Geometric series — convergence condition, proof: Exercises

4.3.4 · D4 · Maths › Calculus III — Sequences & Series › Geometric series — convergence condition, proof

Yeh page ek self-test ladder hai jo parent proof ke upar bani hai. Har problem clearly batata hai ki kya find karna hai. Poora worked solution ek collapsible [!recall]- callout mein hai — pehle problem khud try karo, PHIR unfold karo. Levels recognising a geometric series se lekar synthesising kai ideas ek saath tak chadhte hain.

Yahan sab kuch parent note ke do facts par tikaa hai, jo yahan dobara likhe hain taaki tumhe yeh page kabhi chhodni na pade:

Neeche har problem mein do skills kaam aati hain:

$a$ find karo: starting index plug karo term mein aur pehla actual number compute karo.
$r$ find karo: kisi bhi term ko uske pehle wali term se divide karo — $r = \frac{term}{previous term}$ .

Level 1 — Recognition

(Kya tum $a$ , $r$ spot kar sakte ho, aur bata sakte ho ki converge hoga ya diverge?)

L1.1 — Kya $5 + 10 + 20 + 40 + \dots$ geometric hai? Agar haan, toh $a$ aur $r$ do, aur converge/diverge batao.

Recall Solution L1.1

Geometric test: consecutive terms divide karo. $\frac{10}{5} = 2$ , $\frac{20}{10} = 2$ , $\frac{40}{20} = 2$ . Ratio har baar same number hai, toh haan, yeh geometric hai.

$a = 5$ (pehla term jo likha gaya).
$r = 2$ . Converge karega? $∣ r ∣ = 2 \geq 1$ , toh terms badhte hain aur series diverge karti hai.

L1.2 — $n = 0 \sum \infty 4 (\frac{1}{3})^{n}$ ke liye, pehle teen terms likhao, aur $a$ aur $r$ do.

Recall Solution L1.2

$n = 0, 1, 2$ plug karo: $4 \cdot (\frac{1}{3})^{0} = 4, 4 \cdot (\frac{1}{3})^{1} = \frac{4}{3}, 4 \cdot (\frac{1}{3})^{2} = \frac{4}{9} .$ Toh series hai $4 + \frac{4}{3} + \frac{4}{9} + \dots$ .

$a = 4$ ( $n = 0$ wala term).
$r = \frac{1}{3}$ . Kyunki $∣ r ∣ = \frac{1}{3} < 1$ , yeh converge karta hai.

L1.3 — Inme se kaun converge karta hai? (a) $n = 0 \sum \infty (0.9)^{n}$ , (b) $n = 0 \sum \infty (- 0.9)^{n}$ , (c) $n = 0 \sum \infty (1.1)^{n}$ , (d) $n = 0 \sum \infty (- 1)^{n}$ .

Recall Solution L1.3

Convergence sirf $∣ r ∣ < 1$ par depend karti hai.

(a) $∣ r ∣ = 0.9 < 1$ → converges.
(b) $∣ - 0.9∣ = 0.9 < 1$ → converges ( $r$ ka sign is baat par asar nahi dalta ki converge karega ya nahi, sirf value par dalta hai).
(c) $∣ r ∣ = 1.1 \geq 1$ → diverges (terms badhte hain).
(d) $∣ r ∣ = 1$ → diverges ( $r^{N}$ oscillate karta hai $1, - 1, 1, - 1, \dots$ aur kabhi settle nahi hota).

Level 2 — Application

(Formula mein sahi se plug karo — starting index dhyan se dekho.)

L2.1 — $n = 0 \sum \infty 6 (\frac{1}{4})^{n}$ evaluate karo.

Recall Solution L2.1

$a = 6$ ( $n = 0$ wala term), $r = \frac{1}{4}$ , $∣ r ∣ < 1$ ✓. $S = \frac{a}{1 - r} = \frac{6}{1 - \frac{1}{4}} = \frac{6}{\frac{3}{4}} = 6 \cdot \frac{4}{3} = 8.$

L2.2 — $n = 1 \sum \infty 6 (\frac{1}{4})^{n}$ evaluate karo. (L2.1 jaisi hi series hai lekin $n = 1$ se shuru hoti hai.)

