Sequences as functions

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A sequence can be thought of as a function from the integers to the real numbers. There are two ways to establish whether a sequence has a limit.

Recall that in the previous section, we defined a sequence as an ordered list of numbers, and chose to adopt the notation $\{a_n\}_{n=1}$ to denote the list:

$a_1, a_2, a_3 , \ldots$

In the previous section, we studied different ways to generate the numbers in this list, special types of sequences, and new sequences that we could construct from a given one.

Once we have a sequence, we can turn our attention to two fundamental questions regarding it:

Do the numbers in the list approach a finite value?
Can I sum all of the numbers in the list?

We will address the first question in this section and the second question in the following section. we begin by giving a definition:

Limits of sequences

Since sequences are essentially discrete, meaning that the points are separate and distinct, the notion of a “limit at a point” cannot be made to really make sense. However, limits at infinity are a different story.

In short, given a sequence, it is helpful to be able to say something qualitative about it; we may want to address the question such as “what happens after a while?”

Earlier, you’ve studied a similar question about $\lim _{x\to \infty } f(x)$ when $x$ is a variable taking on real values; now, we simply want to restrict the “input” values to be integers. No significant difference is required in the definition of limit, except that we specify, perhaps implicitly, that the variable is an integer:

Given a sequence $\{a_n\}_{n =n_0}$ , we say that the limit of the sequence is $L$ if $a_n$ becomes arbitrarily close to $L$ as $n$ grows arbitrarily large.

If $\lim _{n\to \infty }a_n=L$ we say that the sequence converges. If there is no finite value $L$ so that $\lim _{n\to \infty }a_n = L$ , then we say that the limit does not exist, or equivalently that the sequence diverges.

Suppose that $\{a_n\}_{n=1}$ is a sequence and that $\lim _{n \to \infty } = L$ . What can we say about $\lim _{n \to \infty } a_{n+1}$ ?

$\lim _{n \to \infty } a_{n+1}$ exists, but we do not know what its value is. $\lim _{n \to \infty } a_{n+1}$ exists, and $\lim _{n \to \infty } a_{n+1}=L$ exists. $\lim _{n \to \infty } a_{n+1}$ could exist but does not have to exist.

Note that the sequence $\{a_n\}_{n=1}$ is the list:

$a_1,a_2,a_3, \ldots$

while the sequence $\{a_{n+1}\}_{n=1}$ is:

$a_2,a_3,a_4, \ldots$

It should be clear that since the first sequences tends to $L$ , the second sequence must also tend to $L$ !

This definition of a limit can be made more precise as follows:

Suppose that $\{a_n\}_{n=n_0}$ is a sequence. We say that $\lim _{n\to \infty }a_n=L$ if for every $\epsilon >0$ , there exists an integer $N$ , such that $|a_n-L|<\epsilon$ for any $n \geq N$ .

Here $\epsilon$ measures how close the terms in the sequence are to the limit $L$ and the above definition really states that no matter how close you want these terms to be, there is a value for the index $N$ so that all subsequent terms in the sequence are at least that close to it!

In the case that $\lim _{n \to \infty } a_n = \pm \infty$ , we say that $\{a_n\}$ diverges. The only time we say that a sequence converges is when the limit exists and is equal to a finite value.

We have previously studied how to compute limits, so the curious young mathematician could certainly ask whether the old techniques for continuous functions can still can be applied to find limits of sequences.

One way to compute the limit of a sequence is to compute the limit of the related function.

Let $f(x)$ be a real-valued function. If $a_n = f(n)$ defines a sequence and if $\lim _{x\to \infty }f(x)=L,$ then $\lim _{n\to \infty } a_n=L$ as well.

It is important to note that the converse of this theorem is not true. There are functions $f(x)$ and sequences $(a_n)$ where $a_n =f(n)$ and $\lim _{x\to \infty }f(x)=\text {DNE} \quad \text {but} \quad \lim _{n\to \infty } a_n = L$ for some number $L$ .

To show the converse is not true, it is enough to provide a single example where it fails. Here is such a counterexample.

Let $(a_n)$ be given by the rule $a_n = f(n)=\sin (n\pi )$ . Show that $\lim _{n\to \infty } a_n \ne \lim _{x\to \infty }f(x).$

The sequence $(a_n)$ is the sequence $\sin (0\pi ),\, \sin (1\pi ),\, \sin (2\pi ),\,\sin (3\pi ),\,\ldots ,$ which is just the sequence $0, 0, 0, 0, \ldots$ since $\sin (n\pi )=0$ whenever $n$ is an integer. Since the sequence is just the constant sequence, we have $\lim _{n\to \infty } f(n)= \lim _{n\to \infty } 0 = 0.$ But $\lim _{x\to \infty }f(x)$ , when $x$ is real, does not exist: as $x$ gets bigger and bigger, the values $\sin (x\pi )$ do not get closer and closer to a single value, but instead oscillates between $-1$ and $1$ .

