Geometric series — convergence condition, proof: Visual walkthrough

This page rebuilds the whole result $n = 0 \sum \infty a r^{n} = \frac{a}{1 - r}$ from nothing but a pile of shrinking (or growing) rectangles. No formula is used before we see why it must be true. If a symbol appears, we point at the picture where it lives.

Parent: 4.3.4 · Geometric series.

Step 1 — WHAT are we even adding up?

WHAT. A geometric series is a list of numbers where you get the next one by multiplying by the same fixed number every single time. Call the very first number $a$ (the "starting height"), and call the fixed multiplier $r$ (the "shrink/grow factor").

Draw every term as a bar of width exactly $1$ , standing on the axis, with height equal to the term's value. Because each bar has width $1$ , its area equals its height — so from the very start, "the value of a term" and "the area of its bar" are the same number. The list is $term 0 a, term 1 a r, term 2 a r^{2}, term 3 a r^{3}, \dots$

Look at each symbol on the picture:

$a$ is the height (= area) of the first bar (the tall burnt-orange one on the left).
$r$ is the ratio "next bar $\div$ this bar". Every neighbouring pair shares the same ratio — that is the whole meaning of the word "geometric".
$a r^{n}$ means "start at $a$ , then multiply by $r$ a total of $n$ times". The little exponent $n$ is just a counter for how many shrinks have happened.

WHY draw unit-width bars. Because "adding numbers" then becomes "stacking areas": with width $1$ everywhere, the total sum is the total shaded area. That turns an algebra question into a picture question, and keeps heights and areas interchangeable throughout the page.

Figure — Geometric series — convergence condition, proof

Step 2 — WHY shrinking is the whole game (and what a negative $r$ looks like)

WHAT. Compare three piles of unit-width bars. In the middle, $∣ r ∣ < 1$ with $r > 0$ (say $r = \frac{1}{2}$ ): each bar is a shrunken copy of the one before, so heights race down toward zero. On the right, $∣ r ∣ > 1$ (say $r = 2$ ): each bar is bigger, so heights explode. And on the far strip, a negative ratio $r = - \frac{1}{2}$ : the bars still shrink in size, but they now flip — one bar points up, the next points down below the axis, alternating forever.

Here $∣ r ∣$ (read "the size of $r$ ", ignoring its sign) is the honest measure of "how fast do terms shrink". We need the size, not $r$ itself, because a negative $r$ still shrinks the bars if its size is under $1$ — it just flips them below the axis, so the up-bars and down-bars partly cancel.

WHY this matters. For $∣ r ∣ < 1$ the total area is finite (bars vanish fast, bounded ink) whether $r$ is positive or negative — with a negative $r$ the below-axis bars subtract area, so the running total wobbles up and down toward its limit. For $∣ r ∣ > 1$ the ink is endless. So the single fact that decides everything is:

$Does ∣ r ∣ < 1 ? ⟺ do the bars die out fast enough?$

PICTURE. Left/middle: bars nose-diving to the axis, area capped. A separate panel shows $r = - \frac{1}{2}$ with bars alternating above and below the axis (shrinking). Right: bars marching off the top.

Step 3 — WHAT is a partial sum $S_{N}$ ?

WHAT. We can never physically add infinitely many bars, so we add the first $N$ of them and call that running total $S_{N}$ : $S_{N} = bar 0 a + bar 1 a r + bar 2 a r^{2} + \dots + bar N - 1 a r^{N - 1} .$

Read the last symbol carefully: the highest exponent is $N - 1$ , not $N$ . Why? Because we started counting at bar $0$ , so "the first $N$ bars" are bars $0, 1, 2, \dots, N - 1$ — that is $N$ bars in total, ending one step below $N$ .

WHY do this. An infinite sum is defined as what $S_{N}$ approaches as $N$ grows. So first we need a clean closed formula for $S_{N}$ — a single expression, no " $\dots$ ". Steps 4–5 hunt for that formula.

PICTURE. The first $N$ bars are shaded solid; the ghostly grey bars beyond bar $N - 1$ are the tail we have not added yet. $S_{N}$ is the solid area only.

Step 4 — WHY multiply the whole sum by $r$ (the telescoping trick)

WHAT. Take the same pile $S_{N}$ and shift it: multiply every term by $r$ . This slides the whole staircase over by one bar: $r S_{N} = was bar 0 a r + was bar 1 a r^{2} + \dots + was bar N - 2 a r^{N - 1} + new tail a r^{N} .$

Line $S_{N}$ and $r S_{N}$ up so that the equal-height bars sit directly above one another.