Recall Solution L2.2

Kya badla: index ab $n = 1$ se shuru hota hai, toh $n = 0$ wala term ( $6$ ) missing hai. Pehla likha hua term $n = 1$ wala term hai: $a = 6 \cdot (\frac{1}{4})^{1} = \frac{6}{4} = \frac{3}{2} .$ Ratio abhi bhi $r = \frac{1}{4}$ hai. $S = \frac{a}{1 - r} = \frac{\frac{3}{2}}{\frac{3}{4}} = \frac{3}{2} \cdot \frac{4}{3} = 2.$ Sanity check: L2.1 minus uska $n = 0$ term L2.2 ke barabar hona chahiye: $8 - 6 = 2$ ✓.

L2.3 — $n = 0 \sum \infty 3 (- \frac{1}{2})^{n}$ evaluate karo.

Recall Solution L2.3

$a = 3$ , $r = - \frac{1}{2}$ , $∣ r ∣ = \frac{1}{2} < 1$ ✓. Denominator hai $1 - r = 1 - (- \frac{1}{2}) = 1 + \frac{1}{2} = \frac{3}{2}$ . Negative $r$ denominator ko bada bana deta hai, toh sum $a$ se neeche aata hai: $S = \frac{3}{\frac{3}{2}} = 2 < 3. ✓$

L2.4 — $0. \overline{45}$ (yaani $0.454545 \dots$ ) ko geometric series use karke fraction mein likho. Dekho Repeating decimals as fractions.

Recall Solution L2.4

Repeating block " $45$ " ko do decimal places ke chunks mein group karo: $0. \overline{45} = \frac{45}{100} + \frac{45}{10000} + \frac{45}{1000000} + \dots = \sum_{n = 1}^{\infty} 45 (\frac{1}{100})^{n} .$ Pehla likha hua term $a = \frac{45}{100}$ , ratio $r = \frac{1}{100}$ . $S = \frac{45/100}{1 - 1/100} = \frac{45/100}{99/100} = \frac{45}{99} = \frac{5}{11} .$ Check: $5 \div 11 = 0.4545 \dots$ ✓.

Level 3 — Analysis

(Tod ke dekho, re-index karo, ya parameters ke baare mein soocho.)

L3.1 — Kin real $x$ ke liye $n = 0 \sum \infty (x - 2)^{n}$ converge karta hai, aur wahaan uska sum kya hai? (Yeh model Power series and radius of convergence ke peeche hai.)

Recall Solution L3.1

Yahan ratio hai $r = x - 2$ aur $a = (x - 2)^{0} = 1$ . Convergence ke liye $∣ r ∣ < 1$ chahiye: $∣ x - 2∣ < 1 ⟺ - 1 < x - 2 < 1 ⟺ 1 < x < 3.$ Us interval par: $S = \frac{a}{1 - r} = \frac{1}{1 - ( x - 2 )} = \frac{1}{3 - x} .$ Centre $x = 2$ hai aur radius $1$ hai — neeche figure dekho. Yeh ek horizontal number line dikhata hai: accent-red open segment $x = 1$ se $x = 3$ tak exactly wahi jagah hai jahan series converge karti hai, black centre mark $x = 2$ par hai aur bahar ke dono regions "diverges" label kiye hain.

Figure — Geometric series — convergence condition, proof

Figure (s01): number line jo convergence sirf open interval $1 < x < 3$ par dikhata hai, $x = 2$ par centred with radius $1$ ; red segment ke bahar series diverge karti hai.

L3.2 — Ek geometric series mein $n = 0 \sum \infty a r^{n} = 9$ aur second term $a r = 2$ hai. $a$ aur $r$ find karo.

Recall Solution L3.2

Do equations: $\frac{a}{1 - r} = 9 (1), a r = 2 (2) .$ (1) se: $a = 9 (1 - r)$ . (2) mein substitute karo: $9 (1 - r) r = 2 ⟹ 9 r - 9 r^{2} = 2 ⟹ 9 r^{2} - 9 r + 2 = 0.$ Factor karo: $(3 r - 1) (3 r - 2) = 0$ , toh $r = \frac{1}{3}$ ya $r = \frac{2}{3}$ . Dono $∣ r ∣ < 1$ satisfy karte hain, toh do valid solutions hain:

$r = \frac{1}{3} \Rightarrow a = 9 (1 - \frac{1}{3}) = 9 \cdot \frac{2}{3} = 6.$
$r = \frac{2}{3} \Rightarrow a = 9 (1 - \frac{2}{3}) = 9 \cdot \frac{1}{3} = 3.$ Check: $(6, \frac{1}{3})$ : $a r = 2$ ✓, sum $= 6/ (2/3) = 9$ ✓. $(3, \frac{2}{3})$ : $a r = 2$ ✓, sum $= 3/ (1/3) = 9$ ✓.