Here is some general advice. If you want to know $\lim _{n\to \infty } a_n$ , you might first think of a function $f(x)$ where $a_n = f(n)$ , and then attempt to compute $\lim _{x\to \infty }f(x)$ . If the limit of the function exists, then it is equal to the limit of the sequence. But, if for some reason $\lim _{x\to \infty }f(x)$ does not exist, it may nevertheless still be the case that $\lim _{n\to \infty }a_n$ exists, you’ll just have to figure out another way to compute it.

Let’s summarize the preceding section in the following formal definition.

A sequence $(a_n)$ is a real-valued function with domain $\{ n \in \Z : n \ge N \},\quad \text {for some integer $N$}$ where $\Z = \{\dots , -5,-4,-3,-2,-1,0,1,2,3,4,5,\dots ,\}$ is the set of integers.

Stated more humbly, a sequence assigns a real number to each of the integers starting with an index $N$ .

When thought of as a function, the “outputs” of a sequence are the elements of the sequence; the “ $n$ th element” is the real number that the sequence associates to the natural number $n$ , and is usually written $a_n$ . The $n$ in the phrase “ $n$ th element” is called an index; the plural of index is either indices or indexes, depending on who you ask. The first index $N$ is called the initial index.

What function corresponds to the sequence given by the explicit formula $a_n = -n/2+3$ for $n=1,2,3,\dots$ ? $a(x) = \answer [given]{-x/2 + 3}$

What function corresponds to the sequence given by the explicit formula $a_n = 6\left (\frac {2}{5}\right )^{n+2}$ for $n=1,2,3,\dots$ ? $a(x) = \answer [given]{6\left (\frac {2}{5}\right )^{x+2}}$

Since sequences can be conceptualized as functions, and calculus is used to study functions, we can now apply our knowledge of calculus to sequences!

Plotting sequences

First, we plot sequences as points. Later, we will see another interpretation.

Plots of arithmetic sequences

Recall that arithmetic sequences are those where the difference between neighboring elements is constant. Arithmetic sequences are analogues of lines. Consider a basic example:

Check out a graph of the sequence:

Here is an arithmetic sequence that decreases as its index increases.

Here we see a graph:

To what type of curve do arithmetic sequences correspond?

lines parabolas polynomials exponential curves impossible to say

Since the average growth rate of a line is constant regardless of the size of the interval chosen, arithmetic sequences correspond to lines.

Plots of geometric sequences

Recall that geometric sequences are those where the ratio between neighboring elements is constant. When this ratio is positive, a geometric sequence corresponds to an exponential function. Consider a basic example:

Let’s see a graph:

If the common ratio of a geometric sequence is between $0$ and $1$ , a geometric sequence will decrease as it progresses.

Let’s see a graph:

When the common ratio between successive elements of a geometric sequence is positive, what type of curves do geometric sequences correspond to?

lines parabolas polynomials exponential curves impossible to say

If a geometric sequence is expressed as $a_n = a_1 \cdot r^{n-1}$ , then in function notation we have $a(x) = a_1 \cdot r^{x-1}$ , an exponential function.

On the other hand, if the common ration between successive elements of a geometric sequence is not positive, then something interesting happens. Check out this example:

Let’s see a graph:

The sign of this sequence alternates!

Monotonicity

We’d like some terminology to describe features we might notice about sequences. Here is some of that terminology, focused on the relationships between the terms of a sequence.

A sequence is called

increasing if $a_n<a_{n+1}$ for all $n$ ,
nondecreasing if $a_n\le a_{n+1}$ for all $n$ ,
decreasing if $a_n>a_{n+1}$ for all $n$ ,
nonincreasing if $a_n\ge a_{n+1}$ for all $n$ .

Lots of facts are true for sequences which are either increasing or decreasing; to talk about this situation without constantly saying “either increasing or decreasing,” we can make up a single word to cover both cases.

If a sequence is

increasing, or
nondecreasing, or
decreasing, or
nonincreasing,

it is said to be monotonic.

If an arithmetic sequence $a_n = m\cdot n + b$ is monotonic, what must be true about $m$ and $b$ ?

The sign of $m$ , and the sign of $b$ is is positiveis negativedoes not matter

An arithmetic sequence is always monotonic, regardless of the values of $m$ and $b$ .

If a geometric sequence $a_n = a_1 \cdot r^{n-1}$ is monotonic, what must be true about $a_1$ and $r$ ?

The sign of $a_1$ , and the sign of $r$ is is positiveis negativedoes not matter

A geometric sequence $a_n = a_1 \cdot r^{n-1}$ is monotonic if and only if the sign of $r$ is positive.

