Geometric series — convergence condition, proof: Worked examples

4.3.4 · D3 · Maths › Calculus III — Sequences & Series › Geometric series — convergence condition, proof

Yeh geometric-series parent note ka practice arena hai. Wahan humne formula derive kiya tha: $\sum_{n = 0}^{\infty} a r^{n} = \frac{a}{1 - r} valid iff ∣ r ∣ < 1.$ Yahan hum har tarah ki situation face karenge jo yeh formula face kar sakta hai: positive ratio, negative ratio, trivial ratio $0$ , decimal ke andar chupi ratio, weird index se shuru hone wali ratio, bahut badi ratio, aur degenerate edges. Har step se pehle guess karo — wahi seekhna hai.

Shuru karne se pehle: do words hain jo line one se tumhare honay chahiye.

Hum parent note se ek idea baar baar use karenge, toh ise abhi pin kar lo:

The scenario matrix

Geometric-series ka har problem jo tum kabhi bhi miloge, in cells mein se kisi ek mein land karta hai. "Twist" column us complication ka naam batata hai jo har cell ko plain textbook case se alag banata hai (ek positive ratio jo $n = 0$ se start hoti hai) — jaise sign flip, shifted starting index, decimal in disguise, ya ratio unknown hona. Neeche ke examples uss cell ke saath labelled hain jo woh cover karte hain, toh aakhir mein koi cell dark nahi bachti.

Cell	Sign / size of $r$	Twist (the complication)	Example
C0	$r = 0$ (trivial)	pehle term ke baad har term $0$ hai	Ex 0
C1	$0 < r < 1$ (positive, shrinking)	$n = 0$ se shuru (baseline case)	Ex 1
C2	$- 1 < r < 0$ (negative, alternating)	terms ka sign flip hota hai	Ex 2
C3	$0 < r < 1$	$n = k$ se shuru (index shift)	Ex 3
C4	$0 < r < 1$	repeating decimal → fraction	Ex 4
C5	$	r	\ge 1$
C6	degenerate: $r = 1$ aur $r = - 1$	limiting/edge cases	Ex 6
C7	$0 < r < 1$	real-world word problem (bouncing ball)	Ex 7
C8	$r$ unknown (ek symbol)	exam twist: $r$ ke liye solve karo	Ex 8

Figure — Geometric series — convergence condition, proof

Abhi master figure padhlo (Figure s01). Yeh ratio $r$ ke liye ek number line hai. Pale-green band jo $- 1$ se $1$ tak stretch karta hai, wahan sum finite value pe settle hota hai; bahar coral regions hain jahan terms ki pile kabhi nahi marti. $r = \pm 1$ par do coral dots boundary cases hain (Ex 6). Har labelled dot — "Ex1", "Ex2", "Ex0", … — ek worked example hai jo apne $r$ par plot kiya gaya hai. Yeh picture apne dimag mein rakho: neeche har example iss line par sirf ek point hai, aur uski convergence decide hoti hai ki woh kis colour mein land karta hai.

Cell C0 — trivial ratio $r = 0$

Worked example Ex 0 (C0):

$n = 0 \sum \infty 9 \cdot 0^{n}$ Forecast: agar ratio $0$ hai, toh pehle term ke baad har term multiply hokar kuch nahi banta — toh yeh "series" really sirf apna pehla term hai. Guess: $S = 9$ .

Step 1 — terms likhte hain. $r = 0$ ke saath: $n = 0$ deta hai $9 \cdot 0^{0} = 9 \cdot 1 = 9$ ; har baad ka term $9 \cdot 0^{n} = 0$ hai. Toh sum hai $9 + 0 + 0 + 0 + \dots$ . Yeh step kyun? Convention $0^{0} = 1$ (power series aur geometric series ke liye standard) $n = 0$ term ko ek maatra survivor banata hai; ise likhne se pata chalta hai ki yahan kuch subtle nahi hai.

Step 2 — check karo ki band ke andar hai aur formula apply karo. $∣ r ∣ = 0 < 1$ ✓ — Figure s01 mein green band ke dead centre mein. Toh formula poori tarah legal hai: $S = \frac{a}{1 - r} = \frac{9}{1 - 0} = 9.$ Yeh step kyun? $r = 0$ excluded case nahi hai ( $r = 1$ excluded case hai); formula ko — aur karta bhi hai — obvious answer $9$ reproduce karna chahiye.

