Limits of a sequence

The limit of a sequence is the number that the sequence approaches at infinity. The limit of a sequence is then a kind of precursor to the limit of a function, which is of great importance in mathematical analysis.

A brief introduction to limits

The limit of a sequence is the number that a given sequence of numbers continually approaches (or eventually reaches). The best picture to use is the one that shows the sequence given by the rule $a_n=\frac1n$:

In the figure we see that the members of the sequence are getting closer and closer to zero. The larger n, the smaller the value of a_n and the closer it gets to zero. The graph never touches the horizontal axis because the equation 1/n = 0 (where n is a natural number) never holds. However, it is worth finding out what number the whole sequence approaches. So the question might be - if we keep increasing n, what number will the value of the sequence approach? If we approach n to infinity, what will be the value of the sequence member? This is exactly what limits solve. We write it down using the command lim. In the subscript we then have which number n approaches, for sequences it will always be infinity. The whole notation might look like this, for example:

$$\lim_{n\rightarrow\infty}\frac{1}{n}=0.$$

We read, "The limit of the sequence one slash n for n approaching infinity is zero." Some other trivial examples: what is the limit of the sequence a_n = n? This is a simple sequence where a₁ = 1, a₄₂ = 42, a₅₈₇₆₆ = 58766, etc. Which number will the members of the sequence approach at infinity? The larger n, the larger a_n, so the limit will be infinity.

$$\lim_{n\rightarrow\infty}n=\infty$$

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

$$\lim_{n\rightarrow\infty}n^2=\infty,\quad\lim_{n\rightarrow\infty}\frac{1}{2n}=0$$

Definition of the intrinsic limit of the sequence

It's time to define the limit of the sequence properly.

A sequence $\left\{a_n\in\mathbb{R}\right\}_{n=1}^\infty$ has an eigenlimit a, (converges to the limit a ∈ ℝ, is convergent) if for every ε > 0 there exists n₀∈ℕ such that for all n>n₀, |a_n − a| < ε holds.

The notations used are:

$$\lim_{n\rightarrow\infty}a_n=a$$

$$a_n\rightarrow a\mbox{ For } n\rightarrow\infty$$

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

That's a nice definition, now let's explain it. To do this, we will need a picture of a 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, so the limit of this sequence is five. Before I start demonstrating the definition of the limit on this sequence, I'll add some color to the graph:

We already know that the limit of this sequence is five. The first part of the definition talks about ε>0, some epsilon that is greater than zero. The epsilon here represents the neighborhood of the limit, the neighborhood of the point. It is greater than zero because the neighborhood cannot be negative, nor zero. In the figure, this neighborhood is represented by the vertical green lines on the axis y. The point here is the value five, and the neighborhood of the point is the green lines on the axis y. As such, the neighborhood is defined in both directions, up and down. We will now need this imaginary strip defined by the green horizontal lines. In the figure, ε = 1.

Our goal is to find a number on the bottom horizontal axis n₀, which will divide the bottom axis in two. It must be the case that all members of the sequence to the right of the number n₀ are in this green strip. Where the numbers to the left of n₀ are located is irrelevant, we do not need to find the smallest n₀ for which this definition holds. In the figure, n₀ = 12 applies.

What does it really mean that a member of the sequence is in the green band? That it is at most an epsilon away from the limit (in this case, the number five). And that's exactly what the definition says: |a_n − a|<ε. Recall that the value a denotes the limit, so in our case it does: a = 5. Let's try to calculate the value of the fifteenth element:

$$a_{15}=\frac{5\cdot15}{15+1}=4{,}6875$$

Now we plug in the relation |a_n − a|<ε:

$$\begin{eqnarray} \left|a_n-a\right|&<&\epsilon\\ \left|4{,}6875-5\right|&<&1\\ 0{,}3125&<&1 \end{eqnarray}$$

We see that the inequality holds. This point satisfies the definition. The number 0.3125 expresses the distance between the fifteenth member of the sequence and the limit of the sequence. In the figure, this distance is highlighted by the short purple line. If we were to test all numbers greater than n₀ in this way, we would find that this inequality still holds. Thus, the number a = 5 satisfies the definition of the limit of this sequence.

