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Section 5.4 The Monotone Convergence Theorem

What if we have a sequence \((a_n)\) that we think converges to some limit, but we don't have a candidate for what that limit is? This isn't such an uncommon situation: we will want to define interesting numbers like \(e\) and \(\pi\) as limits of certain sequences, so we can't very well use these numbers in the course of proving that their defining sequences converge!

This is where good recognition principles for convergent sequences come in. The first of these, the Monotone Convergence Theorem, deals with a very special class of sequences.

Definition 5.4.1.

A sequence \((a_{n})_{n=1}^{\infty}\) is increasing if \(a_{n+1}\geq a_{n}\) for all \(n\text{.}\) It is decreasing if \(a_{n+1}\leq a_{n}\) for all \(n\text{.}\) We say \((a_{n})\) is monotonic if it is either increasing or decreasing.

We begin with the first part. Assume that \((a_n)\) is increasing and bounded above. Notice that we are not told what its limit is going to be. This has to come out of our argument somehow, and the first job is to find a candidate for it.

Since \((a_{n})\) is bounded above, this means that the set 1  \(\{a_{n} : n\in\NN\}\) is bounded above. Thus, by the Completeness Axiom, the set has an supremum, call it \(L\text{.}\) We claim that

\begin{equation*} \lim_{n\rightarrow\infty}a_{n}=L\text{.} \end{equation*}
Note the distinction between the sequence \((a_n)\text{,}\) in which the order of members is important, and the set \(\{ a_n\}\text{,}\) in which it isn't important or even recorded.

We begin the proof of the claim. Let \(\epsilon>0\text{,}\) we wish to show there is \(N\in\NN\) so that

\begin{equation*} |a_{n}-L|\lt \epsilon \end{equation*}

for all \(n> N\text{,}\) or equivalently,

\begin{equation} L-\epsilon\lt a_{n}\lt L+\epsilon\text{.}\label{e_L-e_a_L_e}\tag{5.4.1} \end{equation}

Since \(L\) is an upper bound for \(\{a_{n} : n\in\NN\}\text{,}\) we know \(a_{n}\leq L\lt L+\epsilon\) for all \(n\text{,}\) thus dealing with the second inequality in (5.4.1). Therefore, it suffices to show that there is \(N\in\NN\) so that for all \(n> N\text{,}\)

\begin{equation*} L-\epsilon\lt a_{n}\text{.} \end{equation*}

Since \(L\) is the supremum for \(\{a_{n} : n\in\NN\}\text{,}\) this means that \(L-\epsilon\) is not an upper bound for this set, so there exists an integer \(N\) so that \(a_{N}>L-\epsilon\text{.}\) Since the sequence is increasing, this means that for all \(n> N\text{,}\)

\begin{equation*} a_{n} \geq a_N > L - \epsilon\text{.} \end{equation*}

This shows (5.4.1) holds for all \(n>N\) and completes the proof.

We leave the proof of the second part to the reader. (Why is it an easy consequence of the first part?)

Thus, if you can show that a sequence is bounded above and increasing, you can deduce that it has a limit, and in some cases this, together with the rules for limits that we discussed last week, allows you to figure out what the limit actually is.

Let us look at some examples.

Let \(a_n= \frac{n-1}{n}\text{.}\) Then \((a_n)\) is increasing since \(\frac{n}{n+1} \geq \frac{n-1}{n} \iff n^2 \geq (n+1)(n-1) = n^2 -1\text{,}\) which is true for all \(n\text{.}\) \((a_n)\) is also bounded above by \(1\text{.}\) The MCT tells us that \((a_n)\) converges to some \(L \leq 1\text{.}\) (In fact we don't need the MCT for this example because \(a_n = 1 - 1/n \to 1 - 0 = 1\) by the rules for limits, but it nevertheless illustrates the general point.)

The next example can also be done by bare hands, but we use it to introduce a general technique which will be very useful to us: finding a simple equation which any potential limit must satisfy.

If \(0\lt a\lt 1\text{,}\) show that the sequence \((a^n)_{n=1}^\infty\) converges and \(\lim_{n\rightarrow\infty} a^{n}=0\text{.}\)

Solution.

Since \(a^n\) is a product of positive numbers, it is positive, so the sequence \((a^{n})_{n=1}^{\infty}\) is bounded below by \(0\text{.}\) Moreover, since \(a\lt 1\text{,}\) \(a^{n+1}=a\cdot a^{n}\lt a^{n}\text{,}\) so the sequence is decreasing. By the MCT, it has a limit \(L\text{.}\) To find \(L\text{,}\) we will find a simple equation which \(L\) must satisfy — in this case \(aL = L\text{.}\) Recall that if \((a_{n})\) is a convergent sequence, then \((a_{n+1})\) also converges, and indeed converges to the same limit (see the exercises from last week). Thus, by the rules for limits,

\begin{equation*} L=\lim_n a^{n} = \lim_n a^{n+1}=\lim_n a^{n} =a\lim_n a^{n}=aL\text{,} \end{equation*}

as we claimed. Therefore \(L(a-1) =0\) and since \(a \neq 1\) we must have \(L =0\text{,}\) and we are done.

In this example, we used the rules for limits, together with the fact that if \((x_n)\) converges, then \((x_{n+1})\) also converges to the same limit. In the next example we instead use the fact that if \(x_n \to L\text{,}\) then \(x_{2n} \to L\) too. (Why?)

Let \(a_n = 2^{1/n}\text{.}\) Show that \((a_n)\) converges and find its limit.

Solution.

We notice that \(2^{1/n} \geq 1\) for all \(n\) (since this is equivalent to \(2 \geq 1^n = 1\)) and also that \(2^{1/(n+1)} \leq 2^{1/n}\) for all \(n\) (since this is equivalent to \(2^n \leq 2^{n+1}\)). Therefore \((a_n)\) is a decreasing sequence which is bounded below by \(1\text{.}\) Consequently, by the MCT, it converges to some number \(L\) which satisfies \(L\geq1\) (and \(L\lt 2\)). Consider the sequence \(b_n = 2^{1/2n}\text{.}\) It runs through alternate values of \(a_n\text{,}\) and so \(b_n \to L\) also as \(n \to \infty\text{.}\) But \(b_n^2 = a_n\text{,}\) so by the rules for limits we must have \(L^2 = L\text{,}\) or \(L(L-1) =0\text{,}\) meaning either \(L=0\) or \(L=1\text{.}\) Since \(1 \leq L \lt 2\) the only possibility is \(L = 1\text{.}\)