Series — partial sums, convergence definition: Visual walkthrough

We answer one question, visually: when you add $a + a r + a r^{2} + \dots$ forever, where does the pile end up — and when does it refuse to settle?

Step 1 — What is a "running total"? Draw the dots.

WHAT. Start with a list of numbers we intend to add. Call the list $(a_{n})$ : the first number is $a_{1}$ , the second $a_{2}$ , and so on. The little $n$ underneath is just a counter — it says "which number in the list". So $a_{5}$ means "the fifth number".

WHY. We can never physically add infinitely many numbers — addition only ever combines two things at a time. So instead of adding forever, we add a finite chunk and watch what the chunk-total does. That finite total has a name:

PICTURE. Below, each number $a_{n}$ is a coloured bar. The dot above bar $N$ sits at height $s_{N}$ = the total of all bars up to and including bar $N$ . Watch the dots climb as we include more bars.

Convergence will be nothing more than: do those climbing dots level off at a fixed height?

Step 2 — Our specific list: the geometric one.

WHAT. We pick a particular list where each number is the previous one times a fixed factor $r$ : $a_{1} = a, a_{2} = a r, a_{3} = a r^{2}, \dots, a_{n} = a r^{n - 1} .$ Here $a$ is the starting number and $r$ is the common ratio — the fixed multiplier that takes you from one term to the next.

WHY. This is the one series we can fully conquer by hand, and everything else in the chapter compares back to it (see Geometric series, Ratio & Root Tests, Comparison Test). Get this and you own the model.

PICTURE. With $a = 1$ and $r = \frac{1}{2}$ , each bar is half the height of the one before — a shrinking staircase. The exponent $n - 1$ is annotated on each bar so you see why it starts at $r^{0} = 1$ .

Step 3 — The multiply-and-slide trick.

WHAT. Multiply the entire partial sum by $r$ . Every term gains one factor of $r$ , i.e. its exponent goes up by one: $r s_{N} = was a a r + was a r a r^{2} + \dots + new last term a r^{N} .$

WHY. Look at the two lists $s_{N}$ and $r s_{N}$ : they are almost identical, just shifted by one slot. When two lists share a long middle, subtracting them annihilates that middle. This is the same annihilation idea behind Telescoping series.

PICTURE. Top row = the bars of $s_{N}$ . Bottom row = the same bars of $r s_{N}$ , shifted right by one. The overlapping (shared) bars are greyed; only the two un-matched ends survive.

Step 4 — Subtract, and the middle vanishes.

WHAT. Line up $s_{N}$ over $r s_{N}$ and subtract: $s_{N} - r s_{N} = (a + a r + a r^{2} + \dots + a r^{N - 1}) - (a r + a r^{2} + \dots + a r^{N - 1} + a r^{N}) = a - a r^{N} .$ Every interior term appears once with a $+$ and once with a $-$ : it cancels. Only the first term of the top ( $a$ ) and the last term of the bottom ( $a r^{N}$ ) are left standing.

WHY. We turned a sum of $N$ pieces into a sum of just two pieces. That is the payoff — no " $\dots$ " remains, so we can actually compute.

PICTURE. The cancelled bars fade out; a pale-yellow $+ a$ and a chalk-pink $- a r^{N}$ are all that survive.

Now factor each side: $s_{N} (1 - r) = a (1 - r^{N}) .$ On the left, $s_{N} - r s_{N} = s_{N} (1 - r)$ — factor out the common $s_{N}$ . On the right, factor out $a$ .

Step 5 — Solve for the running total.

WHAT. Provided $1 - r \neq = 0$ (that is, $r \neq = 1$ ), divide both sides by $(1 - r)$ :

WHY. We now have a formula in $N$ with a single moving part, $r^{N}$ . Convergence (Step 1's "do the dots level off?") becomes the sharp question: does $r^{N}$ settle down as $N \to \infty$ ? (See Sequences — limits and convergence for what " $\to$ " means.)

PICTURE. The formula drawn as a target height $\frac{a}{1 - r}$ (dashed line) minus a shrinking correction $\frac{a r ^{N}}{1 - r}$ . The dots approach the dashed line as the correction melts away.

The symbol $lim_{N \to \infty}$ just means: the height the dots head toward as $N$ grows without bound. We now split into every possible $r$ .

Step 6 — Case $0 < r < 1$ : the pile climbs and settles.

WHAT. When $r$ is a positive fraction (between $0$ and $1$ ), every term $a r^{n - 1}$ is positive, so the dots only ever climb. And $r^{N}$ marches steadily to $0$ : $r^{N} \to 0 ⟹ s_{N} = \frac{a ( 1 - r ^{N} )}{1 - r} \to \frac{a ( 1 - 0 )}{1 - r} = \frac{a}{1 - r} .$

WHY. Each new bar is a shrunken positive copy of the last, so the leftover tail after $N$ terms shrinks to nothing from below — the correction $r^{N}$ dies and the dots press up against a definite ceiling.

PICTURE. For $a = 1, r = \frac{1}{2}$ the dots climb $1, 1.5, 1.75, 1.875, \dots$ and press against the ceiling $\frac{1}{1 - \frac{1}{2}} = 2$ from below. The pale-yellow error band — a strip of some small half-width we will call the tolerance $ε$ (a chosen small positive number: "how close is close enough?") — eventually traps every dot. That $ε$ -band is exactly the picture from the parent note's convergence definition.

Step 7 — Case $- 1 < r < 0$ : the pile still settles, but oscillates in.

