Monotonic and Bounded Sequences

A monotonic sequence is simply one that only ever moves in one direction: always up, or always down. The name comes from "monotone," meaning a single tone, and the sequence likewise keeps a single direction and never switches. A bounded sequence is one that stays between a floor and a ceiling. Put the two together and you get a way to prove a limit exists even when you can't compute it: something that only goes up and can't get past a ceiling has to level off somewhere. The statement $a_n \le a_{n+1} \le M$ looks dense, but it's just two facts side by side: $a_n \le a_{n+1}$ (each number is at least as big as the one before) and $\le M$ (none of them ever gets past the ceiling $M$). It's a lot of symbols for a very physical idea, so bear with it.

In the last lesson we found limits by computing them. But some sequences, especially recursive ones like $a_{n+1} = \sqrt{6 + a_n}$, don't have a formula you can just take the limit of. So what can we do? Sometimes you can show a limit has to exist from the shape of the sequence alone, and then go find it. First, some words.

Vocabulary.

• Increasing if $a_n < a_{n+1}$ for every $n$; decreasing if $a_n > a_{n+1}$. Either one is monotonic.
• Bounded above if some $M$ has $a_n \le M$ for all $n$; bounded below if some $m$ has $a_n \ge m$. Both together: bounded.

Checking monotonicity. Three ways, use whichever is easiest:

1. The sign of $a_{n+1} - a_n$.
2. For positive terms, whether $\dfrac{a_{n+1}}{a_n}$ is above or below 1.
3. The sign of $f'(x)$, where $f(n) = a_n$.

The theorem. A bounded monotonic sequence converges. Think about why that should be true. An increasing sequence keeps moving right on the number line, never backing up. If there's a wall at $M$ it can't cross, the terms can't run off to infinity, and they can't bounce around either, because they never go backward. The only thing left is to bunch up against some number at or below $M$. That number is the limit.

Notice what the theorem does and doesn't give you. It tells you a limit exists without telling you what it is. That turns out to be exactly what you need for recursive sequences.

Recursive sequences. Once you know the limit $L$ exists, let $n \to \infty$ on both sides of the recursion. Both $a_n$ and $a_{n+1}$ approach $L$, since they're the same list shifted by one, which gives an equation for $L$.

A common mistake. It's tempting to think a sequence has to be monotonic to converge, since that's what the theorem talks about. It doesn't. The theorem is sufficient, not necessary: $\frac{(-1)^n}{n}$ jumps back and forth across 0 and still converges to it. A second trap: skipping the existence step. If you plug $L$ into the recursion and solve without first knowing a limit exists, you can get an answer for a sequence that actually diverges.

The Babylonian method above is still how computers find square roots (it's Newton's method for $x^2 = 2$). Monotone bounded sequences are how you prove an iterative method settles down before you know the answer.