But there is one important thing in the definition. It says that for every ε>0 we are able to find that number n₀. So for an arbitrarily large epsilon, but especially for an arbitrarily small epsilon. No matter how small we reduce the epsilon to a small (but still positive) number, we must still be able to find n₀.

So let's try to reduce the epsilon to some smaller value. What happens? The green band we have to fit into gets smaller. Which is fine, we'll just increase the value of n₀ - move it to the right. Both the shrinking and the shifting are illustrated in the following figure:

This displacement nicely illustrates the behavior of the limits. We expect the limit to be approached by the members of the sequence, so that their distance from the limit is constantly decreasing. (The distance from the limit is what we calculated a moment ago, the purple line in the figure.) If the distance is continually decreasing and the members are getting closer to the limit, at some point they must get to a state where they are closer to the limit than the epsilon. And thus we have found our value of n₀. Conversely, if we choose epsilon larger, we can choose n₀ smaller:

I mentioned at the beginning that the sequence is constantly approaching its limit, but never touching it. However, a sequence can touch its limit, a sequence can even acquire its limit infinitely many times. For example, the constant sequence a_n = 1 will have a value of one at each point and will also have a limit of one.

Demonstrating the absence of a limit

Let's try to modify the definition of a sequence to fit the following graph: (the rule itself is not relevant for the moment)

The sequence behaves the same as in the previous case except for the tenth, twentieth, thirtieth, etc. member of the sequence (assume that these members are always equal to 3.5 - they do not grow or shrink in any way). These members are somehow out of rhythm. I have drawn two neighborhoods in the figure, epsilon one and two. As we can see, for ε₂ (the larger epsilon neighborhood) we are able to find such n₀, that all the other terms are in the green band. But if we reduce the epsilon value to ε₁, we can no longer find such a n₀ - every tenth member will be outside this neighborhood. This sequence would have no limit at point five (and has no limit anywhere else).

Examples of other limits

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

We can see that the sequence does not converge to any number; the members of the sequence are constantly decreasing and increasing. Such a sequence is called divergent.

For now, we have shown convergent sequences that converge to a given limit from one side. But a sequence can converge to the limit from both sides, as this sequence shows

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

This sequence will also converge to zero.

Non-eigenlimit

So far we have discussed the proper limit, the case where the sequence converges to some real number. However, what about such a sequence a_n = n? What real number does the sequence approach? Take a look at the graph:

It obviously does not approach any real number, but it grows beyond all limits, growing to infinity. Thus such a sequence has a limit at infinity, and we say that such a limit is not proper. The definition of a non-proper limit is simple. What does it mean that a sequence grows to infinity? That if we choose some real number A, arbitrarily large, we always find an index n₀ such that all members of the sequence that are after this member are greater than this number A. In short, if we choose a limit A, then starting from some part of the sequence, all members are greater than this A. Written mathematically:

$$\lim_{n\rightarrow\infty}a_n=\infty\Leftrightarrow\forall A\in\mathbb{R}\exists n_0\in\mathbb{N}\forall n\in\mathbb{N}: n>n_0\Rightarrow a_n>A$$

At the beginning we have for all A, that is, for all the limits we would like to choose. Then we have an existence quantifier which says that there must be an index n₀, for which it holds that if n is greater than this n₀, then the value of the member a_n is greater than the limit we chose at the beginning, i.e. A. A similar definition will hold for the non-proper limit at minus infinity.