Verify: partial sums hain $S_{1} = 9, S_{2} = 9, S_{3} = 9, \dots$ — hamesha $9$ par flat, aur $\frac{9}{1 - 0} = 9$ agree karta hai. Trivial case sanity anchor hai: machinery plain answer wapas deti hai. ✓

Cell C1 — positive shrinking ratio, index from 0

Worked example Ex 1 (C1):

$n = 0 \sum \infty 5 (\frac{2}{3})^{n}$ Forecast: ratio $\frac{2}{3}$ 1 se kam hai, toh yeh converge karta hai. Guess: pehle term $5$ se bada ya chhota? (Positive ratio aur oopar pile karta hai, toh yeh 5 se bada hona chahiye.)

Step 1 — $a$ read karo. $n = 0$ plug karo: $5 \cdot (\frac{2}{3})^{0} = 5 \cdot 1 = 5$ . Toh $a = 5$ . Yeh step kyun? $a$ literal pehla term hai, aur $(anything)^{0} = 1$ , toh $n = 0$ term yahan sirf coefficient hai.

Step 2 — $r$ read karo aur check karo. $r = \frac{2}{3}$ . Size check: $∣ r ∣ = \frac{2}{3} < 1$ ✓ — hum green band ke andar hain, converges. Yeh step kyun? Convergence formula use karne ki permission hai. Kabhi skip mat karo.

Step 3 — $\frac{a}{1 - r}$ apply karo. $S = \frac{5}{1 - \frac{2}{3}} = \frac{5}{\frac{1}{3}} = 15.$ Yeh step kyun? Yeh partial sums ka limit hai, parent note mein telescoping trick se derive kiya gaya.

Verify: kuch terms add karo: $5 + 3.33 + 2.22 + 1.48 + \dots$ . 4 terms ke baad $\approx 12.03$ , 8 terms ke baad $\approx 14.6$ — $15$ ki taraf neeche se badh raha hai, kabhi overshoot nahi karta. Aur $15 > 5$ , hamare forecast se match karta hai. ✓

Figure s02 dekho. Lavender bars individual terms $5 (2/3)^{n}$ hain — har ek visibly pehle se chhota (woh "die" karte hain). Line se joined coral dots partial sums $S_{N}$ hain, step by step $15$ ki dashed mint line ki taraf badh rahe hain par kabhi cross nahi karte. Woh gap jo close ho raha hai wahi convergence in action hai.

Cell C2 — negative alternating ratio

Worked example Ex 2 (C2):

$n = 0 \sum \infty 4 (- \frac{1}{2})^{n}$ Forecast: ratio negative hai, toh terms sign flip karte hain: $+, -, +, -$ . Woh partly cancel karte hain, toh sum $a = 4$ se chhota hona chahiye. 2 aur 4 ke beech value guess karo.

Step 1 — $a$ . $n = 0$ : $4 \cdot (- \frac{1}{2})^{0} = 4$ . Toh $a = 4$ . Yeh step kyun? Wahi rule; $r$ ka sign $n = 0$ term ko nahi badalta kyunki $()^{0} = 1$ .

Step 2 — $r$ aur check. $r = - \frac{1}{2}$ , $∣ r ∣ = \frac{1}{2} < 1$ ✓. Convergence size $∣ r ∣$ decide karta hai, sign nahi. Yeh step kyun? Alternating terms magnitude mein abhi bhi shrink karte hain, toh pile phir bhi marti hai.

Step 3 — formula apply karo. $S = \frac{4}{1 - ( - \frac{1}{2} )} = \frac{4}{1 + \frac{1}{2}} = \frac{4}{\frac{3}{2}} = \frac{8}{3} \approx 2.67.$ Yeh step kyun? Negative $r$ denominator $1 - r$ ko 1 se bada banata hai, jo fraction ko $a$ se chhota shrink karta hai — exactly woh cancellation jo humne forecast kiya tha.

Verify: partial sums $S_{1} = 4, S_{2} = 2, S_{3} = 3, S_{4} = 2.5, S_{5} = 2.75, S_{6} = 2.625, \dots$ — woh $\frac{8}{3} \approx 2.667$ ke around zig-zag inward karte hain, har step tighten hota hai. Yeh oscillating squeeze negative $r$ ka fingerprint hai. Aur $2.67 < 4$ ✓.

Figure s03 dekho. Bars ab alternate colour mein hain — positive terms ke liye mint, negative terms ke liye coral — directly sign flip dikhata hai. Lavender partial-sum line ab monotonically nahi chadhti; woh dashed limit $8/3$ ke oopar aur neeche hops karti hai, har hop pehle se chhota. Woh inward spiral negative ratio ka visual signature hai.