WHY. Because now almost every bar in $r S_{N}$ has an identical twin in $S_{N}$ . If we subtract, all those matched twins cancel to zero. Only the bars with no partner survive:

$S_{N}$ has an unmatched bar on the left: the original $a$ (bar $0$ — $r S_{N}$ never produced it).
$r S_{N}$ has an unmatched bar on the right: the brand-new $a r^{N}$ (it stuck out one past the end of $S_{N}$ ).

This deliberate line-up-then-cancel is what "telescoping" means — a long chain collapses to just its two ends. (Same trick as in Telescoping series.)

PICTURE. Two staircases drawn one above the other; matched bars linked by faint teal ties, the two lonely survivors ( $a$ on the far left, $a r^{N}$ on the far right) circled in plum.

Step 5 — WHAT the algebra hands back: the closed formula

WHAT. Factor both sides of $S_{N} - r S_{N} = a - a r^{N}$ .

Left side: both terms share $S_{N}$ , so $S_{N} - r S_{N} = S_{N} (1 - r)$ . Right side: both terms share $a$ , so $a - a r^{N} = a (1 - r^{N})$ .

Setting them equal and dividing by the common factor $(1 - r)$ : $running total S_{N} = \frac{a first bar ( 1 - r ^{N} tail size )}{cannot be 0 1 - r} .$

Read each piece:

$a$ scales everything (double the first bar, double the total — obvious).
$r^{N}$ is the height of the first bar we did not add; it is the "how much is left over" knob.
$1 - r$ downstairs is where the shrinking magic gets baked in.

WHY exclude $r = 1$ . We divided by $(1 - r)$ . If $r = 1$ that is dividing by zero — illegal. And it should be a separate case: at $r = 1$ every bar has the same height $a$ , so $N$ equal bars give $S_{N} = a N$ , which just climbs forever (unless $a = 0$ ).

PICTURE. The finished formula shown as: (full rectangle of area $\frac{a}{1 - r}$ ) minus (a scaled copy of area $\frac{a r ^{N}}{1 - r}$ representing the un-added tail).

Recall Why

$(1 - r)$ appears downstairs It is the leftover factor after cancelling. Geometrically, $\frac{1}{1 - r}$ is the "amplification" that turns one bar of height $a$ into the whole infinite stack once the tail vanishes.

Step 6 — WHAT happens to the tail $r^{N}$ as we add forever

WHAT. The infinite sum is defined as the limit of $S_{N}$ as $N$ grows without bound (this "limit of a list" idea lives in Sequences — limits and convergence). In the formula only one piece depends on $N$ : the tail $r^{N}$ . So watch it.

Five cases — all of them, no gaps (note that $∣ r ∣ > 1$ splits by sign):

case	what $r^{N}$ does	picture
$∣ r ∣ < 1$	$r^{N} \to 0$ : the leftover bar shrinks to nothing	tail dot slides onto the axis
$r > 1$	$r^{N} \to + \infty$ : leftover bar explodes upward	dots fly off the top
$r < - 1$	$∣ r^{N} ∣ \to \infty$ and sign flips: bars grow and alternate up/down	dots fling apart, alternating sign
$r = 1$	$r^{N} = 1$ forever; $S_{N} = a N$	staircase of equal bars, no ceiling
$r = - 1$	$r^{N}$ flips $1, - 1, 1, - 1, \dots$	dot ping-pongs, never lands

WHY. Multiplying by a number of size under $1$ again and again guarantees decay to zero — each multiply removes a fixed fraction. Multiplying by size above $1$ guarantees blow-up; if that big multiplier is also negative ( $r < - 1$ ), each step both magnifies and flips the sign, so the tail leaps to ever-larger positive and negative values — doubly hopeless. At the two boundary values $r = \pm 1$ nothing shrinks, so the sum can never settle.

PICTURE. One axis showing the sequence $r^{N}$ as dots for each case: teal dots homing in on $0$ ; orange dots escaping upward ( $r > 1$ ); a $r < - 1$ trace flinging alternately far up and far down; plum dots stuck at height $1$ ; plum dots oscillating between $\pm 1$ .

