2.6 Divergence: the boundedness and subsequence tests

So far we have almost exclusively studied convergent sequences: working through explicit examples, and using abstract $\varepsilon$ - $N$ arguments to develop a general theory of limits. However, not all sequences are convergent! It is therefore useful to develop some basic tests to help us to understand whether or not convergence holds. To motivate these tests, we now consider two prototypical examples of divergent sequences.

Example 2.47 (Informal).

The terms of the sequence $(n)_{n\in\mathbb{N}}$ get larger and larger, without approaching any fixed value: see Figure 2.1. Therefore, we expect this sequence to diverge.

Example 2.48 (Informal).

The terms of the sequence $((-1)^{n})_{n\in\mathbb{N}}$ alternate between two distinct values: see Figure 2.1. In particular, the sequence does not approach one fixed value and therefore we expect this sequence to diverge.

The above examples are informal, because they just express our intuition about the behaviour of the sequences. We would like to give a proof that the sequences in Example 2.47 and Example 2.48 do indeed diverge, and one way to do this is to directly apply the formal definition. For this, we need to negate the $\varepsilon$ - $N$ definition of convergence (essentially by swapping the quantifiers). This gives:

(2.6) (2.6) A sequence

(a_{n})_{n\in\mathbb{N}}

diverges if for all

a\in\mathbb{R}

there exists

\varepsilon>0

such that for all

N\in\mathbb{N}

there exists some

n\in\mathbb{N}

with

n>N

such that

|a_{n}-a|\geq\varepsilon

Quite a mouthful! Indeed, it turns out that it is rather clunky to work directly with (2.6) (although it is still possible to do so: see Exercise 2.67 below). Instead, we focus on establishing indirect tests which can be used to show a sequence diverges.

The boundedness test

Our first test deals with sequences which have a similar behaviour to $(n)_{n\in\mathbb{N}}$ from Example 2.47.

Lemma 2.49 (Boundedness test).

A convergent sequence is bounded.

By working with the contrapositive, the boundedness test can be used to show various sequences diverge.

Warning 2.50.

It is not possible to use the boundedness test to show sequences converge, since the logical implication only runs in one direction. The converse of Lemma 2.49 (“A bounded sequence is convergent”) is false. For instance, we shall see that the bounded sequence $((-1)^{n})_{n\in\mathbb{N}}$ diverges.

Example 2.51.

Since the sequence $(n)_{n\in\mathbb{N}}$ is unbounded, it must diverge by the (contrapositive of the) boundedness test.

Proof (of Lemma 2.49).

Suppose $(a_{n})_{n\in\mathbb{N}}$ is a convergent, with $a=\lim_{n\to\infty}a_{n}$ . By the $\varepsilon$ - $N$ definition of a limit, taking $\varepsilon:=1$ , there exists some $N\in\mathbb{N}$ such that $|a_{n}-a|<1$ for all $n\in\mathbb{N}$ with $n>N$ . Thus, by the triangle inequality, $|a_{n}|<|a|+1$ for all $n>N$ . If we define

M:=\max\{|a_{1}|,|a_{2}|,\dots,|a_{N}|,|a|+1\},

then if follows that $-M\leq a_{n}\leq M$ for all $n\in\mathbb{N}$ , as required. ∎

Exercise 2.52.

Illustrate the key ideas of the proof of Lemma 2.49 using a sketch. You should incorporate features of both Figure 2.5(a) (which illustrates the boundedness concept) and Figure 2.7 (which illustrates the limit concept).

Tending to infinity

The following definition identifies a special kind of unbounded sequence.

Definition 2.53.

Let $(a_{n})_{n\in\mathbb{N}}$ be a sequence.

1

We write $a_{n}\to\infty$ as $n\to\infty$ if for all $M>0$ there exists some $N\in\mathbb{N}$ such that $a_{n}>M$ for all $n>N$ .
2

We write $a_{n}\to-\infty$ as $n\to\infty$ if for all $M<0$ there exists some $N\in\mathbb{N}$ such that $a_{n}<M$ for all $n>N$ .

Warning 2.54.

As in Warning 2.24, when we write $a_{n}\to\infty$ , here ‘ $\infty$ ’ is merely part of the notation. It is just a symbol and does not represent a number or any other mathematical object.

