Recursive Examples

Let’s revisit one of the recursively defined sequences that we’ve seen already and then think about a couple of other interesting ones. Before we do that, though, we should recognize why we need to treat recursively defined sequences a bit differently than ones defined explicitly.
In an explicit formula, the terms themselves are a function of \(k\text{,}\) the index. This means that we can simply apply a limit as \(k\to\infty\) to understand whether or not the sequence converges and what it might converge to. These limits could be tricky, but we have the tools to evaluate them! In a recursion relation, though, each term is not a function of the index, which means we can’t easily apply a limit as \(k\to\infty\) to the term definition.
We’ll be able to apply a limit, but it will feel a bit different: we’re going to go into the limit work under the assumption that the limit exists. Let’s see how it goes.

Example 8.1.11.

Let’s revisit the first sequence from Example 8.1.3: \(\{a_k\}_{k=1}^\infty\) where \(a_1 = \frac{1}{3}\) and \(a_{k} = 2(a_{k-1})^2\text{.}\)
(a)
Let’s start by assuming that the sequence converges. That means that there exists some real number \(L\) such that
\begin{equation*} \lim_{k\to\infty} a_k = L \text{.} \end{equation*}
What would this \(L\) be, if it exists? A key thing to note is that if \(\lim_{k\to\infty} a_k\) exists (and we have a symbol, \(L\text{,}\) for it) then we can say that
\begin{equation*} \lim_{k\to\infty} a_{k-1} = L\text{.} \end{equation*}
Whether or not this is obvious to you is not a mark of your understanding, but we need to make sure that this ends up being obvious to you. If it’s not, that’s ok! But it is an indicator that you should take a couple of minutes to think about this. Once you are convinced that these two limits are the same thing, move on to the next part.
(b)
Let’s now apply a limit to the sequence definition:
\begin{align*} a_k \amp = 2(a_{k-1})^2\\ \lim_{k\to \infty} a_k \amp = \lim_{k\to\infty} 2(a_{k-1})^2 \\ \underbrace{\lim_{k\to\infty} a_k}_{L} \amp = 2\left(\underbrace{\lim_{k\to\infty} a_{k-1}}_{L}\right)^2\\ L \amp =2L^2\\ 0 \amp =2L^2-L\\ 0 \amp= L(2L-1) \end{align*}
And so we have two solutions to this equation: \(L=0\) and \(L=\frac{1}{2}\text{.}\) This is strange: how can a sequence have more than one value that it converges to?
It’s because we have yet to take into account the initial term, \(a_1\text{!}\) Depending on this value, the sequence might converge or not, and if it does converge, then there are two options for what the sequence can converge to, based on the value of \(a_1\text{.}\)
(c)
You can do the next part on your own, but I want you to pick different numbers for \(a_1\) and write out some terms of the resulting sequence. You should find that some of them look like they’re converging to \(0\text{,}\) one of them will converge to \(\frac{1}{2}\) (it’s a fun hunt to find which one), and some will diverge.
Solution.
You should find that if \(a_1=\frac{1}{2}\text{,}\) then the sequence is constant and converges to \(\frac{1}{2}\text{.}\) If \(0\leq|a_1|\lt \frac{1}{2}\) then the sequence seems like it’ll converge to \(0\text{.}\) And if \(|a_1|\gt \frac{1}{2}\text{,}\) then it looks like the sequence diverges.
(d)
Now it is up to us to show that this sequence, with \(a_1 = \frac{1}{3}\text{,}\) does converge. Sure, we have some evidence and a good conjecture that it converges to \(\frac{1}{2}\text{,}\) but that is just our good guess based on what we have seen in the first handful of numbers.
We will attempt to convince ourselves that this sequence is both monotonic and bounded. We’ll begin with boundedness.
It should be clear that \(a_k\gt 0\text{,}\) since as long as \(a_{k-1} \neq 0\text{,}\) then \((a_{k-1})^2\gt 0\text{.}\) Since we start with \(a_1 \neq 0\text{,}\) we are guaranteed to get non-zero values from the formula for a new term! Great news, we have a lower bound.
Let’s show that \(\frac{1}{2}\) is an upper bound: \(a_k \lt \frac{1}{2}\) when
\begin{align*} 2(a_{k-1})^2 \amp \lt \frac{1}{2}\\ (a_{k-1})^2 \amp \lt \frac{1}{4}\\ a_{k-1} \amp \lt \frac{1}{2} \end{align*}
Since \(a_1\lt \frac{1}{2}\text{,}\) we know that each successive term will also be less than \(\frac{1}{2}\text{.}\) So we have an upper bound!
So the sequence \(\{a_k\}\) is bounded. Now we just need to convince ourselves that this sequence is monotonic. We know that our terms are bounded above by \(\frac{1}{2}\text{,}\) and I hope that this means we can convince ourselves that since our terms are smaller than this, which would produce a constant sequence, then all of our terms are probably decreasing.
Let’s show this by showing that \(a_{k+1}-a_k \lt 0\text{:}\)
\begin{align*} a_{k+1} - a_k \amp = 2(a_k)^2 - a_k\\ \amp = a_k(2a_k - 1) \end{align*}
We can solve for when this is negative! It shouldn’t be hard to show that \(a_k(2a_k-1)\lt 0\) when \(0\lt a_k \lt \frac{1}{2}\text{.}\) And we’ve already shown this is true in our case!
So \(\{a_k\}\) is bounded and monotonic and must therefore converge because of the Monotone Convergence Theorem. Because \(a_1\lt\frac{1}{2}\text{,}\) we know that this sequence doesn’t converge to \(\frac{1}{2}\text{,}\) and so must converge to the only other option: 0.
There are some other fun ways of doing this same thing for other recursive examples. The argument above is relatively bulky to use, and so we understandably will not think about recursively defined sequences very much: we’ll leave that topic for another course where we have more time to really explore them. If you are interested in trying this same argument with other sequences though, we’ll end this section with two more fun examples.

