Limits of sequences

$\newenvironment {prompt}{}{} \newcommand {\accordion }[0]{} \newcommand {\ungraded }[0]{} \newcommand {\xmDefaultPreamble }[0]{xmPreamble.tex} \newcommand {\xmDefaultPrintstyle }[0]{xmPrintstyle.sty} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{}$

There are two ways to establish whether a sequence has a limit.

In the previous section, we defined a sequence as a function defined on a subset of the natural numbers, and we discussed how we can represent this by an ordered list. We chose the notation $\{a_n\}_{n=1}$ to denote the list below.

\[ a_1, a_2, a_3 , \ldots \]

In the previous section, we found many ways to generate this list. Regardless of how we obtain it, there are two fundamental questions we can ask.

Do the numbers in the list approach a finite value?
Can I add all of the numbers in the list together and obtain a finite result?

As it turns out, the second question will be more important for us. However, as we will see in a future section, we can reduce the second question to the first one. As such, we should examine the first question in detail. We begin by giving an intuitive definition.

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

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.

This intuitive 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 $n \geq N$.

This precise definition captures the same idea as the intuitive definition but makes it more precise. The quantity $\epsilon $ measures how close the terms in the sequence are to the limit $L$. We say that the limit exists and is $L$ if we can choose how far we want the terms to be from $L$ and we know the terms in the sequence eventually become and stay that close to $L$.

The precise definition is extremely important to establish the theoretical foundations of sequences and is used frequently in more theoretically-oriented courses. For our purposes, however, the intuitive definition will be sufficient.

Suppose that $\{a_n\}_{n=1}$ is a sequence and that $\lim _{n \to \infty } a_n = L$. Intuitively, 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. $\lim _{n \to \infty } a_{n+1}$ may or may not exist.

One way to think about this is by noting that the sequence $\{a_n\}_{n=1}$ is represented by the list

\[ a_1,a_2,a_3, \ldots , \]

while the sequence $\{a_{n+1}\}_{n=1}$ is represented by the list below.

\[ a_2,a_3,a_4, \ldots \]

If the first sequence tends to $L$, the second sequence must also tend to $L$.

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.

1 Connections to real-valued functions

Since sequences are functions defined on the integers, the notion of a “limit at a specific $n$” is not very interesting since we can explicitly find $a_n$ for a given $n$. However, limits at infinity are a different story. An important question can now be asked; given a sequence, how do we determine if it has a limit?

There are several techniques that allow us to find limits of real-valued functions, and we have seen that if we have a sequence, we can often find a real-valued function that agrees with it on their common domains. Suppose that we have found a real-valued function $f(x)$ that agrees with $a_n$ on their common domains, i.e. that $f(n)=a_n$. If we know $\lim _{x\to \infty } f(x)$, can we use this to conclude something about $\lim _{n \to \infty } a_n$?

Before answering this question, consider the following cautionary example.

Let $a_n = \sin (n\pi )$ and $f(x) = \sin (\pi x)$. Show that

\[ \lim _{n\to \infty } a_n \ne \lim _{x\to \infty }f(x). \]

The sequence $\{a_n\}_{n=1}^{\infty }$ is represented by the ordered list of numbers below.

\[ \sin (0\pi ),\, \sin (1\pi ),\, \sin (2\pi ),\,\sin (3\pi ),\,\ldots \]

Since $\sin (n\pi )=0$, this list is actually a list of zeroes.

\[ 0,\, 0 , \, 0 ,\,0,\,\ldots \]

Since every term in the sequence is $0$, we have

\[ \lim _{n\to \infty } a_n = 0. \]

But $\lim _{x\to \infty }f(x)$, when $x$ is real, does not exist; as $x$ becomes arbitrarily large, the values $\sin (x\pi )$ do not get closer and closer to a single value, but instead oscillate between $-1$ and $1$.

This is shown graphically below.

What can we conclude from the above example?

If $\lim _{n \to \infty } a_n$ exists, then $\lim _{x \to \infty } f(x)$ exists. If $\lim _{x \to \infty } f(x)$ does not exist, then $\lim _{n \to \infty } a_n$ does not exist. If $\lim _{x \to \infty } f(x)$ does not exist, $\lim _{n \to \infty } a_n$ may still exist.

This might lead us to believe that we need to develop a whole new arsenal of techniques in order to determine if limits of sequences exist, but there is good news.

