Limits of Sequences

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convergence

In the previous section, we defined a sequence as a function defined on a subset of the whole 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?
If so, what is that value?

We begin with an intuitive definition.

Intuitive Limit

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 $\lim \limits _{n\to \infty }a_n=L$ we say that the sequence converges. If there is no finite value $L$ so that $\lim \limits _{n\to \infty }a_n = L$, 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.

Formal Limit

Suppose that $\{a_n\}_{n=n_0}$ is a sequence. We say that $\lim \limits _{n\to \infty }a_n=L$ 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 any distance 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 $\lim \limits _{n \to \infty } a_n = L$. Intuitively, what can we say about $\lim \limits _{n \to \infty } a_{n+1}$?

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

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 $\lim \limits _{x\to \infty } f(x)$, can we use this to conclude something about $\lim \limits _{n \to \infty } a_n$?

Before answering this question, consider the following cautionary example.

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

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

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

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

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

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

\[ \text {If} \, \lim \limits _{x\to \infty }f(x)=L, \, \text { then } \, \lim \limits _{n\to \infty } a_n=L \, \text { as well. } \]

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 $\lim \limits _{x \to \infty } \frac {5x+1}{6x+7} = \answer [given]{\frac {5}{6}}$, $\lim \limits _{n \to \infty } a_n = \answer [given]{\frac {5}{6}}$.

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

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

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.

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more examples can be found by following this link
More Examples of Limits of Sequences

2025-05-17 23:42:10