Geometric series — convergence condition, proof: Worked examples

This is the practice arena for the geometric-series parent note. There we derived the formula $\sum_{n = 0}^{\infty} a r^{n} = \frac{a}{1 - r} valid iff ∣ r ∣ < 1.$ Here we hit every kind of situation that formula can face: positive ratio, negative ratio, the trivial ratio $0$ , ratio hidden inside a decimal, ratio that starts at a weird index, ratio too big, and the degenerate edges. Guess before each step — that's where the learning is.

Before anything: two words you must own from line one.

We will also lean on one idea from the parent note constantly, so pin it down now:

The scenario matrix

Every geometric-series problem you will ever meet lands in one of these cells. The "Twist" column names the complication that makes each cell different from the plain textbook case (a positive ratio starting at $n = 0$ ) — e.g. a sign flip, a shifted starting index, a decimal in disguise, or the ratio being unknown. The examples that follow are labelled with the cell they cover, so by the end no cell is left dark.

Cell	Sign / size of $r$	Twist (the complication)	Example
C0	$r = 0$ (trivial)	every term after the first is $0$	Ex 0
C1	$0 < r < 1$ (positive, shrinking)	starts at $n = 0$ (baseline case)	Ex 1
C2	$- 1 < r < 0$ (negative, alternating)	terms flip sign	Ex 2
C3	$0 < r < 1$	starts at $n = k$ (index shift)	Ex 3
C4	$0 < r < 1$	repeating decimal → fraction	Ex 4
C5	$	r	\ge 1$
C6	degenerate: $r = 1$ and $r = - 1$	limiting/edge cases	Ex 6
C7	$0 < r < 1$	real-world word problem (bouncing ball)	Ex 7
C8	$r$ unknown (a symbol)	exam twist: solve for $r$	Ex 8

Figure — Geometric series — convergence condition, proof

Read the master figure now (Figure s01). It is a number line for the ratio $r$ . The pale-green band stretching from $- 1$ to $1$ is where the sum settles to a finite value; the coral regions outside are where the pile of terms never dies. The two coral dots at $r = \pm 1$ are the boundary cases (Ex 6). Each labelled dot — "Ex1", "Ex2", "Ex0", … — is one worked example plotted at its own $r$ . Keep this picture in your head: every example below is just a point on this line, and its convergence is decided by which colour it lands in.

Cell C0 — the trivial ratio $r = 0$

Worked example Ex 0 (C0):

$n = 0 \sum \infty 9 \cdot 0^{n}$ Forecast: if the ratio is $0$ , every term after the first is multiplied down to nothing — so the "series" is really just its first term. Guess: $S = 9$ .

Step 1 — write the terms out. With $r = 0$ : $n = 0$ gives $9 \cdot 0^{0} = 9 \cdot 1 = 9$ ; every later term $9 \cdot 0^{n} = 0$ . So the sum is $9 + 0 + 0 + 0 + \dots$ . Why this step? The convention $0^{0} = 1$ (standard for power series and geometric series) makes the $n = 0$ term the only survivor; writing it out shows there is nothing subtle here.

Step 2 — check it lives inside the band and apply the formula. $∣ r ∣ = 0 < 1$ ✓ — dead centre of the green band in Figure s01. So the formula is fully legal: $S = \frac{a}{1 - r} = \frac{9}{1 - 0} = 9.$ Why this step? $r = 0$ is not an excluded case ( $r = 1$ is the excluded one); the formula must — and does — reproduce the obvious answer $9$ .

Verify: the partial sums are $S_{1} = 9, S_{2} = 9, S_{3} = 9, \dots$ — flat at $9$ forever, and $\frac{9}{1 - 0} = 9$ agrees. The trivial case is the sanity anchor: the machinery gives back the plain answer. ✓

Cell C1 — positive shrinking ratio, index from 0

Worked example Ex 1 (C1):

$n = 0 \sum \infty 5 (\frac{2}{3})^{n}$ Forecast: the ratio $\frac{2}{3}$ is under 1, so this converges. Guess: bigger or smaller than the first term $5$ ? (Positive ratio piles more on top, so it should be bigger than 5.)

Step 1 — read off $a$ . Plug $n = 0$ : $5 \cdot (\frac{2}{3})^{0} = 5 \cdot 1 = 5$ . So $a = 5$ . Why this step? $a$ is the literal first term, and $(anything)^{0} = 1$ , so the $n = 0$ term is just the coefficient here.

Step 2 — read off $r$ . $r = \frac{2}{3}$ . Check size: $∣ r ∣ = \frac{2}{3} < 1$ ✓ — we're inside the green band, converges. Why this step? Convergence is permission to use the formula. Never skip it.