2 Calculating limits of sequences

Let $\{a_n\}$ be a sequence and suppose that $f(x)$ is a real-valued function for which $f(n) = a_n$ for all integers $n$. If

\[ \lim _{x\to \infty }f(x)=L, \]

then $\lim _{n\to \infty } a_n=L$ as well.

If we think about the theorem a bit further, the conclusion of the theorem and the content of the preceding example should seem reasonable. If the values of $f(x)$ become arbitrarily close to a number $L$ for all arbitrarily large $x$-values, then the result should still hold when we only consider some of these values. However, if we only know what happens for some arbitrarily large $x$-values, we cannot say what happens for all of them!

Let $a_n = \frac {5n+1}{6n+7}$. Determine if the sequence $\{a_n\}_{n=1}^{\infty }$ has a limit.

A function that can be used to generate the sequence is $f(x) = \frac {5x+1}{6x+7}$. Since $\lim _{x \to \infty } \frac {5x+1}{6x+7} = \answer [given]{\frac {5}{6}}$, $\lim _{n \to \infty } a_n = \answer [given]{\frac {5}{6}}$.

Remember that the converse of this theorem is not true. In the example preceding this theorem, we have an explicit example of a function $f(x)$ and a sequence $(a_n)$ where $a_n =f(n)$ and

\[ \lim _{x\to \infty }f(x)=\text {DNE} \quad \text {but} \quad \lim _{n\to \infty } a_n = 0. \]

In practice, we use the above theorem to compute limits without explicitly exhibiting the function of a real variable from which the limit is derived.

3 Computing limits of sequences using dominant term analysis

The last example shows us that for many sequences, we can employ the same techniques that we used to compute limits previously. While algebraic techniques and L’Hopital’s rule are useful, in many of the following sections, being able to determine limits quickly is an important skill.

Let $a_n = \frac {n^3+4n^2-1}{2-4n^4}$. Determine if the limit of the sequence $\{a_n\}_{n=1}^{\infty }$ exists.

The highest degree term in the numerator is $n^3$, while the largest term in the denominator is $-4n^4$. We can factor out the largest terms from both the numerator and denominator and do a little algebra.

\[ \frac {n^3+4n^2-1}{2-4n^4} = \frac {n^3\left (1+\frac {4}{n}-\frac {1}{n^3}\right )}{-4n^4\left (-\frac {1/2}{n^4}+1\right )} = \frac {n^3}{-4n^4} \cdot \frac {1+\frac {4}{n}-\frac {1}{n^3}}{-\frac {1/2}{n^4}+1} \]

The second term becomes arbitrarily close to $1$ as $n$ grows larger and larger, so the limit of the sequence is completely determined by the ratio of the highest degree term in the numerator to the highest degree term in the denominator. In this case, that ratio is $\frac {n^3}{-4n^4} = \frac {1}{-4n}$, so $\lim _{n \to \infty } a_n = 0$.

In the preceding example, we say that the dominant term in the numerator is $n^3$ and that the dominant term in the denominator is $-4n^4$ because these terms are the only ones that are relevant when finding the limit.

The reader may notice that the last example is a special case of the Rational Function Theorem. However, the name given to this result is not as important as the idea it captures. When finding limits of functions, it is only necessary to consider the dominant term. When treating quotients of functions, we only need to consider the dominant terms in the numerator and denominator.

Sometimes, this technique can be used to find limits where L’Hopital’s rule or an algebraic approach would be complicated.

Let $a_n = \frac {n^2(2n+1)(5-3n)}{(1+2n)^4}$. Determine if the limit of the sequence $\{a_n\}_{n=1}^{\infty }$ exists.

While we could perform the multiplication in both the numerator and denominator explicitly, we can spot the dominant term more efficiently.

The highest degree term in the numerator is $n^2 \cdot 2n \cdot (-3n) = \answer {-6n^4}$.
The dominant term in the denominator is $(2n)^4 = \answer {16}n^4$.

By noting that

\[ \lim _{n \to \infty } \frac {n^2(2n+1)(5-3n)}{(1+2n)^4} = \lim _{n \to \infty } \frac {-6n^4}{16n^4} = \answer {-\frac {6}{16}}, \]

we find $\lim _{n \to \infty } a_n = \answer {-\frac {6}{16}}$.

3.1 Growth rates

The preceding examples illustrate that higher positive powers of $n$ grow more quickly than lower positive powers of $n$. We can introduce a little notation that captures the rate at which terms in a sequence grow in a succinct way.