Methods for calculating the limit

We already know some basic methods to compute the limit of a sequence. Various permutations of the sequence a_n = n such as

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

have a limit at plus infinity. In contrast, sequences of the type

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

have a limit of zero. We now review some basic relations for counting with sequences. Suppose that the following holds

$$\lim_{n\rightarrow\infty} a_n=a,\quad\lim_{n\rightarrow\infty}b_n=b,\quad a,b\in\mathbb{R}.$$

Then the following relations also hold:

$$\lim_{n\rightarrow\infty}(c_1a_n+c_2b_n)=c_1a+c_2b\quad\forall c_1,c_2\in\mathbb{R}$$

$$\lim_{n\rightarrow\infty}a_nb_n=a\cdot b;\quad \lim_{n\rightarrow\infty}\frac{a_n}{b_n}=\frac{a}{b}; b\ne 0.$$

We could apply the first formula in the following example:

$$\lim_{n\rightarrow\infty}(\frac{1}{n}+5)=\lim_{n\rightarrow\infty}\frac{1}{n}+\lim_{n\rightarrow\infty}5=0+5=5$$

What if we have such an example?

$$\lim_{n\rightarrow\infty}\frac{3n+1}{2n}$$

Here a problem arises because the expression in the numerator approaches infinity, the expression in the denominator also approaches infinity. But if we have infinity in both the numerator and denominator, we get a so-called indefinite expression. In short, we cannot determine what the expression is equal to:

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

Our task will be to modify the expression so that this indeterminate expression does not arise. If we can remove the variable n from the denominator or the numerator, we can already get an expression with only one infinity.

We first modify the expression by dividing each term by the variable n, or expanding the fraction by the expression 1/n.

$$\lim_{n\rightarrow\infty}\frac{3n+1}{2n}=\lim_{n\rightarrow\infty}(\frac{3n+1}{2n}\cdot\frac{\frac{1}{n}}{\frac{1}{n}})=\lim_{n\rightarrow\infty}\frac{\frac{3n}{n}+\frac{1}{n}}{\frac{2n}{n}}=\lim_{n\rightarrow\infty}\frac{3+\frac{1}{n}}{2}$$

We can see that there is no longer a variable n in the denominator, so the expression in the denominator is not close to infinity. Following the previous formulas, we can decompose the whole limit into several smaller ones by counting the limits of all the expressions separately and adding and dividing the results.

$$\lim_{n\rightarrow\infty}\frac{3+\frac{1}{n}}{2}=\frac{\lim_{n\rightarrow\infty}3+\lim_{n\rightarrow\infty}\frac{1}{n}}{\lim_{n\rightarrow\infty}2}$$

This puts us in a situation where we can already calculate all the limits. Let's do this.

$$\frac{\lim_{n\rightarrow\infty}3+\lim_{n\rightarrow\infty}\frac{1}{n}}{\lim_{n\rightarrow\infty}2}=\frac{3+0}{2}=\frac{3}{2}$$

The limit of this sequence is the number 1.5.

Let's try another example:

$$\lim_{n\rightarrow\infty}\frac{3n^2+1}{2n}$$

Here we can either use the same procedure as last time or we can try to involve logical reasoning. The function n² will pull the sequence up much faster than the function n will pull it down. The n² function is much stronger and so the whole expression will have a limit at infinity. Let's try and do the math. Again, divide each expression by n:

$$\begin{eqnarray} \lim_{n\rightarrow\infty}\frac{3n^2+1}{2n}&=&\lim_{n\rightarrow\infty}\frac{\frac{3n^2}{n}+\frac{1}{n}}{\frac{2n}{n}}\\ &=&\lim_{n\rightarrow\infty}\frac{3n+\frac{1}{n}}{2}\\ &=&\frac{\lim_{n\rightarrow\infty}3n+\lim_{n\rightarrow\infty}\frac{1}{n}}{\lim_{n\rightarrow\infty}2}\\ &=&\frac{\infty+0}{2}=\infty \end{eqnarray}$$

A few more words on why we divided n and not, for example, n², which also occurs in the expression. Our goal, as we know, is to get rid of a variable. However, we know that the value of the denominator must not be zero, or we would again get an indeterminate expression. We will expand the fraction with an expression such that the denominator does not contain zero. In this case, that meant expanding it with the expression with the highest exponent in the denominator.

So we can say that the limit will follow a function that grows an order of magnitude faster. But be careful of the signs, even when counting with infinity we must keep the sign:

$$\lim_{n\rightarrow\infty}\frac{3n^2+1}{-2n}=\ldots=\frac{\infty+0}{-2}=-\infty$$