Step 3 — apply $\frac{a}{1 - r}$ . $S = \frac{5}{1 - \frac{2}{3}} = \frac{5}{\frac{1}{3}} = 15.$ Why this step? This is the limit of the partial sums, derived by the telescoping trick in the parent note.

Verify: add a few terms: $5 + 3.33 + 2.22 + 1.48 + \dots$ . After 4 terms $\approx 12.03$ , after 8 terms $\approx 14.6$ — creeping up toward $15$ from below, never overshooting. And $15 > 5$ , matching our forecast. ✓

Look at Figure s02. The lavender bars are the individual terms $5 (2/3)^{n}$ — each visibly shorter than the last (they "die"). The coral dots joined by a line are the partial sums $S_{N}$ , climbing step by step toward the dashed mint line at $15$ but never crossing it. That gap you see closing is convergence in action.

Cell C2 — negative alternating ratio

Worked example Ex 2 (C2):

$n = 0 \sum \infty 4 (- \frac{1}{2})^{n}$ Forecast: ratio is negative, so terms flip sign: $+, -, +, -$ . They partly cancel, so the sum should be smaller than $a = 4$ . Guess a value between 2 and 4.

Step 1 — $a$ . $n = 0$ : $4 \cdot (- \frac{1}{2})^{0} = 4$ . So $a = 4$ . Why this step? Same rule; the sign of $r$ doesn't change the $n = 0$ term because $()^{0} = 1$ .

Step 2 — $r$ and check. $r = - \frac{1}{2}$ , $∣ r ∣ = \frac{1}{2} < 1$ ✓. It's the size $∣ r ∣$ that decides convergence, not the sign. Why this step? Alternating terms still shrink in magnitude, so the pile still dies.

Step 3 — apply the formula. $S = \frac{4}{1 - ( - \frac{1}{2} )} = \frac{4}{1 + \frac{1}{2}} = \frac{4}{\frac{3}{2}} = \frac{8}{3} \approx 2.67.$ Why this step? Negative $r$ makes the denominator $1 - r$ bigger than 1, which shrinks the fraction below $a$ — exactly the cancellation we forecast.

Verify: partial sums $S_{1} = 4, S_{2} = 2, S_{3} = 3, S_{4} = 2.5, S_{5} = 2.75, S_{6} = 2.625, \dots$ — they zig-zag inward around $\frac{8}{3} \approx 2.667$ , tightening each step. That oscillating squeeze is the fingerprint of negative $r$ . And $2.67 < 4$ ✓.

Look at Figure s03. The bars now alternate colour — mint for positive terms, coral for negative ones — showing the sign flip directly. The lavender partial-sum line no longer climbs monotonically; it hops above and below the dashed limit $8/3$ , each hop smaller than the last. That inward spiral is the visual signature of a negative ratio.

Cell C3 — index shift (does NOT start at $n = 0$ )

Worked example Ex 3 (C3):

$n = 3 \sum \infty (\frac{1}{4})^{n}$ Forecast: starting late means we chopped off the biggest terms ( $n = 0, 1, 2$ ), so this sum should be tiny.

Step 1 — find the ACTUAL first term. Plug in the starting index $n = 3$ : $(\frac{1}{4})^{3} = \frac{1}{64}$ . So $a = \frac{1}{64}$ . Why this step? The parent-note mistake to avoid: $a$ is not the coefficient (which is $1$ here) — it's the first term you truly write, at $n = 3$ .

Step 2 — the ratio is unchanged. $r = \frac{1}{4}$ still (each step multiplies by $\frac{1}{4}$ ), $∣ r ∣ < 1$ ✓. Why this step? Shifting where you start never changes how consecutive terms relate — that's the ratio.

Step 3 — apply the formula with the shifted $a$ . $S = \frac{a}{1 - r} = \frac{1/64}{1 - 1/4} = \frac{1/64}{3/4} = \frac{1}{64} \cdot \frac{4}{3} = \frac{1}{48} .$ Why this step? The formula only needs "first term" and "ratio"; feed it the true first term $\frac{1}{64}$ .

Verify (cross-check). Take the full sum from $n = 0$ : $\frac{1}{1 - 1/4} = \frac{4}{3}$ , then subtract the first three terms $1 + \frac{1}{4} + \frac{1}{16} = \frac{21}{16}$ : $\frac{4}{3} - \frac{21}{16} = \frac{64 - 63}{48} = \frac{1}{48} . ✓$ Two independent routes agree, and $\frac{1}{48}$ is indeed tiny. ✓

Cell C4 — repeating decimal as a fraction

Worked example Ex 4 (C4): write

$0. \overline{45} = 0.454545 \dots$ as a fraction Forecast: it's less than 1, and the block "45" repeats — expect a fraction with $99$ lurking in the denominator (99 = "two nines" for a two-digit block).

