Sum of infinite GP — when it converges, proof: Visual walkthrough

Before anything else, let us pin down the two words we cannot proceed without.

If you have not met the raised-number notation ( $r^{N}$ = " $r$ multiplied by itself $N$ times") or why repeated multiplying can shrink a number, keep Geometric Progression — nth term and Limits of Sequences open — they are the ground floor of this page.

Step 1 — Draw the terms as shrinking bricks

WHAT. Take a concrete GP: $a = 2$ , $r = \frac{1}{2}$ . The terms are $2, 1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \dots$ Lay each term down as a horizontal brick whose length equals its value.

WHY. Before we manipulate symbols, we need to believe that infinitely many pieces can fit inside a finite length. A picture of shrinking bricks makes the finiteness obvious to the eye.

PICTURE. Look at Step 1. Each new brick (burnt orange) is exactly half the length of the one before. Stacked end to end they creep toward a wall — they never pass it. Call the position of that wall the total; in Step 2 we will name it properly once we have defined how to reach it.

Figure — Sum of infinite GP — when it converges, proof

Step 2 — Define the partial sum $S_{N}$ and the wall $S_{\infty}$

WHAT. Instead of the infinite sum all at once, we stop after $N$ terms and add only those. Here $N$ is a counting number ( $N \in N$ , i.e. $N = 1, 2, 3, \dots$ ) — you cannot add "half a term": $S_{N} = 1st a + 2nd a r + 3rd a r^{2} + \dots + N -th a r^{N - 1} .$ Every symbol here: $S_{N}$ is the total of the first $N$ bricks; the last term is $a r^{N - 1}$ because the first brick already used the power $r^{0} = 1$ , so the $N$ -th brick has used $r$ exactly $N - 1$ times.

WHY. An "infinite sum" has no meaning until we say what it approaches. So we watch the growing staircase $S_{1}, S_{2}, S_{3}, \dots$ and ask: does the top of the staircase settle on one height? If it does, we give that height a name — the infinite sum $S_{\infty}$ , defined as the limit the partial sums climb to: $S_{\infty} = def lim_{N \to \infty} S_{N} .$ That is the wall from Step 1: it only exists if the staircase actually settles.

PICTURE. Step 2 shows the staircase of partial sums. Each step adds a thinner slice; the plum dashed line is the height $S_{\infty}$ they are climbing toward.

Recall Why not just "add them all"?

Why do we bother with partial sums? ::: You cannot literally add infinitely many numbers; you can only add finitely many and watch where the totals head. The limit of $S_{N}$ is the definition of the infinite sum $S_{\infty}$ .

Step 3 — The shift-and-subtract trick, drawn

WHAT. Write $S_{N}$ , then write $r$ times $S_{N}$ underneath, shifted one slot to the right:

S_{N} r S_{N} = = a + a r a r + a r^{2} + a r^{2} + \dots + + \dots + a r^{N - 1} a r^{N - 1} + a r^{N}

Multiplying by $r$ turns each term into the next term (since "next $=$ current $\times r$ "), so the whole row slides right by one.

WHY this tool and not another? We want to eliminate the messy middle. Because a GP's rows are identical except for a shift, subtracting them makes every shared term cancel in pairs — a trick that works only because of the constant ratio. No calculus needed; pure alignment.

PICTURE. Step 3 stacks the two rows as coloured tiles. Tiles of the same colour sit directly above one another — those are the ones that will annihilate. Only two lonely tiles have no partner: the leading $a$ (teal, left) and the leaked-out $a r^{N}$ (orange, far right).

Step 4 — Subtract and read off the survivors

WHAT. Subtract the second row from the first. Every matched pair cancels; only the two unmatched tiles remain: $S_{N} - r S_{N} = survivor, left a - survivor, right a r^{N} .$

WHY. This is the payoff of the shift: an infinitely-tedious sum has collapsed to a two-term expression. The right survivor $a r^{N}$ is the piece that leaked past the end of the original list — geometrically, it is the little brick that fell off the far edge in Step 3.

PICTURE. Step 4 fades out every cancelled pair and leaves the two survivors glowing. Notice: $a r^{N}$ is the tiny far-right tile — hold on to it, because whether it shrinks to nothing decides everything.

Step 5 — Factor and solve for $S_{N}$

WHAT. Both sides share common factors. On the left, factor out $S_{N}$ ; on the right, factor out $a$ : $S_{N} (1 - r) = a (1 - r^{N}) .$ Now divide both sides by $(1 - r)$ — allowed only if $r \neq = 1$ , because dividing by $1 - r$ when $r = 1$ means dividing by zero: $S_{N} = \frac{a ( 1 - r ^{N} )}{1 - r} (r \neq = 1) .$ Term by term: $a$ = first brick, $r^{N}$ = the leaked far-right piece, and $\frac{1}{1 - r}$ = the scaling that packs the shrinking bricks together.

WHY. We now have one clean, exact formula for the sum of any finite chunk of a GP — this is the Sum of finite GP result, and it is the launch pad for the infinite case.

PICTURE. Step 5 shows the same brick-stack, but now annotated: the full stack height is $\frac{a ( 1 - r ^{N} )}{1 - r}$ , with the "missing tip" of size proportional to $r^{N}$ marked in plum at the top.