Intuitively, “ $a_{n}\to\infty$ as $n\to\infty$ ” means that for any given threshold $M$ , eventually all the terms of the sequence will lie beyond that threshold. It is clear that if $a_{n}\to\infty$ as $n\to\infty$ , then $(a_{n})_{n\in\mathbb{N}}$ is unbounded and so divergent. Hence, sequences satisfying $a_{n}\to\infty$ as $n\to\infty$ are a special kind of divergent sequence.

Example 2.55.

The sequence $(n^{2})_{n\in\mathbb{N}}$ satisfies $n^{2}\to\infty$ as $n\to\infty$ . We can intuitively see this from a sketch (see Figure 2.5), where eventually all the terms lie above the line $y=M$ . It is not difficult to make this idea precise, using similar arguments to those used in the $\varepsilon$ - $N$ proofs above (but now $M$ plays the role of $\varepsilon$ ).

Proof.

Let $M>0$ be given and choose $N:=\lceil M\rceil$ . If $n>N$ , then $n^{2}\geq n>N\geq M$ . Thus, by definition, $n^{2}\to\infty$ as $n\to\infty$ . ∎

Exercise 2.56.

Arguing from the definition (Definition 2.53), show that the following sequences $(a_{n})_{n\in\mathbb{N}}$ satisfy $a_{n}\to\infty$ as $n\to\infty$ .

(i)

$\displaystyle\Big{(}\frac{n^{3}}{n+1}\Big{)}_{n\in\mathbb{N}}$ ;
(ii)

$\displaystyle\Big{(}\frac{2^{n}}{n}\Big{)}_{n\in\mathbb{N}}$ ;
(iii)

$\displaystyle\Big{(}\frac{2^{2n}}{3^{n}+n^{2}}\Big{)}_{n\in\mathbb{N}}$ .

Hint: given $0<r<1$ , from Exercise 2.31 and Workshop 3, we know that $r^{n}\to 0$ and $nr^{n}\to 0$ and $n^{2}r^{n}\to 0$ as $n\to\infty$ .

It is important to realise that not all unbounded sequences satisfy Definition 2.53.

Exercise 2.57.

Find an example of a sequence $(a_{n})_{n\in\mathbb{N}}$ which is unbounded but does not satisfy either $a_{n}\to\infty$ as $n\to\infty$ or $a_{n}\to-\infty$ as $n\to\infty$ .

The subsequence test

Our next task is to develop a test which deals with sequences which have a similar behaviour to $((-1)^{n})_{n\in\mathbb{N}}$ from Example 2.48. This will require some groundwork.

If we only consider the even terms of $((-1)^{n})_{n\in\mathbb{N}}$ , we arrive at the constant sequence $(1)_{k\in\mathbb{N}}=(1,1,1,\dots)$ which converges to $1$ by Exercise 2.29. On the other hand, if we only consider the odd terms, we arrive at the constant sequence $(-1)_{k\in\mathbb{N}}=(-1,-1,-1,\dots)$ which again converges, but this time to $-1$ . As we shall see, the divergence of $((-1)^{n})_{n\in\mathbb{N}}$ can be explained using the fact that there are two subsequences with distinct limits.

To make the above discussion more precise, we need to define what we mean in general by a subsequence of a sequence.

Definition 2.58.

Suppose $(a_{n})_{n\in\mathbb{N}}$ is a sequence of real numbers. A subsequence of $(a_{n})_{n\in\mathbb{N}}$ is a sequence of real numbers of the form $(a_{n_{k}})_{k\in\mathbb{N}}$ where $(n_{k})_{k\in\mathbb{N}}$ is an increasing sequence of natural numbers:

\text{$n_{k}\in\mathbb{N}$ for all $k\in\mathbb{N}$}\qquad\text{and}\qquad n_{% 1}<n_{2}<n_{3}<\cdots.

Thus, $(a_{n_{k}})_{k\in\mathbb{N}}$ is just a selection of some (possibly all) of the $a_{n}$ taken in order.

Example 2.59.

Let $(a_{n})_{n\in\mathbb{N}}$ be a sequence. If we take $n_{k}:=2k$ , then $(a_{n_{k}})_{k\in\mathbb{N}}=(a_{2k})_{k\in\mathbb{N}}$ is the subsequence consisting of all even terms. For the specific sequence $a_{n}:=(-1)^{n}$ , as discussed in the motivating example above, $(a_{2k})_{k\in\mathbb{N}}=(1)_{k\in\mathbb{N}}$ is the constant sequence where each term is $1$ .