Step 1 — split into a geometric series. Each repeat shifts the block two decimal places, i.e. multiplies by $\frac{1}{100}$ : $0. \overline{45} = \frac{45}{100} + \frac{45}{10000} + \frac{45}{1 0 ^{6}} + \dots$ Why this step? A repeating decimal is by construction a geometric series — that's what "the same digits forever" means.

Step 2 — identify $a, r$ . First term $a = \frac{45}{100}$ ; each next term is $\frac{1}{100}$ of the last, so $r = \frac{1}{100}$ , $∣ r ∣ < 1$ ✓. Why this step? Two decimal places per block ⇒ ratio $\frac{1}{100}$ .

Step 3 — apply the formula. $S = \frac{45/100}{1 - 1/100} = \frac{45/100}{99/100} = \frac{45}{99} = \frac{5}{11} .$ Why this step? The $\frac{1}{100}$ 's cancel, leaving digits over "nines" — the general repeating-decimal shortcut, now proven.

Verify: $5 \div 11 = 0.454545 \dots$ ✓ (long division literally repeats "45"). See Repeating decimals as fractions.

Cell C5 — divergence and the algebra trap

Worked example Ex 5 (C5):

$n = 0 \sum \infty 7 (\frac{5}{4})^{n}$ Forecast: ratio $\frac{5}{4} > 1$ — terms grow. There is NO finite sum. Anticipate a trap number.

Step 1 — check $∣ r ∣$ FIRST. $r = \frac{5}{4}$ , $∣ r ∣ = 1.25 \geq 1$ . We're outside the green band ⇒ diverges. STOP — do not use the formula. Why this step? The formula was derived assuming $r^{N} \to 0$ . Here $(\frac{5}{4})^{N} \to \infty$ , so that derivation is void.

Step 2 — expose the trap. If you illegally plug in anyway: $\frac{a}{1 - r} = \frac{7}{1 - 5/4} = \frac{7}{- 1/4} = - 28.$ Why this step? Seeing the trap value $- 28$ inoculates you: a negative "sum" of positive growing terms is impossible — that absurdity is the alarm bell that you skipped the $∣ r ∣ < 1$ check.

Verify: terms are $7, 8.75, 10.94, \dots$ — each bigger than the last, so partial sums race to $+ \infty$ . A sum of positive numbers can never be $- 28$ . Diverges. ✓ (The Ratio Test would flag this instantly: ratio $= 1.25 \geq 1$ .)

Cell C6 — the degenerate edges $r = 1$ and $r = - 1$

Worked example Ex 6 (C6): behaviour of

$n = 0 \sum \infty 6 \cdot r^{n}$ at $r = 1$ and $r = - 1$ Forecast: both sit exactly on the boundary of the green band — expect no settling in either.

Case $r = 1$ . Step 1. Every term equals $a = 6$ : $6 + 6 + 6 + \dots$ . The partial sum of the first $N$ terms is $S_{N} = 6 N$ . Why this step? When $r = 1$ , the "telescoping" subtraction gives $S_{N} (1 - r) = S_{N} \cdot 0 = 0$ — a useless $0 = 0$ . The formula divides by $1 - r = 0$ , which is forbidden, so we count terms directly instead. Recall $N$ is just how many terms we added, so $S_{N} = 6 N$ literally means " $N$ copies of $6$ ". Step 2. $S_{N} = 6 N \to \infty$ . Diverges.

Case $r = - 1$ . Step 1. Terms flip: $6 - 6 + 6 - 6 + \dots$ . Partial sums are $S_{1} = 6, S_{2} = 0, S_{3} = 6, S_{4} = 0, \dots$ Why this step? $r^{N} = (- 1)^{N}$ never settles to one value — it oscillates, so $S_{N}$ oscillates too. Step 2. The sequence $6, 0, 6, 0, \dots$ has no single limit ⇒ diverges (oscillating).

Verify: at $r = 1$ , $S_{100} = 600$ and $S_{1000} = 6000$ — growing without bound ✓. At $r = - 1$ , $S_{100} = 0$ but $S_{101} = 6$ — two different accumulation values, so no limit ✓. Both boundary points are excluded by $∣ r ∣ < 1$ , exactly as the theory demands.

Look at Figure s04. Two side-by-side pictures of the partial sum $S_{N}$ against $N$ . On the left ( $r = 1$ ) the coral dots march upward in a straight line — that runaway is divergence to infinity. On the right ( $r = - 1$ ) the lavender dots bounce forever between $6$ and $0$ , never converging on a single height. Both live on the coral boundary dots of the master figure s01.

Cell C7 — real-world word problem

Worked example Ex 7 (C7): a bouncing ball

A ball is dropped from height $10 m$ . After each bounce it returns to $\frac{3}{5}$ of its previous height. What total vertical distance does it travel before coming to rest?