WHAT. Now let $r$ be a negative fraction, e.g. $r = - \frac{1}{2}$ . Its size $∣ r ∣$ is still under $1$ , so the SAME limit holds — $r^{N} \to 0$ and $s_{N} \to \frac{a}{1 - r}$ . But the sign of $r^{N}$ now flips every step ( $r^{1} < 0, r^{2} > 0, r^{3} < 0, \dots$ ), so the correction $\frac{a r ^{N}}{1 - r}$ alternates sign.

WHY. Because the terms alternate positive/negative, the dots overshoot the ceiling, then undershoot, then overshoot by less — zig-zagging onto the limit rather than climbing from below. It still converges (the swings shrink to nothing), just from both sides.

PICTURE. For $a = 1, r = - \frac{1}{2}$ : dots go $1, 0.5, 0.75, 0.625, \dots$ bouncing above and below the ceiling $\frac{1}{1 - ( - \frac{1}{2} )} = \frac{2}{3}$ , with the bounces shrinking into the same $ε$ -band. This is the case buried inside " $∣ r ∣ < 1$ " that Step 6 did not show.

Step 8 — Case $∣ r ∣ > 1$ : the pile explodes.

WHAT. If $r$ 's size exceeds $1$ , multiplying by $r$ grows things, so $∣ r^{N} ∣ \to \infty$ . The correction term dominates and $s_{N}$ has no finite ceiling — it diverges.

WHY. Each bar is bigger than the last, so the running total accelerates upward (or swings wider and wider if $r < - 1$ ). There is nothing to level off toward.

PICTURE. With $r = 2$ : bars $1, 2, 4, 8, \dots$ and dots $1, 3, 7, 15, \dots$ shooting off the top — no ceiling exists. Contrast this directly with Steps 6–7.

Step 9 — The exact-boundary cases $r = 1$ and $r = - 1$ .

WHAT. These sit exactly on the edge $∣ r ∣ = 1$ , where our Step 5 division was banned ( $r = 1$ ) or where $r^{N}$ never settles ( $r = - 1$ ). Handle each by hand:

$r = 1$ : every term equals $a$ , so $s_{N} = a + a + \dots + a = a N$ . As $N$ grows this marches to $\pm \infty$ (sign of $a$ ). Diverges. (This is why Step 5 needed $r \neq = 1$ — dividing by $1 - r = 0$ is illegal.)
$r = - 1$ : terms are $a, - a, a, - a, \dots$ , so the running total flip-flops $a, 0, a, 0, \dots$ It stays bounded but never picks one value. No limit ⇒ diverges.

WHY. "Diverges" does not always mean "shoots to infinity" — the $r = - 1$ case is bounded yet still has no sum. A sequence must settle on one number to converge; bouncing forever fails that.

PICTURE. Left: $r = 1$ dots climb a straight ramp $a, 2 a, 3 a, \dots$ Right: $r = - 1$ dots ping-pong between two heights forever.

The one-picture summary

Everything on one number line for $r$ : a convergence window $(- 1, 1)$ (chalk-blue, the only zone with a finite sum $\frac{a}{1 - r}$ ) — climbing-in on the right half $0 < r < 1$ , oscillating-in on the left half $- 1 < r < 0$ — and outside it, divergence: explosion for $∣ r ∣ > 1$ , a straight ramp at $r = 1$ , a flip-flop at $r = - 1$ .

Recall Feynman retelling — the whole walkthrough in plain words

I want the total of $a$ , then $a$ times $r$ , then that times $r$ again, forever. I can't add forever, so I add just the first $N$ and call it $s_{N}$ — the height of my pile after $N$ steps. Clever trick: I make a copy of the pile and multiply the whole copy by $r$ , which slides every block over by one slot. When I subtract the copy from the original, all the middle blocks match up and cancel — only the very first block $a$ and a leftover block $a r^{N}$ survive. That gives me a clean formula $s_{N} = \frac{a ( 1 - r ^{N} )}{1 - r}$ with just one moving piece, $r^{N}$ . Now I ask what $r^{N}$ does. If $r$ is a small positive fraction, the dots climb up and settle at $\frac{a}{1 - r}$ . If $r$ is a small negative fraction, they still settle there but zig-zag in from both sides. If $r$ is big ( $∣ r ∣ > 1$ ), each block is bigger than the last, the pile explodes. Right on the edge, $r = 1$ just stacks equal blocks into a ramp to infinity, and $r = - 1$ makes the pile flip between two heights forever, never choosing one. So the pile has a real total only when $r$ lives strictly inside $(- 1, 1)$ .

Active recall

Rebuild the closed form

Multiply

s_{N}

r

, subtract to cancel the middle, get

s_{N} (1 - r) = a (1 - r^{N})

, divide by

1 - r

Why does

∣ r ∣ < 1

give convergence

Then

r^{N} \to 0

s_{N} \to \frac{a}{1 - r}

; the correction melts away.

How does

- 1 < r < 0

approach the limit

It oscillates — dots zig-zag above and below

\frac{a}{1 - r}

with shrinking swings.

Why exclude

r = 1

when dividing

Because

1 - r = 0

there — division is illegal; separately

s_{N} = a N \to \pm \infty

What happens at

r = - 1

Partial sums flip-flop

a, 0, a, 0, \dots

— bounded but no limit, so it diverges.

Connections

4.3.03 Series — partial sums, convergence definition (Hinglish) — the parent topic
Geometric series — the result derived here in full
Sequences — limits and convergence — what " $s_{N} \to S$ " means
Telescoping series — same "middle cancels" idea as Steps 3–4
n-th Term Test for Divergence — the $r = \pm 1, ∣ r ∣ > 1$ cases fail this
Ratio & Root Tests — measure the " $r$ " of a general series
Power series and radius of convergence — geometric series with $r = x$