Sequences — convergence, divergence, boundedness, monotonicity: Visual walkthrough

This walkthrough leans on the completeness axiom and closes the loop the parent opened. It is the visual companion to the parent topic.

Step 1 — What is a sequence, as dots on a number line?

WHAT. A sequence is an infinite ordered list $a_{1}, a_{2}, a_{3}, \dots$ — one real number for each counting number $n = 1, 2, 3, \dots$ . Instead of drawing it as a graph, we drop each value as a dot on a horizontal number line. The label on the dot tells you its position in the list; the place on the line tells you its value.

WHY this picture. Convergence is a statement about where the values pile up, not about how the graph rises. Dots-on-a-line makes "piling up near one point" literally visible.

PICTURE. Below, $a_{1}$ is far left, and each later dot sits a little to its right. The label $n$ above each dot is just its index — "I am the $n$ -th term."

Figure — Sequences — convergence, divergence, boundedness, monotonicity

Step 2 — What does "monotone increasing" look like?

WHAT. Increasing means each new term is at least as far right as the one before it: $a_{n + 1} \geq a_{n} for every n .$

$a_{n + 1}$ — the next term (the dot to come).
$a_{n}$ — the current term.
$\geq$ — "sits to the right of, or on top of" — the dots never move left.

WHY it matters. If the dots may only march rightward, the only thing that can stop them is a wall on the right. That single restriction is what will trap the limit.

PICTURE. Every arrow points right. There is no backtracking — the hallmark of monotone increasing.

Recall Why not just handle both directions at once?

Increasing and decreasing are mirror images. We prove increasing carefully; flipping the number line (replace $a_{n}$ by $- a_{n}$ ) gives the decreasing case free. We show that mirror in Step 8.

Step 3 — What does "bounded above" put into the picture?

WHAT. Bounded above means there is a number $M$ that no term ever passes: $a_{n} \leq M for every n .$

$M$ — an upper bound, a wall on the right.
$a_{n} \leq M$ — every dot stays on or left of the wall, forever.

WHY. Step 2 gave rightward-only motion; Step 3 gives a wall. A dot that must always move right but can never pass a wall has nowhere to go but to bunch up — that intuition becomes the whole proof.

PICTURE. The red wall at $M$ is a barrier. Notice $M$ is some wall, not necessarily the closest one — there are infinitely many walls (any bigger number is also a wall).

Step 4 — The tightest wall: the supremum

WHAT. Among all the walls $M$ that work, there is a least (leftmost) one. Call it $L$ : $L = sup {a_{1}, a_{2}, a_{3}, \dots} .$

$sup$ ("supremum") — the least upper bound: the leftmost wall that still no term crosses.

WHY this exists — and why we need an axiom. Why should a smallest wall exist at all? On the rationals it can fail: the terms $1, 1.4, 1.41, 1.414, \dots$ climb toward $2$ , and every rational wall can be nudged tighter with no rational wall being the smallest. The completeness of $\mathbb{R}$ is exactly the promise that on the real line the least wall always exists. This is the deep fact MCT rests on — no completeness, no theorem.

WHY $sup$ and not, say, the average or the max? Because we are chasing the point the dots approach. A monotone-increasing pile can only approach its ceiling; the ceiling is its least upper bound. Max might not exist (dots may never reach it), but $sup$ always does.

PICTURE. Push the wall left as far as it will go before some dot pokes through — that limiting position is $L$ .

Step 5 — Turn "least wall" into a captured term

WHAT. Fix any tolerance $ε > 0$ — think of it as a tiny gap. Because $L - ε$ is not a wall (Step 4, property 2), some term must have slipped past it: $\exists N : a_{N} > L - ε .$

$ε$ — the challenger's tolerance, a positive gap however small.
$L - ε$ — a point just left of the ceiling.
$a_{N}$ — the first witness term that lies to the right of $L - ε$ .

WHY. This is the engine. "Least upper bound" is not decoration — it guarantees a witness. If no term passed $L - ε$ , then $L - ε$ would itself be a wall smaller than $L$ , contradicting that $L$ is the least wall.

PICTURE. The green band $(L - ε, L]$ is the target zone. The dot $a_{N}$ is the first that lands inside it.

Step 6 — Monotonicity drags the entire tail into the band

WHAT. From Step 5, $a_{N} > L - ε$ . Now use Step 2: the sequence never moves left, so every term after $a_{N}$ is at least $a_{N}$ : $n > N ⟹ a_{n} \geq a_{N} > L - ε .$ And from Step 4 (property 1), no term passes the ceiling: $a_{n} \leq L < L + ε$ . Combine: $n > N ⟹ L - ε < a_{n} \leq L < L + ε .$

Left inequality: monotonicity (dots can't retreat past the witness).
Right inequality: the wall $L$ (dots can't pass the ceiling).

