Proving Monotonicity And Boundedness Together

Why Prove Two Properties Together?

A recursive sequence is defined by an initial value and a rule for producing each next term. To prove that such a sequence converges using the monotone convergence theorem, it is often necessary to establish two facts: the terms move in one direction, and they remain within a fixed bound.

These facts can support one another. A bound may identify an interval on which the recurrence behaves predictably, while comparing consecutive terms can help show that the sequence stays in that interval. Rather than treating monotonicity and boundedness as unrelated tasks, we can put both claims into one induction hypothesis.

Definitions and the Monotone Convergence Theorem

A sequence \((x_n)\) is nondecreasing if \(x_n\le x_{n+1}\) for every \(n\in\mathbb{N}\). It is nonincreasing if \(x_{n+1}\le x_n\) for every \(n\in\mathbb{N}\). A sequence is bounded if there are real numbers \(L\) and \(U\) such that \(L\le x_n\le U\) for every \(n\).

For a recurrence \(x_{n+1}=f(x_n)\), an interval \(I\) is invariant if \(f(x)\in I\) whenever \(x\in I\). The function \(f\) is order-preserving on \(I\) if \(u,v\in I\) and \(u\le v\) imply \(f(u)\le f(v)\).

An invariant interval supplies both an upper and a lower bound. Thus, once induction establishes monotonicity as well as membership in that interval, the theorem applies immediately.

The Joint Induction Argument

Suppose \(x_{n+1}=f(x_n)\), the initial value \(x_1\) belongs to an invariant interval \(I\), and \(f\) is order-preserving on \(I\). The crucial induction property includes two pieces of information: the current term lies in \(I\), and the pair of consecutive terms has the desired order.

Here \(x_2=f(x_1)\), so the initial comparison can be checked directly. To prove the nondecreasing case, use the induction property \(P(n)\): \(x_n\in I\), \(x_{n+1}\in I\), and \(x_n\le x_{n+1}\).

By induction, \(P(n)\) holds for every \(n\). Consequently, all terms lie in \(I\), and \(x_n\le x_{n+1}\) for every \(n\). The nonincreasing case uses the same property with \(x_{n+1}\le x_n\): order preservation gives \(x_{n+2}=f(x_{n+1})\le f(x_n)=x_{n+1}\). This is the complete reason one induction can prove both claims: the interval membership makes the recurrence applicable within the chosen interval, and the order comparison propagates through \(f\).

Worked Examples

A Common Pitfall: Proving Only One Half

It is not enough to show that a recurrence preserves an interval and then assume the sequence is monotone. Nor does observing several consecutive terms in increasing order establish the inequality for every index. For example, in the nonlinear example, the computations \(0<1/2<5/8<89/128\) check only the first few terms.

The induction step must explicitly prove both parts. In the nondecreasing case, it needs to show that \(x_{n+2}\) stays in the interval and that \(x_{n+1}\le x_{n+2}\). Leaving out either inequality leaves a gap: without interval membership, the order-preserving property may not be available; without the comparison, monotonicity has not been proved.

Why This Method Matters

For a recursive sequence, each new term depends on the preceding one. An induction hypothesis that tracks both a bound and a comparison matches that dependence directly. It can also prevent circular reasoning: rather than presuming that the sequence remains in the region where the recurrence behaves well, the induction proves that it does.

Once the two properties are established, convergence follows from a standard theorem rather than from numerical evidence. The method does not require calculating a limit or solving an equation for one; it establishes the hypotheses needed to know that a limit exists.