Cell C3 — index shift ( $n = 0$ se shuru nahi)

Worked example Ex 3 (C3):

$n = 3 \sum \infty (\frac{1}{4})^{n}$ Forecast: late shuru karne ka matlab hai ki humne sabse bade terms ( $n = 0, 1, 2$ ) chop kar diye, toh yeh sum tiny hona chahiye.

Step 1 — ACTUAL pehla term nikalo. Starting index $n = 3$ plug karo: $(\frac{1}{4})^{3} = \frac{1}{64}$ . Toh $a = \frac{1}{64}$ . Yeh step kyun? Parent note ki galti jo avoid karni hai: $a$ coefficient nahi hai (jo yahan $1$ hai) — yeh woh pehla term hai jo tum actually likhte ho, $n = 3$ par.

Step 2 — ratio unchanged hai. $r = \frac{1}{4}$ abhi bhi hai (har step $\frac{1}{4}$ se multiply karta hai), $∣ r ∣ < 1$ ✓. Yeh step kyun? Shuru karne ki jagah shift karna kabhi consecutive terms ke relation ko nahi badalta — wahi ratio hai.

Step 3 — shifted $a$ ke saath formula apply karo. $S = \frac{a}{1 - r} = \frac{1/64}{1 - 1/4} = \frac{1/64}{3/4} = \frac{1}{64} \cdot \frac{4}{3} = \frac{1}{48} .$ Yeh step kyun? Formula ko sirf "first term" aur "ratio" chahiye; ise true pehla term $\frac{1}{64}$ do.

Verify (cross-check). $n = 0$ se poora sum lo: $\frac{1}{1 - 1/4} = \frac{4}{3}$ , phir pehle teen terms subtract karo $1 + \frac{1}{4} + \frac{1}{16} = \frac{21}{16}$ : $\frac{4}{3} - \frac{21}{16} = \frac{64 - 63}{48} = \frac{1}{48} . ✓$ Do independent routes agree karte hain, aur $\frac{1}{48}$ indeed tiny hai. ✓

Cell C4 — repeating decimal as a fraction

Worked example Ex 4 (C4):

$0. \overline{45} = 0.454545 \dots$ ko fraction mein likho Forecast: yeh 1 se kam hai, aur block "45" repeat hota hai — expect karo ek fraction jisme denominator mein $99$ ho ( $99$ = do digit block ke liye "two nines").

Step 1 — geometric series mein split karo. Har repeat block ko do decimal places shift karta hai, yaani $\frac{1}{100}$ se multiply karta hai: $0. \overline{45} = \frac{45}{100} + \frac{45}{10000} + \frac{45}{1 0 ^{6}} + \dots$ Yeh step kyun? Ek repeating decimal by construction ek geometric series hai — "wahi digits hamesha ke liye" ka yahi matlab hai.

Step 2 — $a, r$ identify karo. Pehla term $a = \frac{45}{100}$ ; agla har term pichle ka $\frac{1}{100}$ hai, toh $r = \frac{1}{100}$ , $∣ r ∣ < 1$ ✓. Yeh step kyun? Ek block mein do decimal places ⇒ ratio $\frac{1}{100}$ .

Step 3 — formula apply karo. $S = \frac{45/100}{1 - 1/100} = \frac{45/100}{99/100} = \frac{45}{99} = \frac{5}{11} .$ Yeh step kyun? $\frac{1}{100}$ cancel ho jaate hain, digits over "nines" bachte hain — general repeating-decimal shortcut, ab proven.

Verify: $5 \div 11 = 0.454545 \dots$ ✓ (long division literally "45" repeat karta hai). Dekho Repeating decimals as fractions.

Cell C5 — divergence aur algebra trap

Worked example Ex 5 (C5):

$n = 0 \sum \infty 7 (\frac{5}{4})^{n}$ Forecast: ratio $\frac{5}{4} > 1$ — terms grow karte hain. Koi finite sum nahi hai. Ek trap number anticipate karo.

Step 1 — PEHLE $∣ r ∣$ check karo. $r = \frac{5}{4}$ , $∣ r ∣ = 1.25 \geq 1$ . Hum green band ke bahar hain ⇒ diverges. RUKO — formula use mat karo. Yeh step kyun? Formula $r^{N} \to 0$ assume karke derive kiya gaya tha. Yahan $(\frac{5}{4})^{N} \to \infty$ , toh woh derivation void hai.

