Monotone and Cauchy Sequences

An increasing sequence blocked from above climbs to the very top. The scene also watches the tail squeeze into a single point.

Bounded above, the sequence has nowhere to go but its highest allowed spot. That spot is the supremum, and the sequence converges to it. Cut to the tail and the remaining terms spread less and less. A sequence whose spread dies to 0 is a Cauchy sequence. Every convergent sequence is Cauchy.

For example, drop coins into a 🐷 piggy bank: half the goal, then half of what remains, and so on. The total creeps toward the goal without ever jumping to it. The coins added later also differ from each other less and less.

Examples & Exercises

Show $a_n = 1 - \frac{1}{n}$ is increasing with $\sup = 1$ , and conclude $a_n \to 1$ from the Monotone Convergence Theorem.

a_{n+1} - a_n = \frac{1}{n(n+1)} > 0, \qquad \sup a_n = 1

n = 8

terms: for this problem's sequence,

a_8 = 1 - \frac{1}{8} = 0.875

. The scene displays

1 - 2^{-n}

(

0.99609375

n = 8

), which approaches

1

the same way.

Show proof

(Monotone) $a_{n+1} - a_n = \left(1 - \frac{1}{n+1}\right) - \left(1 - \frac{1}{n}\right) = \frac{1}{n} - \frac{1}{n+1} = \frac{1}{n(n+1)} > 0$ , so the sequence increases.
(Bounded) $a_n = 1 - \frac{1}{n} < 1$ for every $n$ , so $1$ is an upper bound. For any real $u < 1$ , $1 - \frac{1}{n} > u \iff n > \frac{1}{1-u}$ , and such an $n$ exists by the Archimedean property, so $u$ is not an upper bound.
Hence $\sup a_n = 1$ .
By the Monotone Convergence Theorem (cited), an increasing bounded-above sequence converges to its supremum, so $a_n \to 1$ .
Numerically: at $n = 40$ , $a_{40} = 1 - \frac{1}{40} = 0.975$ , already within $0.025$ of the limit $1$ .

Prove that every Cauchy sequence is bounded.

m \geq N \Rightarrow \lvert a_m\rvert \leq \lvert a_N\rvert + 1

Tail start

6

— past

N

the tail is trapped in a band of radius

1

Show proof

Let $(a_n)$ be Cauchy. For $\varepsilon = 1$ there is $N$ with $\lvert a_m - a_n\rvert < 1$ whenever $m, n \geq N$ .
Fix this $N$ .
For $m \geq N$ , the triangle inequality gives $\lvert a_m\rvert \leq \lvert a_N\rvert + \lvert a_m - a_N\rvert < \lvert a_N\rvert + 1$ — the whole tail is bound by one number.
The head $a_1, \dots, a_{N-1}$ has finitely many terms, so with $M = \max(\lvert a_1\rvert, \dots, \lvert a_{N-1}\rvert, \lvert a_N\rvert + 1)$ we have $\lvert a_n\rvert \leq M$ for every $n$ . Hence $(a_n)$ is bounded.
Numerically: for this scene's $a_n = 1 - 2^{-n}$ , $N = 1$ works for $\varepsilon = 1$ (since $\lvert a_m - a_1\rvert < \frac12 < 1$ ), and $M = \lvert a_1\rvert + 1 = 1.5$ bounds everything.

Prove that the partial sums $s_N = \sum_{k=1}^{N} 2^{-k}$ form a Cauchy sequence.

s_N = 1 - 2^{-N}, \qquad \lvert s_m - s_n\rvert = 2^{-n} - 2^{-m} < 2^{-n}

The partial sums are exactly this scene's sequence:

\sum_{k=1}^{n} 2^{-k} = 1 - 2^{-n}

Show proof

Geometric sum: $(1 - \frac12)\sum_{k=1}^{N} 2^{-k} = \sum_{k=1}^{N}(2^{-k} - 2^{-(k+1)}) = \frac12 - 2^{-(N+1)}$ (telescoping), so $s_N = 1 - 2^{-N}$ .
For $m > n$ , the same telescoping gives $s_m - s_n = \sum_{k=n+1}^{m} 2^{-k} = 2^{-n} - 2^{-m}$ , so $\lvert s_m - s_n\rvert = 2^{-n} - 2^{-m} < 2^{-n}$ .
Given $\varepsilon > 0$ , choose $N$ with $2^{-N} < \varepsilon$ ( $N > \log_2 \frac{1}{\varepsilon}$ ). Let $m, n \geq N$ .
If $m = n$ then $\lvert s_m - s_n\rvert = 0 < \varepsilon$ . Otherwise, by symmetry $\lvert s_m - s_n\rvert = \lvert s_n - s_m\rvert$ , so assume without loss of generality $m > n$ ; step 2 then gives $\lvert s_m - s_n\rvert < 2^{-n} \le 2^{-N} < \varepsilon$ . Hence $(s_N)$ is Cauchy.
Numerically: with tail start $4$ and indices $\leq 12$ , the tail diameter is $2^{-4} - 2^{-12} = 0.0625 - 0.000244140625 = 0.062255859375$ — matching the scene's readout.