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 $(a_n)$.

Suppose that $a_n = 4n^2+3n$ and $b_n = 5n^{3/2}+2n$. Then, we can compute $\lim _{n \to \infty } \frac {a_n}{b_n} = \answer {\infty }$ and $\lim _{n \to \infty } \frac {b_n}{a_n} = \answer {0}$. Using the notation we just introduced, we have that $a_n \ll b_n$$b_n \ll a_n$

Many sequences of interest involve terms other than powers of $n$. It is often useful to understand how different types of functions grow relative to each other.

Growth rates of sequences Let $p,q$ be positive real numbers, and let $b> 1$. We have the following relationships.

\[ \ln ^p(n)\ll n^q \ll b^n \ll n! \ll n^n \]

The first inequality in this theorem essentially guarantees that any power of $\ln (n)$ grows more slowly than any power of $n$. For example:

Let $a_n = \frac {\ln ^{9}(n)}{n^{1/2}}$. What is $\lim _{n \to \infty } a_n$?

The notation $\ln ^p(n)\ll n^q$ means that $\lim _{n \to \infty } \frac {\ln ^p(n)}{n^q} = 0$ for any positive numbers $p$ and $q$. In this example, $p=9$ and $q=1/2$, so by the growth rates result, $\lim _{n\to \infty }a_n =\answer [given]{0}$.

This allows us to extend the dominant term idea to more complicated expressions.

Let $a_n = \frac {n^{100} + n^n}{n!+5^n}$. What is $\lim _{n \to \infty } a_n$?

By growth rates, the dominant term in the numerator is $n^n$, and the dominant term in the denominator is $n!$. We thus will know if $\lim _{n \to \infty } a_n$ exists by considering $\lim _{n \to \infty } \frac {n^n}{n!}$. By growth rates, this limit is infinite, so $\lim _{n \to \infty } a_n = \infty $.

This can be made more explicit by the following computation, which shows exactly how the growth rates results are used. As with a previous example, it relies on factoring the dominant term in the numerator and the denominator. These terms are determined by the growth rates results.

\[ \lim _{n \to \infty } \frac {n^{100} + n^n}{n!+5^n} = \lim _{n \to \infty } \frac {n^n \left (\frac {n^{100}}{n^n} + 1\right )}{n!\left (1+\frac {5^n}{n!}\right )} \]

By the growth rates results, $\lim _{n \to \infty } \frac {n^{100}}{n^n} =0$ and $\lim _{n \to \infty }\frac {5^n}{n!} =0$, so we have:

\[ \lim _{n \to \infty } \frac {n^n \left (\frac {n^{100}}{n^n} + 1\right )}{n!\left (1+\frac {5^n}{n!}\right )} = \lim _{n \to \infty } \frac {n^n \left (0 + 1\right )}{n!\left (1+0\right )} = \lim _{n \to \infty } \frac {n^n}{n!}. \]

3.2 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, $\lim _{n\to \infty } b_n = \answer [given]{0}$.

The squeeze theorem is helpful in establishing a more general result about geometric sequences.

Given a geometric sequence $\{a_n\}_{n=n_0}$ where $a_n = a \cdot r^{n}$,

\[ \lim _{n\to \infty } a_n = \begin{cases} 0 &\text {if $|r|<1$,}\\ 1 &\text {if $r=1$,}\\ \text {DNE} &\text {if $|r|>1$ or $r=-1$.} \end{cases} \]

Of course, when $r$ is positive, the squeeze theorem is not necessary, but it is useful when establishing the convergence results for $1<r<0$ as in the preceding example.

4 Existence results for limits

Perhaps the best way to determine whether the limit of a sequence exists is to compute it. Even though we’ve been working with sequences that are generated by an explicit formula in this section thus far, not all sequences are defined this way. Sometimes, we’ll only have a recursive description of a sequence rather than an explicit one, and sometimes we will have neither. In many of the coming sections, we will only have a recursive description of a sequence, so we want to determine a good approach for determining whether a limit exists without having to compute it directly. To do this, we introduce some 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 introduce a single word to cover both cases.

If a sequence is always increasing, or always nondecreasing, or always decreasing, or always nonincreasing, it is said to be monotonic.

If $a_n = 2n^2+1$, then $\{a_n\}_{n=1}^{\infty }$ is increasingdecreasing, so it is monotonicnot monotonic

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

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

We can model an arithmetic sequence $a_n = m\cdot n + b$ with the line $f(x) = mx+b$. Can a line ever increase then decrease or vice-versa?