Step 7 — WHAT we land on: the convergence condition

WHAT. In the only surviving good case $∣ r ∣ < 1$ , the tail $r^{N} \to 0$ , so $lim_{N \to \infty} S_{N} = \frac{a ( 1 - 0 )}{1 - r} = \frac{a}{1 - r} .$

WHY it's iff. If $∣ r ∣ \geq 1$ (with $a \neq = 0$ ) the tail does not vanish, the limit does not exist, and there is no finite sum. So the formula is valid exactly when $∣ r ∣ < 1$ — not one value more: $\sum_{n = 0}^{\infty} a r^{n} = \frac{a}{1 - r} ⟺ ∣ r ∣ < 1.$

Sanity checks on the picture (all cases of sign):

$r$ tiny positive ⇒ $1 - r \approx 1$ ⇒ sum $\approx a$ : just the first bar plus crumbs. ✓
$r \to 1^{-}$ ⇒ $1 - r \to 0^{+}$ ⇒ sum $\to + \infty$ : bars barely shrink, ink piles up. ✓
$r < 0$ ⇒ $1 - r > 1$ ⇒ sum $< a$ : bars alternate above/below the axis and partly cancel. ✓

Step 8 — WHAT if the first bar has zero height? ( $a = 0$ , the trivial edge)

WHAT. Suppose the very first bar has height $a = 0$ . Then $a r^{n} = 0$ for every $n$ : there are no bars at all, just a flat line on the axis. The sum is $0 + 0 + 0 + \dots = 0$ , no matter what $r$ is — even a huge growing $r$ multiplies zero and stays zero.

WHY mention it. So no reader wonders whether the rules break when $a$ vanishes. They don't — they just become trivial:

The formula still agrees: $\frac{a}{1 - r} = \frac{0}{1 - r} = 0$ for any $r \neq = 1$ . ✓
The convergence condition " $∣ r ∣ < 1$ " is stated for $a \neq = 0$ ; when $a = 0$ the series converges to $0$ for all $r$ (including $∣ r ∣ \geq 1$ ), because there is literally nothing to blow up.
The lone genuinely-excluded spot $r = 1$ still gives $S_{N} = a N = 0$ when $a = 0$ — still $0$ , still fine.

PICTURE. A flat axis with all bars of height $0$ ; the running total stays pinned at $0$ regardless of $r$ .

The one-picture summary

Everything at once: the shrinking bars, the shifted copy that telescopes, the leftover tail $\frac{a r ^{N}}{1 - r}$ melting away, and the full rectangle $\frac{a}{1 - r}$ that is left standing — valid only inside the band $∣ r ∣ < 1$ (with the trivial $a = 0$ case always summing to $0$ ).

Recall Feynman retelling — the walkthrough in plain words

Picture bars standing in a row, each one unit wide, so a bar's height is its area. The first is height $a$ , and each next bar is $r$ times as tall as the one before it. Adding the numbers is the same as measuring the total shaded area. If the bars keep shrinking (size of $r$ under one), there's only a limited amount of ink — a finite total. If $r$ is negative but still small in size, the bars flip above and below the axis and partly cancel, but they still shrink to a finite total. If they stay the same or grow, the ink never ends — and if that growing $r$ is also negative, the bars both blow up and flip, doubly hopeless. To find the total of the first few bars, I make a second copy of the whole staircase but nudged over by one bar (that's "multiply by $r$ "). When I subtract the shifted copy from the original, every middle bar has a twin and cancels — only the first bar $a$ and one leftover tail bar $a r^{N}$ survive. Tidying up gives $S_{N} = \frac{a ( 1 - r ^{N} )}{1 - r}$ . Now imagine adding bars forever. The only piece that changes is the leftover tail $r^{N}$ . If the bars shrink, that tail melts to zero and the total settles on $\frac{a}{1 - r}$ . If they don't shrink, the tail never dies and there is no total. And if the very first bar had height zero to begin with, the whole thing is just zero. So the one rule that runs the whole show is: the multiplier's size must be under one.

Connections

Geometric series — convergence condition, proof (parent — the full result and examples)
Sequences — limits and convergence (the $lim r^{N}$ behaviour in Step 6)
Partial sums and series convergence (a series is the limit of its $S_{N}$ )
Telescoping series (the shift-and-cancel move of Step 4)
Ratio Test (generalises the constant-ratio idea)
Power series and radius of convergence (the model case $\sum x^{n} = \frac{1}{1 - x}$ )
Repeating decimals as fractions (a favourite application)