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The Concept of a Limit

For some infinite sequences, the terms get closer and closer to a certain number, for example:

$$\frac{3}{2}, \frac{4}{3}, \frac{5}{4}, \ldots$$

It's best not to use the phrase, "get closer to," since it sounds vague. But there are a handful of accepted ways to describe what's going on, listed below. Some require names for the terms of the sequence, so call the sequence $a_{1}, a_{2}, a_{3}, \ldots$.

$\bullet$ $a_{n}$ approaches 1 as $n$ approaches $\infty$.

$\bullet$ $a_{n}\rightarrow 1$ as $n\rightarrow\infty$. (the arrow means "approaches")

$\bullet$ The sequence approaches 1.

$\bullet$ The limit of $a_{n}$, as $n$ approaches $\infty$, is 1.

$\bullet$ $\displaystyle\lim_{n\rightarrow\infty}a_{n} = 1$. ("lim" means "limit," and you read this equation exactly as the bullet above)

$\bullet$ The limit of the sequence is 1.

Consider the sequence $0.3, 0.33, 0.333, \ldots$, and let $x_{n}$ be the $n$th term of the sequence.

It's possible for a sequence to approach a number but never reach it. This is because any number has infinitely many other numbers close to it. (More precisely, any two different real numbers have infinitely many real numbers between them.)

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For either of the sequences we've considered, we say that the sequence has a limit, meaning that the sequence approaches some number.

One simple example is $1, 2, 3, \ldots$.

Finally, we need to establish definitions of "approaches" and "limit." The idea behind the definition is this: if a sequence approaches a number $x$, then we should be able to pick any small number $\epsilon$ we want, and the sequence should eventually remain within $\epsilon$ of $x$.

Definition. A sequence $a_{1}, a_{2}, a_{3}, \ldots$ approaches $x$ (or has limit x) if, for any $\epsilon > 0$, there exists a number $N$, such that $x - \epsilon < a_{k} < x + \epsilon$ for all $k\ge N$.

(You could replace the inequality with $|a_{k} - x| < \epsilon$.)

Proving a Sequence Doesn't Have a Limit

Consider the following sequence:

$$1, 2, 3, \ldots$$

To explain why the limit doesn't exist, we first need to remember the definition of a limit of a sequence. Take a moment to write down the definition if you remember it.

(If you don't remember it, scroll up a bit.)

To prove the limit doesn't exist, we need to show that this definition doesn't hold up for any $x$. This suggests that we use a proof by contradiction. Click continue to see a proof; then you'll have a chance to come up with your own proof for another sequence.

Here's a proof:

Suppose for the sake of contradiction that the limit of the sequence is $x$. Then the definition says (taking $\epsilon = 1$) that we can find a number $N$ such that all terms after the $N$th term are within 1 of $x$. But the $(N + 1)$th term and the $(N + 4)$th term differ by 3, so they cannot both be within 1 of $x$. This is a contradiction, so the limit doesn't exist.

Here's a more informal explanation; see whether you find it more or less convincing then the formal proof above:

Suppose for the sake of contradiction that the limit of the sequence is $x$, so that the terms of the sequence eventually all get close to $x$. But all the terms after the $\lceil x\rceil$th term are greater than $x$ by at least 1, so we have a contradiction. Therefore, the limit doesn't exist.

($\lceil x\rceil$ is the "ceiling function of $x$"; it's the smallest integer that's at least as large as $x$, or in other words, $x$ rounded up.)

Now consider a new sequence:

$$1, -1, 1, -1, \ldots$$

Try to prove that this sequence doesn't have a limit yourself. Use a proof by contradiction. Click continue when you've written down a proof.

Here's a proof, modeled after the proof above:

Suppose for the sake of contradiction that the limit is $x$. Then the definition says (taking $\epsilon = 1/2$) that we can find a number $N$ such that all terms after the $N$th term are within 1/2 of $x$. But the $(N + 1)$th term and the $(N + 2)$th term differ by 2, so they cannot both be within 1/2 of $x$. This is a contradiction, so the limit doesn't exist.

Here's an informal explanation, modeled after the informal explanation above:

Suppose for the sake of contradiction that the limit is $x$, so that the terms of the sequence eventually all get close to $x$. But if $x\ne 1$, then the 1's don't get close to $x$, and if $x\ne -1$, then the -1's don't get close to $x$. Thus the terms can't all get close to $x$. This is a contradiction, so the limit doesn't exist.

How'd you do? Compare your work with the proof and explanation above.

This post presents a graphical visualization of sequence convergence.

OK, so we sort of know what it means mathematically for a sequence to converge. Now let's figure out what it looks like!

When we try to picture function convergence, it makes sense to visualize it using a graph, as the graph is a very natural way to picture the behavior of functions.

As you figured out, one of the best ways to plot real numbers is on a number line.

Let's see what happens to sequences on the number line. We'll start with $1,2,...$

Now draw this sequence on a number line. What happens to the dots?

That's correct, as $n \to \infty$, the sequence goes infinitely far to the right.

Pause for a bit and think about what happens with convergent sequences. Then, once you have a guess (or are having trouble picturing it), draw the sequence $a_n=\frac{1}{n^2}$ on the number line.

So now we have another way of thinking about convergence, this time graphically: when plotted on a number line, the dots of a convergent sequence will all converge on the point corresponding to the number the sequence converges to!

To conclude, try coming up with some other convergent or divergent sequences, and see what they look like when plotted on a number line!

A formal definition for the convergence behavior of a sequence.

Intuitively, one would say that a sequence is just a set of numbers that we put them in order. For instance, the first term is $x$, the second term is $2x$......

However, in pure math, we regard a sequence as a function whose domain is a set of the form $\{b\in\mathbb{N}:n\ge m\}$ or $\{n\in\mathbb{Z}:n\ge m\}$. Usually, we would like to set $m$ to be equal to $1$ or $0$, and the sequence would be denoted by $(s_n)_{n=m}^\infty$, or more frequently,$(s_n)_{n\in\mathbb{N}}$.

Limits and Convergence

The "Limit" of a sequence, say $(x_n)$, is a real number (we usually consider the sequence in the set of real number, but more generally, we should consider the sequence in a metric space) that the values of $(x_n)$ get (arbitrarily) close to a number $x$ as $n$ goes larger.

Formal Definition

If for each $\epsilon>0$, $\exists N$ s.t. (such that) $$n>N\implies |x_n-x|<\epsilon$$ then we say that the sequence $(x_n)$ converges to the limit $x$.

If a sequence doesn't satisfy this condition, then we say it's divergent.

This is quite abstract, so let's consider an example.

$(s_n)_{n\in\mathbb{N}}=\frac{1}{n}$, it is quite obvious that $(s_n)\to0$. Choose an $\epsilon>0$, say $\epsilon=1$, then we can find an $N=1$ (in this case the term will be $\frac{1}{N}=1$) s.t. $n>N\implies |\frac{1}{n}-0|<1$, right? If you set $\epsilon=\frac{1}{1000}$ or even closer to $0$, still, you can find ALL terms s.t. $n>N$ that lies in the interval $(0,\frac{1}{1000})$.

If you can't understand this definition, try to draw a graph of this sequence so that you can visualize the convergent behavior in terms of this $\epsilon$ condition.

Questions

Determine whether the sequence $(k_n)=\dfrac{4n^3+3n}{n^3-6}$ converges using the condition above. (You need to first find the limit and then apply the $\epsilon$ condition)