Limit of a sequence

The limit of a sequence is the number that a given sequence approaches at infinity. The limit of a sequence is a kind of forerunner of the limit of a function, which is hugely important in mathematical analysis.

A short introduction to limits

The limit of a sequence is the number that a given sequence of numbers keeps getting closer to (and possibly even reaches in the end). It is best to start with a picture showing the sequence given by the formula $a_n=\frac{1}{n}$:

In the picture we can see that the terms of the sequence get closer and closer to zero. The larger n is, the smaller the value of a_n is, and the closer it gets to zero. The graph never touches the horizontal axis, because the equation 1/n = 0 never holds (where n is a positive integer). Still, it is worth finding out which number the whole sequence approaches. So the question can be: if we keep increasing n, which number will the value of the sequence approach? If we let n go to infinity, what value will the term of the sequence have? This is exactly what limits deal with. We write it using the symbol $lim$. In the subscript we put the number that n approaches, which for sequences is always infinity. The whole notation can look like this, for example:

$$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}=0.$$

We read it: “The limit of the sequence one over n as n tends to infinity is zero.” A few more trivial examples: what is the limit of the sequence a_n = n? This is a simple sequence where a₁ = 1, a₄₂ = 42, a₅₈₇₆₆ = 58766 and so on. Which number will the terms of the sequence approach at infinity? The larger n is, the larger a_n is, so the limit will be infinity.

$$\underset{n\to \mathrm{\infty }}{lim}n=\mathrm{\infty }$$

You can probably guess that the limit of the sequence a_n = n² will also be infinity. This sequence will grow even faster than the previous one. On the other hand, the sequence $a_n=\frac{1}{2n}$ will fall to zero even faster than the first example sequence.

$$\underset{n\to \mathrm{\infty }}{lim}{n}^2=\mathrm{\infty },\underset{n\to \mathrm{\infty }}{lim}\frac{1}{2n}=0$$

Definition of a finite limit of a sequence

It is time to define the limit of a sequence properly.

A sequence ${\{a_n\in \mathbb{R}\}}_{n=1}^{\mathrm{\infty }}$ has a finite limit a (converges to the limit a ∈ ℝ, is convergent) if for every ϵ > 0 there exists n₀ ∈ ℤ⁺ such that for all n>n₀ we have |a_n − a| < ϵ.

The notations in use are:

$$\underset{n\to \mathrm{\infty }}{lim}{a}_n=a$$

$$a_n\to a\text{ for }n\to \mathrm{\infty }$$

$${\left\{a_n\right\}}_{n=1}^{\mathrm{\infty }}\to a$$

That is a nice definition, so now let's explain it. For that we need a picture of some sequence, so let's choose, for example, the following sequence:

$$a_n=\frac{5n}{n+1}$$

And its graph:

From the graph we can see nicely that the sequence converges to five, that is, that the limit of this sequence is five. Before I start demonstrating the definition of a limit on this sequence, I'll add a bit of colour to the graph:

We already know that the limit of this sequence is five. The first part of the definition talks about ϵ>0, that is, some epsilon that is greater than zero. Epsilon here represents the neighbourhood of the limit, the neighbourhood of a point. It is greater than zero because a neighbourhood can be neither negative nor zero. In the picture this neighbourhood is shown by the horizontal green lines on the y-axis. The point here is the value five, and the neighbourhood of the point is those green lines on the y-axis. The neighbourhood itself is defined in both directions, up and down. We will now need this imaginary strip bounded by the green horizontal lines. In the picture ϵ = 1.

Our aim is to find a number n₀ on the bottom horizontal axis that splits the axis into two parts, such that all the terms of the sequence to the right of n₀ lie in this green strip. It doesn't matter where the terms to the left of n₀ are, and we don't have to look for the smallest n₀ for which the definition holds. In the picture n₀ = 12.

What does it actually mean for a term of the sequence to be in the green strip? It means that its distance from the limit (in this case from the number five) is at most epsilon. And that is exactly what the definition says: |a_n − a|<ϵ. Remember that the value a stands for the limit, so in our case a = 5. Let's try working out the value of the fifteenth term:

$$a_{15}=\frac{5\cdot 15}{15+1}=4.6875$$

Now we substitute into the relation |a_n − a|<ϵ:

$$\begin{array}{rcl}|a_n-a|& <& ϵ\\ |4.6875-5|& <& 1\\ 0.3125& <& 1\end{array}$$

We can see that the inequality holds. This point satisfies the definition. The number 0.3125 is the distance between the fifteenth term of the sequence and the limit of the sequence. In the picture this distance is highlighted by a short purple line segment. If we tested all the numbers greater than n₀ in this way, we would find that the inequality still holds. So the number a = 5 satisfies the definition of the limit of this sequence.