Example 2.60.

Every possible sequence of $-1$ and $1$ ’s can be realised as a subsequences of $((-1)^{n})_{n\in\mathbb{N}}$ . For instance, the sequence $(1,-1,-1,1,1,1,-1,-1,-1-1,\dots)$ can be realised as a subsequence of $((-1)^{n})_{n\in\mathbb{N}}$ by taking

(n_{1},n_{2},n_{3},\dots)=(2,3,5,6,8,10,11,13,15,17,\dots).

Lemma 2.61 (Subsequence test).

If $(a_{n})_{n\in\mathbb{N}}$ is sequence which converges to a limit $a\in\mathbb{R}$ and $(a_{n_{k}})_{k\in\mathbb{N}}$ is a subsequence, then $(a_{n_{k}})_{k\in\mathbb{N}}$ also converges to $a$ .

Proof.

The proof is essentially a matter of unravelling the definitions. We leave it as an exercise: see Exercise 2.62 below. ∎

Exercise 2.62.

Prove Lemma 2.61.

Similar to the boundedness test from Lemma 2.49, the subsequence test can be used to show various sequences diverge. If we can find two subsequences which converge to different limits, then the subsequence test tells us that the parent sequence cannot converge. We formulate this observation as a corollary of Lemma 2.61.

Corollary 2.63.

Suppose $(a_{n})_{n\in\mathbb{N}}$ is a sequence and $(a_{n_{k}})_{k\in\mathbb{N}}$ , $(a_{m_{k}})_{k\in\mathbb{N}}$ are subsequences such that $a_{n_{k}}\to b$ and $a_{m_{k}}\to c$ as $k\to\infty$ where $b$ , $c\in\mathbb{R}$ with $b\neq c$ . Then $(a_{n})_{n\in\mathbb{N}}$ is divergent.

Proof.

We argue by contradiction. Suppose $(a_{n})_{n\in\mathbb{N}}$ satisfies the hypotheses of the corollary and converges to some limit $a\in\mathbb{R}$ . Since $(a_{n_{k}})_{k\in\mathbb{N}}$ and $(a_{m_{k}})_{k\in\mathbb{N}}$ are both subsequences of $(a_{n})_{n\in\mathbb{N}}$ , by the subsequence test, we must have $a_{n_{k}}\to a$ as $k\to\infty$ and $a_{m_{k}}\to a$ as $k\to\infty$ . However, by uniqueness of limits from Lemma 2.37, we must have $a=b$ and $a=c$ . Thus, $b=c$ , which contradicts the hypothesis $b\neq c$ . ∎

Example 2.64.

We can use the subsequence test to give a proof that the sequence $((-1)^{n})_{n\in\mathbb{N}}$ diverges. Indeed, as noted above, the subsequence of even terms is the constant sequence $(1)_{n\in\mathbb{N}}$ which converges to $1$ and the subsequence of odd terms is the constant sequence $(-1)_{n\in\mathbb{N}}$ which converges to $-1$ . Thus, we have found two subsequences of $((-1)^{n})_{n\in\mathbb{N}}$ which converge to different limits and therefore the subsequence test tells us that $((-1)^{n})_{n\in\mathbb{N}}$ cannot converge.

Exercise 2.65.

Show that the sequence $\big{(}\sin^{2}(\frac{\pi n}{50}\big{)}\big{)}_{n\in\mathbb{N}}$ diverges.

Exercise 2.66.

As in Exercise 2.5, for $n\in\mathbb{N}$ let $q_{n}\in\mathbb{N}_{0}$ and $r_{n}\in\{0,1,2,3,4\}$ be the unique values such that $n=5q_{n}+r_{n}$ . Show that $(q_{n})_{n\in\mathbb{N}}$ and $(r_{n})_{n\in\mathbb{N}}$ both diverge. What about $\big{(}\frac{r_{n}}{q_{n}+1}\big{)}_{n\in\mathbb{N}}$ ?

Exercise 2.67.

Arguing directly from the $\varepsilon$ - $N$ definition of divergence from (2.6), show that the following sequences diverge:

(i)

$(n)_{n\in\mathbb{N}}$ ;
(ii)

$((-1)^{n})_{n\in\mathbb{N}}$ .