Let’s see some examples:

Describe the growth of $a_n = \frac {2^n-1}{2^n}$ for $n=1,2,3,\dots$ .

To do this, check out the difference between two sequential terms, write with me: $\begin{align*} a_{n+1} - a_n &= \answer[given]{\frac{2^{n+1}-1}{2^{n+1}}} - \frac{2^n-1}{2^n}\\ &=\frac{2^{n+1}-1 - 2\left(\answer[given]{2^n-1}\right)}{2^{n+1}}\\ &=\frac{2^{n+1}-1 - 2^{n+1}+2}{2^{n+1}}\\ &=\frac{\answer[given]{1}}{2^{n+1}}. \end{align*}$ Since $\frac {1}{2^{n+1}}$ is always positivenegativezero , we see that $(a_n)$ is increasingnondecreasingdecreasingnonincreasing , and hence is monotonicnot monotonic .

Describe the growth of $a_n = \frac {n+1}{n}$ for $n=1,2,3,\dots$ .

To do this, check out the difference between two sequential terms. Write with me. $\begin{align*} a_{n+1} - a_n &= \answer[given]{\frac{(n+1)+1}{n+1}} - \frac{n+1}{n}\\ &=\frac{n(n+2) - (n+1)(\answer[given]{n+1})}{n+1}\\ &=\frac{n^2+2n - n^2 -2n-1}{n+1}\\ &=\frac{\answer[given]{-1}}{n+1}. \end{align*}$ Since $\frac {-1}{n+1}$ is always positivenegativezero for the values of $n$ we are considering, we see that $(a_n)$ is increasingnondecreasingdecreasingnonincreasing , and hence is monotonicnot monotonic .

Sometimes we can say that the sequence doesn’t get too big or too small, in this case we say the sequence is bounded.

A sequence $(a_n)$ is bounded if there is some number $M$ so that for all $n$ , we have $|a_n|\le M$ .

True or False: If a sequence $(a_n)_{n=0}^\infty$ is nondecreasing it is bounded below by $a_0$ .

true false

If a sequence is nondecreasing, then its smallest value is its first element.

True or False: If a sequence $(a_n)_{n=0}^\infty$ is nonincreasing it is bounded above by $a_0$ .

true false

If a sequence is nonincreasing, then its largest value is its first element.

Monotone convergence

We can now state an important theorem.

Bounded-monotone convergence If the sequence $a_n$ is bounded and monotonic, then $\lim _{n \to \infty } a_n = L$ for some real-number $L$ .

Consider the sequence $a_{n}$ . Suppose you know that for all $n > 1$ , $-6 \le a_{n} \le 0$ and $a_{1} = 2$ , and $a_{2} = -1$ , and that the sequence is nonincreasing. Does the sequence converge?

Since the sequence is nonincreasing, the sequence is monotone.

Since for all $n \ge 1$ , we have $a_{n} \ge -6$ , the sequence is bounded below.

So by the Monotone Convergence Theorem, the sequence converges to some value; let us call it $L$ .

Now consider the direction in which the sequence is heading.

Since the sequence is nonincreasing, for all $n \ge 2$ , we have $-6 \le a_{n} \le -1$ .

The limit $L$ must be in that interval as well.

Therefore the sequence converges to a value $L$ so that $-6 \le L \le -1$ .

Yes, with limit between $-6$ and $-1$ . No, the sequence does not converge. Yes, with limit between $-1$ and $0$ .

In short, bounded monotonic sequences always converge, though we can’t necessarily describe the number to which they converge. Let’s try some examples!

Given the sequence $a_n=\frac {2^n-1}{2^n}$ for $n=1,2,3,\dots$ , explain how you know that $\lim _{n\to \infty } a_n$ converges to a finite value.

To start, from previous work we know that $a_n$ is monotonicnot monotonic . Moreover, all of the elements $a_n = (2^n-1)/2^n$ are less than $2$ , and greater than zero. Hence by the bounded-monotone convergence theorem, $\lim _{n\to \infty } a_n$ converges to a finite value.

Given the sequence $a_n=\frac {n+1}{n}$ for $n=1,2,3,\dots$ , explain how you know that $\lim _{n\to \infty } a_n$ converges to a finite value.

To start, from previous work we know that $a_n$ is monotonicnot monotonic . Moreover, all of the elements $a_n = (n+1)/n$ are less than $2$ , and greater than zero. Hence by the bounded-monotone convergence theorem, $\lim _{n\to \infty } a_n$ converges to a finite value.

We don’t actually need to know that a sequence is monotonic to apply the bounded-monotone convergence theorem. It is enough to know that the sequence is “eventually” monotonic, that is, that some point it becomes increasing or decreasing.