Example 8.1.12.

(a)
Consider the sequence defined by \(b_{k}=\sqrt{2+b_{k-1}}\) with \(b_1 = \sqrt{2}\text{.}\) Does this sequence converge? To what?
Hint.
Write out some terms to get a feel for things! Then, assuming that the sequence converges to some real number, \(L\text{,}\) think about what happens when you apply a limit as \(k\to\infty\text{:}\) we should get the equation \(L=\sqrt{2+L}\text{.}\)
(b)
Consider the sequence defined by \(c_k = \frac{1}{2(c_{k-1})+1}\) with \(c_1 = 1\text{.}\) Does this sequence converge? To what?
Hint.
Write out some terms to get a feel for things! Then, assuming that the sequence converges to some real number, \(L\text{,}\) think about what happens when you apply a limit as \(k\to\infty\text{:}\) we should get the equation \(L=\frac{1}{2L+1}\text{.}\)

Practice Problems Practice Problems

1.

What is the index of an infinite sequence? How does the index relate to terms of the infinite sequence?

2.

Explain the differences between how an explicit formula and a recursion relation can define an infinite sequence.

3.

Why does a recursion relation defining a sequence rely on an initial value?

4.

Use the explicit formulas write out the first 8 terms of each sequence.
(a)
\(\{a_k\} = \left\{\dfrac{1}{k}\right\}_{k=1}^\infty\)
(b)
\(\{b_k\} = \left\{\dfrac{k+2}{n=k^2+1}\right\}_{k=1}^\infty\)
(c)
\(\{c_k\} = \left\{\sin\left(\dfrac{k\pi}{4}\right)\right\}_{k=0}^\infty\)
(d)
\(\{d_k\} = \left\{\dfrac{1}{2^k}\right\}_{k=0}^\infty\)
(e)
\(\{j_k\} = \left\{\dfrac{(-1)^k}{k^2}\right\}_{k=1}^\infty\)
(f)
\(\{\ell_k\} = \left\{\dfrac{k+1}{k!}\right\}_{k=0}^\infty\)

5.

For each of the following recursion relations, write out the first 8 terms of the sequence.
(a)
\(a_k = \dfrac{a_{k-1}+1}{2}\text{;}\) \(a_1 = 3\)
(b)
\(b_k = \dfrac{2}{b_{k-1}}\text{;}\) \(b_1 = 4\)
(c)
\(c_k = 2c_{k-1} - 2\text{;}\) \(c_1 = 3\)
(d)
\(d_k = \dfrac{d_{k-1}}{3}\text{;}\) \(d_1 = 2\)
(e)
\(j_k = 3(j_{k-1})^2 - 1\text{;}\) \(j_1 = 1\)
(f)
\(\ell_k = \dfrac{1}{(\ell_{k-1})^2}\text{;}\) \(\ell_1 = 2\)

6.

For each of the following sequences, write down a possibility for an explicit formula defining the sequence as well as a recursion relation (with initial value) defining the sequence.
(a)
\(\{a_k\} = \{4, 5, 6, 7, ...\}\)
(b)
\(\{b_k\} = \{0, 2, 4, 6, ...\}\)
(c)
\(\{c_k\} = \left\{ \dfrac{3}{4}, \dfrac{1}{2}, \dfrac{1}{3}, \dfrac{2}{9}, ... \right\}\)
(d)
\(\{d_k\} = \left\{\dfrac{1}{8}, \dfrac{3}{16}, \dfrac{1}{4}, \dfrac{5}{16}, ...\right\}\)

7.

For each of the following sequences, describe whether they converge or diverge. If they converge, show what they converge to using limit work.
(a)
\(\{a_k\} = \left\{ 3 + \dfrac{(-1)^{k+1}}{k} \right\}_{k=1}^\infty\)
(b)
\(\{b_k\} = \left\{ \sqrt[k]{k} \right\}_{k=1}^\infty\)
(c)
\(\{c_k\} = \left\{ \dfrac{k!}{e^k}\right\}_{k=0}^\infty\)
(d)
\(\{d_k\} = \left\{ \left(1+\dfrac{2}{k}\right)^k\right\}_{k=1}^\infty\)
(e)
\(\{j_k\} = \left\{ \dfrac{k^3 + 1}{k^2+\ln(k)}\right\}_{k=1}^\infty\)
(f)
\(\{\ell_k\} = \left\{ k\ln\left(1/k\right)\right\}_{k=1}^\infty\)

8.

Compare a sequence to a function. What are some similarities? What are some differences? Carefully discuss this in terms of different representations of sequences/functions.

9.

Choose through the sequences in any of the problems above to find one to fit each of the descriptions below. Explain your choices for each.
(c)
Bounded. What are the upper and lower bounds?