Step 2 — trap expose karo. Agar tum illegally plug in karo: $\frac{a}{1 - r} = \frac{7}{1 - 5/4} = \frac{7}{- 1/4} = - 28.$ Yeh step kyun? Trap value $- 28$ dekhna tumhe inoculate karta hai: positive growing terms ka negative "sum" impossible hai — woh absurdity alarm bell hai ki tumne $∣ r ∣ < 1$ check skip kiya.

Verify: terms hain $7, 8.75, 10.94, \dots$ — har ek pehle se bada, toh partial sums $+ \infty$ ki taraf race karte hain. Positive numbers ka sum kabhi $- 28$ nahi ho sakta. Diverges. ✓ (Ratio Test ise instantly flag karta: ratio $= 1.25 \geq 1$ .)

Cell C6 — degenerate edges $r = 1$ aur $r = - 1$

Worked example Ex 6 (C6):

$n = 0 \sum \infty 6 \cdot r^{n}$ ka behaviour $r = 1$ aur $r = - 1$ par Forecast: dono green band ki exactly boundary par hain — kisi mein bhi settling ki expect nahi.

Case $r = 1$ . Step 1. Har term $a = 6$ ke equal hai: $6 + 6 + 6 + \dots$ . Pehle $N$ terms ka partial sum $S_{N} = 6 N$ hai. Yeh step kyun? Jab $r = 1$ , "telescoping" subtraction deta hai $S_{N} (1 - r) = S_{N} \cdot 0 = 0$ — ek useless $0 = 0$ . Formula $1 - r = 0$ se divide karta hai, jo forbidden hai, toh hum directly terms count karte hain. Yaad karo $N$ simply kitne terms add kiye hain, toh $S_{N} = 6 N$ literally matlab hai " $N$ copies of $6$ ". Step 2. $S_{N} = 6 N \to \infty$ . Diverges.

Case $r = - 1$ . Step 1. Terms flip karte hain: $6 - 6 + 6 - 6 + \dots$ . Partial sums hain $S_{1} = 6, S_{2} = 0, S_{3} = 6, S_{4} = 0, \dots$ Yeh step kyun? $r^{N} = (- 1)^{N}$ kabhi ek value par settle nahi karta — oscillate karta hai, toh $S_{N}$ bhi oscillate karta hai. Step 2. Sequence $6, 0, 6, 0, \dots$ ka koi single limit nahi ⇒ diverges (oscillating).

Verify: $r = 1$ par, $S_{100} = 600$ aur $S_{1000} = 6000$ — bina bound ke badh raha hai ✓. $r = - 1$ par, $S_{100} = 0$ lekin $S_{101} = 6$ — do alag accumulation values, toh koi limit nahi ✓. Dono boundary points $∣ r ∣ < 1$ se excluded hain, exactly jaisa theory demand karta hai.

Figure s04 dekho. $S_{N}$ against $N$ ki do side-by-side pictures. Left side ( $r = 1$ ) par coral dots ek straight line mein upar jaate hain — woh runaway infinity ki taraf divergence hai. Right side ( $r = - 1$ ) par lavender dots hamesha $6$ aur $0$ ke beech bounce karte hain, kabhi ek single height par converge nahi karte. Dono master figure s01 ke coral boundary dots par rehte hain.

Cell C7 — real-world word problem

Worked example Ex 7 (C7): ek bouncing ball

Ek ball $10 m$ height se drop ki gayi. Har bounce ke baad woh apni previous height ki $\frac{3}{5}$ tak wapas jaati hai. Total vertical distance kitna travel karta hai woh aaram karne se pehle?

Forecast: woh 10 girta hai, phir baar baar oopar-neeche bounce karta hai, har baar chhota — finite total (ratio $\frac{3}{5} < 1$ ). Kuch tens of metres ki expect karo.

Step 1 — down aur up journeys alag karo. Initial drop $10$ m hai (sirf down). Uske baad, har bounce ek height tak oopar jaata hai phir utni hi height neeche — toh har bounce-height do baar travel hoti hai. Yeh step kyun? Distance har metre count karta hai, oopar ya neeche; pehle drop ka koi matching "up" pehle nahi hai.

Step 2 — geometric series banao. Bounce peaks hain $10 \cdot \frac{3}{5}, 10 \cdot (\frac{3}{5})^{2}, \dots$ . Unke round-trips sum hote hain: $2 \sum_{n = 1}^{\infty} 10 (\frac{3}{5})^{n} .$ Yahan $a = 10 \cdot \frac{3}{5} = 6$ ( $n = 1$ term), $r = \frac{3}{5}$ , $∣ r ∣ < 1$ ✓. Yeh step kyun? Hum $n = 1$ se start karte hain kyunki pehla drop alag handle hota hai; peaks pehle bounce se shuru hote hain.