L3.3 — $n = 0 \sum \infty \frac{2 ^{n} + 3 ^{n}}{6 ^{n}}$ ko do geometric series mein split karo aur evaluate karo.

Recall Solution L3.3

Kyun split karein: $\frac{2 ^{n} + 3 ^{n}}{6 ^{n}} = \frac{2 ^{n}}{6 ^{n}} + \frac{3 ^{n}}{6 ^{n}} = (\frac{2}{6})^{n} + (\frac{3}{6})^{n} = (\frac{1}{3})^{n} + (\frac{1}{2})^{n}$ . Convergent series ka finite sum term by term add kiya ja sakta hai. $\sum_{n = 0}^{\infty} (\frac{1}{3})^{n} = \frac{1}{1 - \frac{1}{3}} = \frac{3}{2}, \sum_{n = 0}^{\infty} (\frac{1}{2})^{n} = \frac{1}{1 - \frac{1}{2}} = 2.$ Total: $\frac{3}{2} + 2 = \frac{7}{2}$ .

Level 4 — Synthesis

(Geometric series ko kisi aur idea ke saath combine karo.)

L4.1 — Ek ball $h = 10$ m height se giraayi jaati hai aur har bounce par apni pichli height ka $\frac{3}{5}$ tak uthti hai. Woh rukne se pehle kitna total vertical distance travel karti hai?

Recall Solution L4.1

Imagine karo: pehli drop $10$ m hai (sirf neeche). Uske baad, har bounce upar peak tak aur wapas neeche jaata hai — toh har subsequent height do baar count hoti hai.

Figure (s02): ball ka path — ek accent-red vertical "first drop" of $10$ m ek baar count hua, phir black rebound arcs $6, 3.6, \dots$ heights ke, har ek upar aur neeche (do baar); total $40$ m nikalta hai.

Pehli drop: $10$ .
Peaks: $10 \cdot \frac{3}{5}, 10 \cdot (\frac{3}{5})^{2}, \dots$ , har ek upar aur neeche travel hua. $D = 10 + 2 \sum_{n = 1}^{\infty} 10 (\frac{3}{5})^{n} .$ Tail geometric hai jiska pehla likha hua term $a = 10 \cdot \frac{3}{5} = 6$ , ratio $r = \frac{3}{5}$ : $\sum_{n = 1}^{\infty} 10 (\frac{3}{5})^{n} = \frac{6}{1 - \frac{3}{5}} = \frac{6}{\frac{2}{5}} = 15.$ $D = 10 + 2 (15) = 40 m .$

L4.2 — $x$ ke liye solve karo: $n = 0 \sum \infty (\frac{x}{x + 3})^{n} = 4$ , aur confirm karo ki ratio actually $∣ r ∣ < 1$ satisfy karta hai.

Recall Solution L4.2

Ratio $r = \frac{x}{x + 3}$ , $a = 1$ . Sum condition: $\frac{1}{1 - \frac{x}{x + 3}} = 4.$ Denominator simplify karo: $1 - \frac{x}{x + 3} = \frac{( x + 3 ) - x}{x + 3} = \frac{3}{x + 3}$ . Toh $\frac{1}{\frac{3}{x + 3}} = \frac{x + 3}{3} = 4 ⟹ x + 3 = 12 ⟹ x = 9.$ Ratio check: $r = \frac{9}{12} = \frac{3}{4}$ , aur $∣ r ∣ = \frac{3}{4} < 1$ ✓ — sum genuinely valid hai.

Level 5 — Mastery

(Re-derive karo, ya koi subtle/degenerate case handle karo.)

L5.1 — Partial sum $S_{N} = a + a r + \dots + a r^{N - 1}$ ko scratch se telescoping subtraction use karke re-derive karo, phir ek line mein explain karo ki limit ke liye $∣ r ∣ < 1$ kyun chahiye.