Together every tail dot is squeezed into $(L - ε, L + ε)$ .

WHY this is the finish. Squeezing the whole tail inside a band of half-width $ε$ around $L$ is word-for-word the $ε$ – $N$ definition of convergence from the parent note.

PICTURE. Once one dot enters the green band, monotonicity forces every later dot to stay in it — trapped between the witness on the left and the ceiling on the right.

Step 7 — Degenerate & edge cases (never leave a gap)

WHAT & WHY. A careful walkthrough must survive every weird input.

Constant sequence $a_{n} = c$ . It is increasing ( $a_{n + 1} \geq a_{n}$ with equality) and bounded. $sup = c$ , and every term already sits at $c$ . Converges to $c$ . ✓
Eventually monotone. Only the tail decides convergence. If the sequence wobbles for the first million terms and then climbs monotonically under a wall, chop off the finite head — the argument runs on the tail unchanged. ✓
The witness $a_{N}$ equals $L$ . Allowed: if some term actually reaches the ceiling, monotonicity + wall pin all later terms to exactly $L$ . Still inside every band. ✓
Terms hug $L$ but never touch it (e.g. $a_{n} = 1 - \frac{1}{n}$ , $sup = 1$ , never reached). The proof never needed the sup to be attained — only approached. ✓ This is why we chose $sup$ , not $max$ .

PICTURE. Three rows: a flat constant line, a wobble-then-climb, and dots crowding a ceiling they never reach. All three land.

Recall What if the sequence is

not bounded above? Then no wall exists, no $sup$ , and an increasing sequence marches to $+ \infty$ — it diverges to infinity. MCT does not apply, and correctly so: remove the wall and the trap vanishes.

Step 8 — The decreasing mirror (case closed)

WHAT. For a sequence that only moves left (decreasing) and is bounded below by a floor $m$ , reflect the whole line: study $b_{n} = - a_{n}$ . Then $b_{n}$ is increasing and bounded above by $- m$ , so Step 6 gives $b_{n} \to sup b_{n}$ . Flipping back: $a_{n} \to - sup {- a_{n}} = in f {a_{n}} .$

$in f$ ("infimum") — the greatest lower bound, the mirror of $sup$ ; the highest floor no term drops below.

WHY. No new work: reflection turns "least wall on the right" into "greatest floor on the left." Both halves of MCT now hold.

PICTURE. Top: decreasing dots dropping toward a floor at $in f$ . Bottom: the reflected increasing copy rising to $- in f$ . Same proof, viewed in a mirror.

The one-picture summary

Everything at once: rightward-only dots (monotone), a red ceiling (bound), the least wall $L = sup$ , a green $ε$ -band, the witness $a_{N}$ that pierces $L - ε$ , and the whole tail crushed inside the band.

Recall Feynman retelling — the whole proof in plain words

Imagine beads sliding on a wire, each new bead placed to the right of the last, but a wall stops any bead passing a certain point. The beads keep going right yet can never break the wall — so they must crowd up somewhere. Where? Slide the wall left as far as it goes before a bead would poke through; that limiting spot is $L$ . Now play a game: a challenger names a tiny gap $ε$ and paints a green strip of that width just left of the wall. Because $L$ is the tightest wall, backing off by $ε$ is no longer a wall — so at least one bead, call it the $N$ -th, has already entered the strip. And since beads only move right and never break the wall, every bead after the $N$ -th is stuck between that entering bead and the wall — inside the strip. No matter how thin the challenger makes the strip, the whole tail ends up inside it. That "whole tail inside every strip" is exactly what "converges to $L$ " means. The one fact we borrowed from outside: that a tightest wall exists at all — that is the completeness of the real numbers.

Recall check

Why must the least upper bound

L

exist?

The completeness axiom guarantees every nonempty set bounded above has a supremum in

R

In Step 5, why does some term exceed

L - ε

Because

L

is the least wall, so

L - ε

is not a wall — some term must have passed it.

What does monotonicity contribute in Step 6?

It drags every term after the witness

a_{N}

to the right of

a_{N}

, so the whole tail stays above

L - ε

Why

sup

and not

max

The limit may be approached but never attained (e.g.

1 - \frac{1}{n} \to 1

);

sup

always exists,

max

may not.

How is the decreasing case handled?

Reflect via

b_{n} = - a_{n}

; it becomes increasing bounded above, then flip back to get convergence to

in f

If an increasing sequence is unbounded above, what happens?

No wall, no

sup

— it diverges to

+ \infty