Step 3 — evaluate karo. $\sum_{n = 1}^{\infty} 10 (\frac{3}{5})^{n} = \frac{6}{1 - 3/5} = \frac{6}{2/5} = 15.$ Total distance $= pehla drop 10 + har bounce ka up+down 2 \times 15 = 10 + 30 = 40 m .$ Yeh step kyun? Doubling har bounce ki rise aur fall account karta hai; akele pehla drop add karna count complete karta hai.

Verify (units + magnitude): saare quantities metres mein, ratio dimensionless ✓. Pehle kuch distances: $10 + 2 (6) + 2 (3.6) + 2 (2.16) \approx 33.5$ , $40$ ki taraf badh raha hai ✓. Finite hai kyunki $∣ r ∣ < 1$ .

Cell C8 — exam twist: $r$ ke liye solve karo

Worked example Ex 8 (C8): sum se ratio nikalo

Ek convergent geometric series ka first term $a = 8$ aur total sum $S = 12$ hai. $r$ nikalo, aur verify karo ki yeh valid (convergent) ratio hai.

Forecast: sum $12$ pehle term $a = 8$ se bada hai, toh terms pehle term ke aage add up karte hain ⇒ $r$ positive aur moderate hai.

Step 1 — formula invert karo. $S = \frac{a}{1 - r}$ se: $12 = \frac{8}{1 - r} \Rightarrow 1 - r = \frac{8}{12} = \frac{2}{3} \Rightarrow r = \frac{1}{3} .$ Yeh step kyun? Formula ek unknown $r$ mein ek equation hai; bas algebraically solve karo.

Step 2 — convergence check optional nahi hai. $∣ r ∣ = \frac{1}{3} < 1$ ✓ — ek legal answer. Agar solving se $∣ r ∣ \geq 1$ milta, toh hum ise spurious reject karte (koi convergent series woh produce nahi karta). Yeh step kyun? Algebraic solution tabhi meaningful hai jab green band ke andar land kare; warna "convergent series" ki premise contradict hoti hai.

Verify: $a = 8, r = \frac{1}{3}$ se sum rebuild karo: $\frac{8}{1 - 1/3} = \frac{8}{2/3} = 12$ ✓. Partial terms $8 + 2.67 + 0.89 + \dots \to 12$ , aur $12 > 8$ jaisa forecast ✓.

Recall Main kis cell mein hoon? (self-quiz)

Ratio exactly 0 hai ::: sum sirf pehla term $a$ hai; formula deta hai $\frac{a}{1 - 0} = a$ (C0). Ratio positive aur 1 se kam ::: converges, sum $a$ se bada (C1). Ratio negative, size 1 se kam ::: converges, inward oscillate karta hai, sum $a$ se chhota (C2). Sum $n = k$ se shuru hota hai, 0 se nahi ::: $a$ ko $n = k$ par term ke roop mein recompute karo (C3). Ek repeating decimal ::: geometric with $r = 1 0^{- (block length)}$ (C4). $∣ r ∣ \geq 1$ with $a \neq = 0$ ::: diverges; $\frac{a}{1 - r}$ ki value ek trap hai (C5). $r = 1$ ::: $S_{N} = a N \to \infty$ ; $r = - 1$ ::: oscillates $a, 0, a, 0$ — dono diverge (C6). Ek bouncing ball / word problem ::: peaks ko geometric series model karo, up+down ke liye double karo, koi un-paired pehla drop add karo (C7). Sum diya hai, $r$ nikalna hai ::: $S = \frac{a}{1 - r}$ invert karo phir $∣ r ∣ < 1$ check karo (C8).

Connections

Geometric series — convergence condition, proof (index 4.3.4) — parent proof jise yeh examples exercise karte hain.
Sequences — limits and convergence — kyun $r^{N} \to 0$ band ke andar (C0–C4) aur bahar fail karta hai (C5–C6).
Partial sums and series convergence — "partial sums ka settling" jo har Verify use karta hai.
Ratio Test — Cell C5 ke liye instant divergence flag.
Power series and radius of convergence — Cell C8 ka " $r$ ke liye solve karo" ek radius ka seed hai.
Repeating decimals as fractions — Cell C4 poora.
Telescoping series — formula ke peeche cancellation.