Forecast: it falls 10, then bounces up-and-down repeatedly, each smaller — a finite total (ratio $\frac{3}{5} < 1$ ). Expect something like a few tens of metres.

Step 1 — separate the down and up journeys. The initial drop is $10$ m (down only). After that, each bounce goes up to a height then down the same height — so each bounce-height is travelled twice. Why this step? Distance counts every metre moved, up or down; the first drop has no matching "up" before it.

Step 2 — build the geometric series. Bounce peaks are $10 \cdot \frac{3}{5}, 10 \cdot (\frac{3}{5})^{2}, \dots$ . Their round-trips sum to $2 \sum_{n = 1}^{\infty} 10 (\frac{3}{5})^{n} .$ Here $a = 10 \cdot \frac{3}{5} = 6$ (the $n = 1$ term), $r = \frac{3}{5}$ , $∣ r ∣ < 1$ ✓. Why this step? We start at $n = 1$ because the first drop is handled separately; the peaks begin at the first bounce.

Step 3 — evaluate. $\sum_{n = 1}^{\infty} 10 (\frac{3}{5})^{n} = \frac{6}{1 - 3/5} = \frac{6}{2/5} = 15.$ Total distance $= first drop 10 + up+down of every bounce 2 \times 15 = 10 + 30 = 40 m .$ Why this step? Doubling accounts for each bounce's rise and fall; adding the lone first drop completes the count.

Verify (units + magnitude): all quantities in metres, ratio dimensionless ✓. First few distances: $10 + 2 (6) + 2 (3.6) + 2 (2.16) \approx 33.5$ , tending up toward $40$ ✓. Finite because $∣ r ∣ < 1$ .

Cell C8 — exam twist: solve for $r$

Worked example Ex 8 (C8): find the ratio from the sum

A convergent geometric series has first term $a = 8$ and total sum $S = 12$ . Find $r$ , and verify it's a valid (convergent) ratio.

Forecast: the sum $12$ is bigger than $a = 8$ , so terms add up beyond the first term ⇒ $r$ is positive and moderate.

Step 1 — invert the formula. From $S = \frac{a}{1 - r}$ : $12 = \frac{8}{1 - r} \Rightarrow 1 - r = \frac{8}{12} = \frac{2}{3} \Rightarrow r = \frac{1}{3} .$ Why this step? The formula is one equation in one unknown $r$ ; just solve algebraically.

Step 2 — the convergence check is not optional. $∣ r ∣ = \frac{1}{3} < 1$ ✓ — a legal answer. If solving had given $∣ r ∣ \geq 1$ , we'd reject it as spurious (no convergent series produces that). Why this step? An algebraic solution is only meaningful if it lands inside the green band; otherwise the premise "convergent series" is contradicted.

Verify: rebuild the sum with $a = 8, r = \frac{1}{3}$ : $\frac{8}{1 - 1/3} = \frac{8}{2/3} = 12$ ✓. Partial terms $8 + 2.67 + 0.89 + \dots \to 12$ , and $12 > 8$ as forecast. ✓

Recall Which cell am I in? (self-quiz)

Ratio is exactly 0 ::: sum is just the first term $a$ ; formula gives $\frac{a}{1 - 0} = a$ (C0). Ratio positive and under 1 ::: converges, sum bigger than $a$ (C1). Ratio negative, size under 1 ::: converges, oscillates inward, sum smaller than $a$ (C2). Sum starts at $n = k$ not 0 ::: recompute $a$ as the term at $n = k$ (C3). A repeating decimal ::: geometric with $r = 1 0^{- (block length)}$ (C4). $∣ r ∣ \geq 1$ with $a \neq = 0$ ::: diverges; the $\frac{a}{1 - r}$ value is a trap (C5). $r = 1$ ::: $S_{N} = a N \to \infty$ ; $r = - 1$ ::: oscillates $a, 0, a, 0$ — both diverge (C6). A bouncing ball / word problem ::: model peaks as a geometric series, double for up+down, add any un-paired first drop (C7). Given the sum, find $r$ ::: invert $S = \frac{a}{1 - r}$ then check $∣ r ∣ < 1$ (C8).

Connections

Geometric series — convergence condition, proof (index 4.3.4) — the parent proof these examples exercise.
Sequences — limits and convergence — why $r^{N} \to 0$ inside the band (C0–C4) and fails outside (C5–C6).
Partial sums and series convergence — the "settling of partial sums" every Verify uses.
Ratio Test — instant divergence flag for Cell C5.
Power series and radius of convergence — Cell C8's "solve for $r$ " is the seed of a radius.
Repeating decimals as fractions — Cell C4 in full.
Telescoping series — the cancellation behind the formula.