SUITES — L'Atelier

The progression

Add the same step, or multiply by the same ratio — the two royal families.

An arithmetic sequence climbs stairs: $u_{n+1}=u_n+r$, always the same step. A geometric one takes percentages: $u_{n+1}=q\,u_n$, always the same ratio. Everything you will ever compute about them — the $n$-th term, the sum — falls out of that one difference.

The family

u₀ where it starts1

r / q step or ratio0.7

u₁₀—

behaviour—

each dot is one term — the walk, not the road

The cobweb

uₙ₊₁ = f(uₙ): bounce between the curve and the mirror, and watch where you land.

Feed the output back into the machine: $u_{n+1}=f(u_n)$. On the picture that is a staircase — up to the curve, across to the diagonal $y=x$, again and again. Babylon computed $\sqrt2$ this way four thousand years ago; the same spider’s web still does it in four steps.

The machine

u₀3

steps8

r the chaos dial2.8

last term—

last move |Δ|—

verdict—

the target—

camellia staircase: the sequence · dashed: the mirror y = x

The chaos diallogistic with $r<3$: one destination. Past $3$: two, four, eight… past $3.57$: none at all. One slider, the edge of chaos.

The limit

Convergence has a contract: pick any ε, I name the day all terms move in.

To say $u_n\to L$ is to sign this contract: for every $\varepsilon>0$, past some rank $N$ every single term lives inside the band $L\pm\varepsilon$. Tighten the band and $N$ retreats — but it always exists. That is the whole definition, drawn.

The contract

ε the band0.3

the limit L—

N(ε) — the move-in day—

gold band: L ± ε · from N on, no term ever leaves

The series

Add infinitely many numbers and, sometimes, get a finite answer.

Stack the partial sums $S_n=1+q+q^2+\dots+q^n$. If $|q|<1$ the stack stops growing: it leans on the wall $\dfrac{1}{1-q}$ and stays there. Half a cake, then a quarter, then an eighth… you eat forever and never pass one whole cake.

The stack

q the ratio0.5

S₃₀—

the wall 1/(1−q)—

verdict—

distance left—

dots: the partial sums · line: the wall they lean on

Infinitely many ≠ infinitean endless list of positive numbers can have a finite sum — provided they die fast enough. $|q|\ge 1$ and the wall is gone.

Fibonacci

Add the last two — and a number nobody invited walks in.

$1,\,1,\,2,\,3,\,5,\,8,\,13\dots$ each term the sum of the previous two. Divide neighbours, $F_{n+1}/F_n$, and the ratios stop caring about the start: they settle on $\varphi=\frac{1+\sqrt5}{2}\approx 1.618$, the golden number — the most irrational number there is, and the reason sunflowers count in Fibonacci.

The rabbits

n how far down the list10

Fₙ—

ratio Fₙ₊₁/Fₙ—

ϕ1.6180339…

distance to ϕ—

the ratios, hunting ϕ from both sides — one above, one below, forever

Épreuves the page checks by itself—

Six commissions, one walk at a time — and an endless examen. The page counts every step.

Pitfalls

The six traps sequences set for the hurried.

Arithmetic ≠ geometric+3 each step and ×3 each step are different universes. Check which one the recurrence actually says.

Convergent ≠ monotone$(-1)^n/n$ zigzags forever and still converges. The band wins, not the direction.

Bounded ≠ convergent$(-1)^n$ stays politely in $[-1,1]$ and never settles anywhere.

Slow is still sure$q=0.99$ converges — after seven hundred terms. Patience is not divergence.

Induction needs a basethe domino argument moves nothing until the first domino actually falls. Verify $n=0$.

The limit obeys the recurrenceif $u_n\to L$ and $u_{n+1}=f(u_n)$, then $L=f(L)$ — find candidates by solving the fixed point, then prove the walk gets there.