Show that the sequence $(a_n)$ given by $a_n = n^{1/n}$ converges.

We might first show that this sequence is decreasing, that is, we show that for all $n$ , $n^{1/n} > (n+1)^{1/(n+1)}.$ But this isn’t true! Take a look $\begin{align*} a_1 &= 1, \\ a_2 &= \sqrt{2} \approx 1.4142, \\ a_3 &= \sqrt[3]{3} \approx 1.4422, \\ a_4 &= \sqrt[4]{4} \approx 1.4142, \\ a_5 &= \sqrt[5]{5} \approx 1.3797, \\ a_6 &= \sqrt[6]{6} \approx 1.3480, \\ a_7 &= \sqrt[7]{7} \approx 1.3205, \\ a_8 &= \sqrt[8]{8} \approx 1.2968, \\ a_9 &= \sqrt[9]{9} \approx 1.2765. \end{align*}$ But it does seem that this sequence perhaps is decreasing after the first few terms. Can we justify this?

Yes! Consider the real function $f(x)=x^{1/x}$ when $x\ge 1$ . We compute the derivative, perhaps via logarithmic differentiation, to find $f'(x)=\answer [given]{\frac {x^{1/x} (1-\ln (x))}{x^2}}.$ Note that when $x\ge 3$ , the derivative $f'(x)$ is negative. Since the function $f$ is decreasing, we can conclude that the sequence is decreasing—well, at least for $n \geq 3$ .

Since all terms of the sequence are positive, the sequence is decreasing and bounded $0\le a_n \le a_3$ when $n \ge 3$ , and so the sequence converges.

The squeeze theorem

Previously, when considering limits, one of our techniques was to replace complicated functions by simpler functions. The Squeeze Theorem tells us one situation where this is possible.

Squeeze Theorem Suppose that $(a_n)$ , $(b_n)$ , and $(c_n)$ are sequences with $a_n \le b_n \le c_n$ for all $n$ greater than some index $N$ . If $\lim _{n\to \infty } a_n = L = \lim _{n\to \infty } c_n,$ then $\lim _{n\to \infty } b_n = L$ .

Let’s see an example.

Consider the sequence $(b_n)_{n=1}^{\infty }$ where $b_n = \left (\frac {-1}{2}\right )^n$ . Compute: $\lim _{n\to \infty }b_n$

To compute this limit we will use the squeeze theorem. Write with me: $-\left (\frac {1}{2}\right )^n\le \left (\frac {-1}{2}\right )^n \le \left (\frac {1}{2}\right )^n$ but we know $\lim _{n\to \infty }\left (-\left (\frac {1}{2}\right )^n\right ) = \answer [given]{0}$ and $\lim _{n\to \infty }\left (\frac {1}{2}\right )^n =\answer [given]{0}.$ Hence by the squeeze theorem, $\lim _{n\to \infty } b_n = \answer [given]{0}$ .

This last result actually gives us a general theorem about geometric sequences:

Given a geometric sequence $a_n = a_1 \cdot r^{n-1}$ , $\lim _{n\to \infty } a_n = \begin {cases} 0 &\text {if $|r|<1$,}\\ 1 &\text {if $r=1$,}\\ \text {DNE} &\text {if $|r|>1$ or $r=-1$.} \end {cases}$

Growth rates

When dealing with functions of real numbers, derivatives tell us the instantaneous growth rate of a function. Since sequences are essentially discrete, derivatives cannot really be used. Nevertheless, the idea of a growth rate is still very important.

Given two sequences $(a_n)$ and $(b_n)$ , the notation $(a_n) \ll (b_n)$ means that $\lim _{n\to \infty } \frac {a_n}{b_n} = 0\qquad \text {and}\qquad \lim _{n\to \infty } \frac {b_n}{a_n} =\infty .$

In essence, writing $a_n \ll b_n$ says that the sequence $(b_n)$ grows much faster than $(a_n)$ .

Growth rates of sequences Let $p,q$ be positive real numbers, and let $b> 1$ . We have the following relationship: $(\ln ^p(n))\ll (n^q) \ll (b^n) \ll (n!) \ll (n^n)$

Let $a_n = \frac {n^3\ln ^{9}(n)}{n^4}$ . Compute: $\lim _{n\to \infty }a_n$

Let’s simplify this a bit, write with me: $\begin{align*} a_n &= \frac{n^3\ln^{9}(n)}{n^4}\\ &= \frac{\ln^{9}(n)}{\answer[given]{n}}. \end{align*}$ Since $(\ln ^{9}(n))\ll (n)$ we see that $\lim _{n\to \infty }a_n =\answer [given]{0}$ . In short, the denominator “wins out” over the numerator, so the terms of the sequence get closer to zero.