Limits of sequences

We often encounter sequences in our day-to-day life without even noticing them. For example, if you wish to give your money to the bank for a return on interest, a monthly or yearly balance can be represented by a sequence. To study their long-term behavior, the concept of limit, which is the value that the sequence approaches as the number of terms goes to infinity is very useful. Let's get acquainted with this concept.

Intuitive introduction to limits

Recall that a sequence is an infinite list of terms. Usually, we talk about sequences of numbers (although that's not the only possible case). Let's look at several examples of number sequences.

Our first sequence is defined by the formula $a_n = \frac{1}{2^{n}}$, where n is a natural number, so it looks like

$$\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \ldots$$Suppose you want to compute the first 50 terms of this sequence. If you do this in your favorite programming language, setting precision to five decimal places, then, after the 17th term, you'll see only zeros. But you know your arithmetic, so you won't believe in obtaining 0 dividing non-zero numbers. The subsequent terms are just very small, so the precision isn't enough to distinguish them from zero. You increase the precision to six decimal places. Now only the terms up to 20th are large enough to be displayed as non-zero. With the precision of seven decimal places, zeros begin after the 24th term.

We say that this sequence tends to zero. No matter what precision you choose, starting at some position all the terms become numerically indistinguishable from zero.

Let's look at another sequence, $b_n = \frac{n - 1}{n}$:

$$0, \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \ldots$$This sequence tends to 1. The difference between 1 and $b_{n}$ is $\frac{1}{n}$, so it eventually becomes "negligible" no matter what precision you choose.

Does every sequence have some number (which we'll call its limit) that it converges to? No. For example, the sequence

$$c_n = (-1)^n$$consisting of alternating 1 and -1, has no limit. There is no single number such that all elements (starting at some position) of the sequence will eventually approach. Sequences having a limit are called convergent, while all others are called divergent.

The definition of a limit

Let's now define the limit in a formal way.

Definition: a number $A$ is the limit of the sequence $(a_{n})$ if and only if for any $\varepsilon > 0$ there exists $N$ such that for any $n > N$ the inequality $|a_{n} -A| < \varepsilon$ holds.

This definition, with all its notation, looks a bit intimidating, but let's see how it reflects the intuition we developed in the previous section. The elements $a_{n}$ with large indices (greater than some $N$) form an "$N$-tail" of the sequence. For the sequence to have a limit $A$, it should have some tail with elements "numerically indistinguishable from $A$" no matter what precision we choose. And what does this indistinguishability between $a_{n}$ and $A$ means for precision $\varepsilon$? It means that the difference $a_{n} -A$ is smaller than $\varepsilon$ in absolute value. For any precision, there exists a tail indistinguishable from the limit, just so.

Some more examples

You may think that if $A$ is the limit of the sequence $(a_{n})$, every next term should be closer to $A$ than the previous one. That's not a necessary condition. Let's look at the sequence

$$a_n = \begin{cases} \frac{1}{n} \text { if } n \text { is even}\\ \frac{1}{2n} \text { if } n \text { is odd} \end{cases}$$

Each odd term is smaller than both of its even neighbors (the only exception is $a_{1} =\frac{1}{2} = a_{2}$), but nevertheless, the sequence tends to zero. How can we prove that? Suppose we chose some "precision" $\varepsilon > 0$. We must find $N$ such that for any $n > N$

$$\begin{cases} \frac{1}{n} < \varepsilon \text { if } n \text { is even}\\ \frac{1}{2n} < \varepsilon \text { if } n \text { is odd} \end{cases}$$We can see it holds true if we take $N = \frac{1}{\varepsilon}$.

And now let's prove that the sequence

$$b_n = \begin{cases} \frac{1}{10} \text { if } n \text { is divisible by } 10\\ \frac{1}{n} \text { otherwise} \end{cases}$$doesn't have a limit. How can we do that? Let's show first that $\frac{1}{10}$ cannot be the limit. We should find some $\varepsilon > 0$ such that for any $N$ some term $b_{n}$ in the "$N$-tail" is far from $\frac{1}{10}$ (that is, $|b_{n} -\frac{1}{10}| > \varepsilon$). But it's obvious that for all n greater than 20 and not divisible by 10, $b_n < \frac{1}{20}$, so $|b_{n} -\frac{1}{10}| > \frac{1}{20}$. Okay, but can some other number $B \neq \frac{1}{10}$ be the limit? No: we just take $\varepsilon < |\frac{1}{10} - B|$, and see that any tail of the sequence has terms (equal to $\frac{1}{10}$) that are too far away from $B$.

The uniqueness of a limit

Can a sequence $(a_{n})$ have two different limits $A$ and $A'$? Let's prove that such a situation is impossible. Let $\varepsilon = |A - A'|/2$. Then there exist $N_{1}$ such that for $n > N_{1}$ $|a_{n} - A| < \varepsilon$ and there exist $N_{2}$ such that for $n > N_{2}$ $|a_{n} - A'| < \varepsilon$. So, for $n > \max(N_{1}, N_{2})$, the triangle inequality gives us

$$|A - A'| < |a_{n} - A| + |a_{n} - A'| < 2 \varepsilon = |A - A'|$$We obtained a contradiction, so, our assumption about two different limits was incorrect. So any sequence either has only one limit or no limit at all.

Conclusion

In this topic, we have discussed the notion of sequences as well as the formal definition of the limit of a sequence. Also, we went over a few examples to familiarize ourselves with the topic further. We have learned that sequences can only have either one limit or none at all.