Monotone and Cauchy Sequences

a_n = 1 - 2^{-n}

increases and is bounded above, so it converges to its supremum

1

(monotone convergence). Watch the tail diameter shrink to

0

: every convergent sequence is Cauchy.

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:

a_8 = 1 - 2^{-8} = 0.99609375

. The problem's

1 - 1/n

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 $2^{-N} < \varepsilon$ ( $N > \log_2 \frac{1}{\varepsilon}$ ). Then $m, n \geq N$ implies $\lvert s_m - s_n\rvert < 2^{-N} < \varepsilon$ , so $(s_N)$ 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.