Step 6 — Let $N \to \infty$ : watch $r^{N}$ decide the fate

WHAT. Send $N$ to infinity. The only part of $S_{N}$ that still depends on $N$ is $r^{N}$ . Everything else ( $a$ and $1 - r$ ) is frozen. So the whole question reduces to: what does $r^{N}$ do as $N$ grows?

WHY. This is exactly the "closing gap" from Step 1. If that gap $r^{N}$ vanishes, the stack reaches a definite wall; if it doesn't, there is no wall to reach.

PICTURE. Step 6 plots $r^{N}$ against $N$ for four cases on one warm-paper grid:

$r = \frac{1}{2}$ (orange): dives to $0$ — sum settles.
$r = - \frac{1}{2}$ (teal): bounces above/below zero but shrinks to $0$ — still settles.
$r = 2$ (plum): rockets up — no settling.
$r = 1$ (grey): stuck at $1$ forever — no settling.

Step 7 — The convergent case: $∣ r ∣ < 1$

WHAT. When $∣ r ∣ < 1$ we have $r^{N} \to 0$ . Drop that term: $S_{\infty} = lim_{N \to \infty} \frac{a ( 1 - r ^{N} )}{1 - r} = \frac{a ( 1 - 0 )}{1 - r} = \frac{a}{1 - r} .$ Symbol check: $a$ = first term, $1 - r$ = "how much room is left after one step". Small $r$ ⇒ denominator near 1 ⇒ sum near $a$ ; $r$ near $1$ ⇒ denominator tiny ⇒ sum huge.

WHY. This is the parent's headline result — and now you have watched the $r^{N}$ term physically disappear, so the formula is no longer a rule to memorise but a picture you can rebuild.

PICTURE. Step 7 returns to the brick wall: the leaked tip $a r^{N}$ has shrunk to a dot, and the full stack sits flush against the wall at height $\frac{a}{1 - r}$ . For our example $a = 2, r = \frac{1}{2}$ : $\frac{2}{1 - \frac{1}{2}} = 4$ — the wall is at 4.

Step 8 — Every edge case, drawn once and for all

WHAT. The formula $\frac{a}{1 - r}$ is only licensed when $∣ r ∣ < 1$ . We must show what happens at every boundary so you never hit a surprise. (Throughout, $∣ r ∣$ means the magnitude of $r$ — its distance from zero, ignoring sign.)

case	what $r^{N}$ does	partial sums	verdict
$∥ r ∥ < 1$	$\to 0$	climb to a wall	converges to $\frac{a}{1 - r}$
$r = 1$	stuck at $1$	$a, 2 a, 3 a, \dots$	diverges ( $\to \pm \infty$ )
$r = - 1$	flips $1, - 1, 1, \dots$	$a, 0, a, 0, \dots$	diverges (oscillates)
$∥ r ∥ > 1$	blows up	run off to $\pm \infty$	diverges

The trivial subcase $a = 0$ . If the first term is zero, then every term is zero ( $0, 0, 0, \dots$ ), so $S_{N} = 0$ for all $N$ and $S_{\infty} = 0$ whatever $r$ is — even for $∣ r ∣ \geq 1$ . It converges only because there is nothing to add. We flag it so the coverage is truly exhaustive, but for a genuine GP we assume $a \neq = 0$ .

WHY. A proof is only complete when no scenario is left uncovered. The naive "just plug into $\frac{a}{1 - r}$ " gives a number even when it shouldn't — the picture is how we catch the lie.

PICTURE. Step 8 shows three mini staircases side by side: (a) $r = \frac{1}{2}$ climbing to a wall, (b) $r = - 1$ jumping between two heights forever, (c) $r = 2$ exploding off the top. Only (a) settles.

The one-picture summary

Everything above, compressed into a single diagram: the shrinking bricks (Step 1) become the climbing staircase (Step 2); the shift-and-subtract (Steps 3–4) collapses them to $S_{N} = \frac{a ( 1 - r ^{N} )}{1 - r}$ (Step 5); sending $N \to \infty$ (Step 6) kills $r^{N}$ only if $∣ r ∣ < 1$ (Steps 7–8), leaving the wall at $\frac{a}{1 - r}$ .

Recall One-line recap per step (hide, then rebuild the picture yourself)

Step 1 — bricks shrink by $r$ each time and stop at a wall ::: the wall is the total we are hunting
Step 2 — $S_{N}$ is the first $N$ bricks; $S_{\infty} = lim S_{N}$ ::: the wall exists only if the staircase settles
Steps 3–4 — slide by $r$ , subtract, twins cancel ::: survivors are $a$ and the leaked $a r^{N}$
Step 5 — rearrange to $S_{N} = \frac{a ( 1 - r ^{N} )}{1 - r}$ ::: valid for $r \neq = 1$
Step 6 — only $r^{N}$ still moves as $N \to \infty$ ::: its size is the whole verdict
Steps 7–8 — $∣ r ∣ < 1$ : tail dies, $S_{\infty} = \frac{a}{1 - r}$ ; else diverges ::: shrink to sink

Connections

↑ Parent: full proof & examples
3.3.03 Sum of infinite GP — when it converges, proof (Hinglish)
Geometric Progression — nth term (what $a r^{N - 1}$ means)
Sum of finite GP (the Step 5 result on its own)
Limits of Sequences (why $r^{N} \to 0$ when $∣ r ∣ < 1$ )
Convergence of Series
Recurring Decimals as Fractions
Present Value & Perpetuities