There is one important thing in the definition, though. It says that for every ϵ>0 we are able to find that number n₀. That is, for an arbitrarily large epsilon, but above all for an arbitrarily small epsilon. However small (but still positive) we make epsilon, we must still be able to find n₀.

So let's try making epsilon smaller. What happens? The green strip that we have to fit into gets narrower. That doesn't bother us at all – we simply increase the value of n₀, moving it to the right. The shrinking and the shift are illustrated in the following picture:

This shift nicely illustrates the behaviour of a limit. We expect the terms of the sequence to get gradually closer to the limit, and their distance from the limit to keep decreasing. (The distance from the limit is what we worked out a moment ago, the purple line in the picture.) If the distance keeps decreasing and the terms approach the limit, at some point they must get closer to the limit than epsilon. And that is how we have found our value n₀. If, on the other hand, we choose a larger epsilon, we can choose a smaller n₀:

At the beginning I mentioned that a sequence keeps approaching its limit but never touches it. However, a sequence can touch its limit, and can even take the value of its limit infinitely many times. For example, the constant sequence a_n = 1 has the value one at every point, and its limit is also one.

Showing that a limit does not exist

Let's change the sequence so that it fits the following graph (the formula itself is not important for now):

The sequence behaves in the same way as in the previous case, except for the tenth, twentieth, thirtieth (and so on) terms of the sequence (let's assume that these terms are always equal to the number 3.5 – they neither rise nor fall). These terms are somehow out of rhythm. I have drawn two neighbourhoods in the picture, epsilon one and epsilon two. As we can see, for ϵ₂ (the larger epsilon neighbourhood) we can find an n₀ such that all later terms are in the green strip. But when we reduce the value of epsilon to ϵ₁, we can no longer find such an n₀ – every tenth term will be outside this neighbourhood. This sequence would not have a limit at the point five (and it has no limit anywhere else either).

Examples of other limits

An example of a sequence that has no limit at all is a_n = sin n, shown in the following picture:

We can see that the sequence does not converge to any number: its terms keep falling and rising. We call such a sequence divergent.

So far we have shown convergent sequences that converged to the given limit from one side. But a sequence can also approach a limit from both sides, as this sequence shows

$$a_n=(-1{)}^n\cdot \frac{1}{n}$$

This sequence also converges to zero.

Infinite limit

So far we have talked about a finite limit, that is, the case where a sequence approaches some real number. But what about a sequence such as a_n = n? Which real number does the sequence approach? Take a look at the graph:

It clearly doesn't approach any real number, but grows without bound, grows to infinity. Such a sequence has the limit infinity, and we say that such a limit is infinite. The definition of an infinite limit is simple. What does it mean for a sequence to grow to infinity? It means that if we choose any real number A, however large, we can always find an index n₀ such that all the terms of the sequence after this term are greater than the number A. In short, if we choose a bound A, then from some point in the sequence onwards all the terms are greater than this A. Written mathematically:

$$\underset{n\to \mathrm{\infty }}{lim}{a}_n=\mathrm{\infty }\iff \mathrm{\forall }A\in \mathbb{R}\mathrm{\exists }{n}_0\in {\mathbb{Z}}^{+}\mathrm{\forall }n\in {\mathbb{Z}}^{+}:n>n_0\Rightarrow a_n>A$$

At the beginning we have “for all A”, that is, for all the bounds we might want to choose. Then we have the existential quantifier, which says that there must exist an index n₀ such that if n is greater than this n₀, then the value of the term a_n is greater than the bound we chose at the start, that is, A. There is a similar definition for the infinite limit minus infinity.

