4.3 Monotone Sequences

So far, in all the examples of sequences we have seen so far we have had to know what the limit of the sequence is in order to prove the existence of it, but sometimes this is tricky to know beforehand if we are dealing with a novel sequence. We will now introduce some tools and theorems that will enable us to prove the existence of a limit for a sequence without knowing what the limit is.

Definition 4.3.

Let $(x_{n})_{n}$ be a sequence of real numbers.

•

$(x_{n})_{n}$ is increasing if $x_{1}\leq x_{2}\leq x_{3}\leq\dots$
•

$(x_{n})_{n}$ is decreasing if $x_{1}\geq x_{2}\geq x_{3}\geq\dots$
•

$(x_{n})_{n}$ is monotone if it is either increasing or decreasing.

Example 4.8.

Here are some examples of sequences with their characterisation by this definition.

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$(a_{n})_{n}=(n)_{n}$ is increasing.
•

$(b_{n})_{n}=\left(\frac{1}{n}\right)_{n}$ is decreasing.
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$(c_{n})_{n}=(p,p,p,\dots),\ p\in\mathbb{R}$ is both increasing and decreasing.
•

$(d_{n})_{n}=(-2n)_{n}$ is decreasing.
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$(e_{n})_{n}=((-1)^{n})_{n}$ is not monotone.
•

$(f_{n})_{n}=\left(\frac{(-1)^{n}}{n}\right)_{n}$ is not monotone.

4.3.1 The Monotone Convergence Theorem

We can make an important observation from these examples. Of the bounded sequences in the previous example ( $(b_{n})_{n}$ , $(c_{n})_{n}$ , $(e_{n})_{n}$ , $(f_{n})_{n}$ ), the ones which are also monotone (just $(b_{n})_{n}$ and $(c_{n})_{n}$ ) are all convergent, whereas none of the unbounded sequences are convergent (which we would expect by theorem 4.3). Although a sequence does not have to be monotone to converge (look at the last example above), it is a general result that if a monotone sequence is bounded then it does converge, which we will now prove.

Theorem 4.5 (Monotone Convergence Theorem (MCT))

Let $(x_{n})_{n}$ be a monotone sequence. Then $(x_{n})_{n}$ is convergent if and only if $(x_{n})_{n}$ is bounded.

Proof.

(Forward direction $\implies$ ): This follows from theorem 4.3.
(Reverse direction $\impliedby$ ): Suppose $(x_{n})_{n}$ is bounded and without meaningful loss of generality assume $(x_{n})_{n}$ is increasing. Recall that since $\mathbb{R}$ is complete, every bounded set of real numbers has a supremum that is also a real number, i.e. if we let

X\coloneqq\{x_{n}:n\in\mathbb{N}\}

(4.32)

then $X$ is a bounded subset of the real numbers and we can let $x=\sup X\in\mathbb{R}$ . Now we want to show that $(x_{n})_{n}$ converges to $x$ . Let $\varepsilon>0$ and note that $x-\varepsilon$ is not an upper bound since $x$ is the supremum. Therefore, there exists $N\in\mathbb{N}$ such that $x_{N}\geq x-\varepsilon$ . Now since $(x_{n})_{n}$ is increasing, for $n\geq N$ we have

\lvert x_{n}-x\rvert=x-x_{n}\leq x-x_{N}\leq\varepsilon,

(4.33)

and hence $x_{n}\to x$ as $n\to\infty$ . ∎

This theorem makes intuitive sense, if we have a sequence which is always increasing or decreasing and is bounded then it must reach a limit before it surpasses the bound, and in fact this limit is the least upper/greatest lower bound of the set of elements of the sequence.

Corollary 4.5.1

Suppose $(x_{n})_{n}$ is a bounded sequence, then

(i)

If $(x_{n})_{n}$ is increasing, $\lim_{n\to\infty}x_{n}=\sup_{n}x_{n}$ .
(ii)

If $(x_{n})_{n}$ is decreasing, $\lim_{n\to\infty}x_{n}=\inf_{n}x_{n}$ .

Proof.

The proof follows directly from the proof of MCT. ∎

Example 4.9.

Let $(x_{n})_{n}$ be a sequence given by

x_{n}=\frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots+\frac{1}{n^{2}}=% \sum_{k=1}^{n}\frac{1}{k^{2}}.

(4.34)

Show that this sequence is convergent.
First notice that $(x_{n})_{n}$ is increasing,

x_{n}=\frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots+\frac{1}{n^{2}}<% \frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots+\frac{1}{n^{2}}+\frac{1% }{(n+1)^{2}}=x_{n+1},

(4.35)

and also the elements of $(x_{n})_{n}$ are bounded between 0 and 2.

$\displaystyle 0$	$\displaystyle<\frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+% \cdots+\frac{1}{n^{2}}$	(4.36)
	$\displaystyle\leq 1+\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+% \cdots+\frac{1}{(n-1)n}$	(4.37)
	$\displaystyle=1+\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1% }{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots+\left(\frac{1}{n-1}-% \frac{1}{n}\right)$	(4.38)
	$\displaystyle=1+1-\frac{1}{n}<2.$	(4.39)

Hence by MCT, this sequence must converge. In fact, it can be shown that this sequence converges to $\frac{\pi^{2}}{6}$ , but the proof of that is a bit more involved.