Monotone Convergence

Definition 3.5   A sequence {a_n} is an increasing sequence if a_n+1$\ge$a_n for every n.

• If we need precision, we can distinguish between a strictly increasing sequence, and a (not necessarily strictly) increasing sequence. This is sometime called a non-decreasing sequence.
• There is a similar definition of a decreasing sequence.
• What does it mean to say that a sequence is eventually increasing?
• A sequence which is either always increasing or always decreasing is called a monotone sequence. Note that an ``arbitrary'' sequence is not monotone (it will usually sometimes increase, and sometimes decrease).

Nevertheless, monotone sequences do happen in real life. For example, the sequence

a₁ = 3        a₂ = 3.1        a₃ = 3.14        a₄ = 3.141        a₅ = 3.1415...

is how we often describe the decimal expansion of $\pi$. Monotone sequences are important because we can say something useful about them which is not true of more general sequences.

Example 3.6   Show that the sequence a_n = $${\frac{{n}}{{2n+1}}}$$ is increasing.

Solution. One way to check that a sequence is increasing is to show a_n+1 - a_n$\ge$ 0, a second way is to compute a_n+1/a_n, and we will see more ways later. Here,

$${\frac{{n+1}}{{2(n+1) + 1}}} {\frac{{n}}{{2n+1}}}$$ a_n+1 - a_n =

$${\frac{{(2n^2+3n+1) - (2n^2+3n)}}{{(2n+3)(2n+1)}}}$$ =

$${\frac{{1}}{{(2n+3)(2n+1)}}}$$ =

and the sequence is increasing.

Exercise 3.7   Show that the sequence a_n = $${\frac{{1}}{{n}}}$$ - $${\frac{{1}}{{n^2}}}$$ is decreasing when n > 1.

If a sequence is increasing, it is an interesting question whether or not it is bounded above. If an upper bound does exist, then it seems as though the sequence can't help converging -- there is nowhere else for it to go.

Figure 3.1: A monotone (increasing) sequence which is bounded above seems to converge because it has nowhere else to go!

In contrast, if there is no upper bound, the situation is clear.

Example 3.8   An increasing sequence which is not bounded above tends to $\infty$ (see definition 2.30).

Solution. Let the sequence be {a_n}, and assume it is not bounded above. Pick K; we show eventually a_n > K. Since K is not an upper bound for the sequence, there is some witness to this. In other words, there is some a_N with a_N > K. But in that case, since the sequence is increasing monotonely, we have a_n$\ge$a_N$\ge$K for all n$\ge$N. Hence a_n$\to$$\infty$ as n$\to$$\infty$.

Theorem 3.9 (The monotone convergence principle)   Let {a_n} be an increasing sequence which is bounded above; then {a_n} is a convergent sequence. Let {a_n} be a decreasing sequence which is bounded below; then {a_n} is a convergent sequence

Proof. To prove this we need to appeal to the completeness of $\mathbb {R}$, as described in Section 1.2. Details will be given in third year, or you can look in (Spivak, 1967) for an accurate deduction from the appropriate axioms for $\mathbb {R}$. $\qedsymbol$

This is a very important result. It is the first time we have seen a way of deducing the convergence of a sequence without first knowing what the limit is. And we saw in 2.12 that just knowing a limit exists is sometimes enough to be able to find its value. Note that the theorem only deduces an ``eventually'' property of the sequence; we can change a finite number of terms in a sequence without changing the value of the limit. This means that the result must still be true of we only know the sequence is eventually increasing and bounded above. What happens if we relax the requirement that the sequence is bounded above, to be just eventually bounded above? (Compare Proposition 2.27).

Example 3.10   Let a be fixed. Then aⁿ$\to$ 0 as n$\to$$\infty$ if | a| < 1, while if a > 1, aⁿ$\to$$\infty$ as n$\to$$\infty$.

Solution. Write a_n = aⁿ; then a_n+1 = a.a_n. If a > 1 then a_n+1 < a_n, while if 0 < a < 1 then a_n+1 < a_n; in either case the sequence is monotone.