Methods for working out limits

We already know a few basic methods for working out the limit of a sequence. Variations of the sequence a_n = n, such as

$$a_n=2n,a_n=\frac{n}{2}$$

have the limit plus infinity. On the other hand, sequences such as

$$a_n=\frac{1}{n},a_n=\frac{15}{4n}$$

have a limit of zero. Now we'll state some basic rules for working with sequences. Suppose that

$$\underset{n\to \mathrm{\infty }}{lim}{a}_n=a,\underset{n\to \mathrm{\infty }}{lim}{b}_n=b,a,b\in \mathbb{R}.$$

Then the following rules also hold:

$$\underset{n\to \mathrm{\infty }}{lim}(c_1{a}_n+c_2{b}_n)=c_{1}a+c_{2}b\mathrm{\forall }{c}_1,c_2\in \mathbb{R}$$

$$\underset{n\to \mathrm{\infty }}{lim}{a}_n{b}_n=a\cdot b;\underset{n\to \mathrm{\infty }}{lim}\frac{a_n}{b_n}=\frac{a}{b};b\ne 0.$$

We could use the first rule in the following example:

$$\underset{n\to \mathrm{\infty }}{lim}(\frac{1}{n}+5)=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}+\underset{n\to \mathrm{\infty }}{lim}5=0+5=5$$

And what about when we have an example like this?

$$\underset{n\to \mathrm{\infty }}{lim}\frac{3n+1}{2n}$$

Here there is a problem, because the expression in the numerator tends to infinity at infinity, and the expression in the denominator also tends to infinity. But if we have infinity in both the numerator and the denominator, we get what is called an indeterminate form. In short, we can't tell what this expression is equal to:

$$\frac{\mathrm{\infty }}{\mathrm{\infty }}$$

Our task will be to rearrange the expression so that this indeterminate form doesn't arise. If we manage to remove the variable n from the denominator or the numerator, we may get an expression with only one infinity in it.

First we rearrange the expression by dividing every term by the variable n, that is, by multiplying the top and bottom of the fraction by 1/n.

$$\underset{n\to \mathrm{\infty }}{lim}\frac{3n+1}{2n}=\underset{n\to \mathrm{\infty }}{lim}(\frac{3n+1}{2n}\cdot \frac{\frac{1}{n}}{\frac{1}{n}})=\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{3n}{n}+\frac{1}{n}}{\frac{2n}{n}}=\underset{n\to \mathrm{\infty }}{lim}\frac{3+\frac{1}{n}}{2}$$

We can see that the denominator no longer contains the variable n, so the expression in the denominator no longer tends to infinity at infinity. Using the previous rules we can split the whole limit into several smaller ones by working out the limits of all the expressions separately, then adding the results and dividing.

$$\underset{n\to \mathrm{\infty }}{lim}\frac{3+\frac{1}{n}}{2}=\frac{\underset{n\to \mathrm{\infty }}{lim}3+\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}}{\underset{n\to \mathrm{\infty }}{lim}2}$$

We have now reached a situation where we can work out all the limits. Let's do so.

$$\frac{\underset{n\to \mathrm{\infty }}{lim}3+\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}}{\underset{n\to \mathrm{\infty }}{lim}2}=\frac{3+0}{2}=\frac{3}{2}$$

The limit of this sequence is the number 1.5.

Let's try another example:

$$\underset{n\to \mathrm{\infty }}{lim}\frac{3n^2+1}{2n}$$

Here we can either use the same method as before or try some logical reasoning. The function n² will pull the sequence up far faster than the function n pulls it down. The function n² is far stronger, so the whole expression will have the limit infinity. Let's work it out as well. Again we divide every term by n:

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim}\frac{3n^2+1}{2n}& =& \underset{n\to \mathrm{\infty }}{lim}\frac{\frac{3n^2}{n}+\frac{1}{n}}{\frac{2n}{n}}\\ & =& \underset{n\to \mathrm{\infty }}{lim}\frac{3n+\frac{1}{n}}{2}\\ & =& \frac{\underset{n\to \mathrm{\infty }}{lim}3n+\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}}{\underset{n\to \mathrm{\infty }}{lim}2}\\ & =& \frac{\mathrm{\infty }+0}{2}=\mathrm{\infty }\end{array}$$

A few more words on why we divided by n and not, for example, by n², which also appears in the expression. Our aim, as we know, is to get rid of a variable. However, we know that the value of the denominator must not be zero, as we would again get an indeterminate form. We multiply the top and bottom of the fraction by an expression chosen so that no zero appears in the denominator. In this case that meant using the expression with the highest power in the denominator.

So we can say that the limit is governed by the function that grows faster by orders of magnitude. But watch out for signs: even when working with infinity we must keep the sign:

$$\underset{n\to \mathrm{\infty }}{lim}\frac{3n^2+1}{-2n}=\dots =\frac{\mathrm{\infty }+0}{-2}=-\mathrm{\infty }$$