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

From our examples earlier in the section, a geometric sequence $a_n = a_1 \cdot r^{n-1}$ can be modeled by an exponential function (which is always increasing or always decreasing) if the sign of $r$ is positive. If $r$ is negative, the signs of each successive term is different from the last.

Sometimes we want to classify sequences for which the terms do not get too big or too small.

A sequence $\{a_n\}$ is:

bounded above if there is some number $M$ so that for all $n$, we have $a_n\le M$.
bounded below if there is some number $m$ so that for all $n$, we have $a_n\ge m$.
bounded if it is both bounded above and bounded below.

So what does this definition actually say? Essentially, we say that a sequence is bounded above if its terms cannot become too large, bounded below if its terms cannot become too large and negative, and bounded if the terms cannot become too large and positive or too large and negative.

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.

So, what do these previous definitions have to do with the idea of a limit? Essentially, there are three reasons that a sequence may diverge:

the terms eventually are either always positive or always negative but become arbitrarily large in magnitude.
the terms are never eventually monotonic.
the terms are never eventually monotonic and become arbitrarily large in magnitude.

Let’s think about the terminology we introduced.

Think about the following statements and choose the correct option.

If we know that a sequence is monotonic and its limit does not exist, then the terms become too large in magnitudethe terms are never eventually monotonicthe terms are never eventually monotonic and become arbitrarily large in magnitude.
If we know that a sequence is bounded and its limit does not exist, then the terms become arbitrarily large in magnitude.the terms are never eventually monotonicthe terms are never eventually monotonic and become arbitrarily large in magnitude.

We can now state an important theorem:

Bounded-monotone convergence theorem If the sequence $a_n$ is bounded and monotonic, then $\lim _{n \to \infty } a_n $ exists.

To think about the statement of the theorem, if we have a sequence that is bounded, the only way it could diverge is if the terms are never eventually monotonic. However, if we know the sequence is also monotonic, this cannot happen! Thus, the series cannot diverge, so it must have a limit.

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 without computing its limit.

To start, note that $a_n=\frac {2^n-1}{2^n} = \frac {2^n}{2^n} - \frac {1}{2^n} = 1 - \frac {1}{2^n} $. Thus, $\{a_n\}$ is monotonicnot monotonic. Moreover, all of the elements $a_n = (2^n-1)/2^n$ are less than $2$ and greater than zero. So, the sequence is bounded above by $2$ and below by $0$, so it is bounded. By the bounded-monotone convergence theorem, $\lim _{n\to \infty } a_n$ must converge to a finite value.

We don’t actually need to know that a sequence is always monotonic to apply the bounded-monotone convergence theorem. It is enough to know that the sequence is eventually monotonic. More formally, this means that there is some integer $N$ for which the sequence $\{a_n\}_{n=N}$ is always increasing or always decreasing.

In the previous examples, we could write down a function $f(x)$ corresponding to each series and apply the theorem from earlier in the section. However, this is not always possible.

Suppose that $a_n = 2 + \sin (n)$, and let $s_n = \sum _{k=1}^{n} a_k$. Determine whether the sequences $\{a_n\}$ and $\{s_n\}$ are bounded or monotonic, and explain whether either has a limit.

The sequence $a_n$ is certainly not monotonic, but it is bounded since $1 \geq a_n \geq 3$ for all $n$. We also can see that $\lim _{n \to \infty } a_n$ does not exist since the terms oscillate.

For $s_n$, note that since $a_n \geq 1$ for all $n$, each term $a_n$ is positive. Thus, $s_n$ is increasing and hence monotonic.

However, since $a_n\geq 1$ for all $n$, we have the following inequality.

\[s_n = \sum _{k=1}^n a_n \geq \sum _{k=1}^n 1 = n\]

Hence, $\{s_n\}$ is not bounded and $\lim _{n \to \infty } s_n$ does not exist.

We had a way to analyze $\{a_n\}$ in the above example because we had an explicit formula for $a_n$, but how would we find such a formula for $s_n$? As it turned out, we didn’t have to do so in order to determine that $\lim _{n \to \infty } s_n$ does not exist. We will make arguments in the coming sections that allow us to determine whether limits of sequences exist without relying on having explicit formulas for the sequences.

2025-01-06 20:18:04