Recall Solution L5.1

$S_{N}$ aur $r S_{N}$ aligned likho: $S_{N} = a + a r + a r^{2} + \dots + a r^{N - 1},$ $r S_{N} = a r + a r^{2} + \dots + a r^{N - 1} + a r^{N} .$ Subtract karo — har middle term $a r^{k}$ dono mein appear karta hai aur cancel ho jaata hai, sirf $S_{N}$ ka pehla term aur $r S_{N}$ ka aakhri term bachta hai: $S_{N} - r S_{N} = a - a r^{N} ⟹ S_{N} (1 - r) = a (1 - r^{N}) ⟹ S_{N} = \frac{a ( 1 - r ^{N} )}{1 - r} (r \neq = 1) .$ $∣ r ∣ < 1$ limit ke liye kyun: jab $N \to \infty$ , $r^{N} \to 0$ tabhi hota hai jab $∣ r ∣ < 1$ ; phir $S_{N} \to \frac{a ( 1 - 0 )}{1 - r} = \frac{a}{1 - r}$ . Agar $∣ r ∣ \geq 1$ , toh $r^{N}$ $0$ par nahi jaata (badhta hai ya oscillate karta hai), toh limit fail ho jaati hai. Dekho Sequences — limits and convergence.

L5.2 — Shifted series $n = 2 \sum \infty 5 (\frac{2}{5})^{n}$ do tareekon se evaluate karo: (a) seedha correct first term use karke, (b) "full sum minus pehle do terms" ke roop mein.

Recall Solution L5.2

$r = \frac{2}{5}$ , $∣ r ∣ < 1$ ✓. (a) Direct. Pehla likha hua term $n = 2$ par hai: $a = 5 (\frac{2}{5})^{2} = 5 \cdot \frac{4}{25} = \frac{4}{5}$ . $S = \frac{\frac{4}{5}}{1 - \frac{2}{5}} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{5} \cdot \frac{5}{3} = \frac{4}{3} .$ (b) Full minus early terms. $n = 0$ se full sum: $\frac{5}{1 - \frac{2}{5}} = \frac{5}{\frac{3}{5}} = \frac{25}{3}$ . $n = 0$ aur $n = 1$ terms subtract karo: $5 + 5 \cdot \frac{2}{5} = 5 + 2 = 7$ . $S = \frac{25}{3} - 7 = \frac{25 - 21}{3} = \frac{4}{3} . ✓$ Dono raaste agree karte hain — re-indexing ka ek clean check.

L5.3 (Degenerate case) — $n = 0 \sum \infty a r^{n}$ kya hoga jab $a = 0$ ho? Saare $r$ ke liye discuss karo, $∣ r ∣ \geq 1$ bhi include karo.

Recall Solution L5.3

Agar $a = 0$ hai, toh har term $0$ hai chahe $r$ kuch bhi ho: $0 + 0 + 0 + \dots = 0$ . Partial sums sab $0$ hain, toh woh trivially $0$ par converge karte hain chahe $∣ r ∣ \geq 1$ ho. Isliye parent note mein divergence statement ke saath caveat hai "with $a \neq = 0$ ." Convergence gate $∣ r ∣ < 1$ tabhi matter karta hai jab actually kuch nonzero sum karna ho.

Connections

Geometric series — convergence condition, proof (parent — formula aur proof)
Partial sums and series convergence (L5.1 re-derivation)
Sequences — limits and convergence (kyun $r^{N} \to 0$ ke liye $∣ r ∣ < 1$ chahiye)
Ratio Test (constant-ratio idea ka generalization)
Power series and radius of convergence (L3.1 ka interval of convergence)
Repeating decimals as fractions (L2.4)
Telescoping series (L5.1 mein cancellation)

Skill Map

Neeche diagram woh decision path hai jo tum kisi bhi candidate series par run karte ho. Words mein: ek given series se shuru karo; consecutive terms divide karo taaki test karo ki ratio constant hai ya nahi (kya yeh geometric hai?). Agar haan, toh $r$ aur $a$ find karo. Phir gate $∣ r ∣ < 1$ check karo. Agar gate fail ho aur $a \neq = 0$ , toh series diverge karti hai; agar gate fail ho lekin $a = 0$ (degenerate case), toh sum simply $0$ hai. Agar gate pass ho, toh sum $\frac{a}{1 - r}$ hai — lekin phir bhi tumhe (i) $a$ actual starting index se padhna hai, aur (ii) parameter problems ke liye apna answer substitute karke confirm karna hai ki $∣ r ∣ < 1$ hai.

Recall Koi bhi answer submit karne se pehle ek-line self-check

Kya maine (1) confirm kiya ki ratio constant hai, (2) actual starting index se $a$ compute kiya, (3) $∣ r ∣ < 1$ check kiya, aur (4) — parameters ke liye — substitute karke $∣ r ∣ < 1$ confirm kiya? Answer ::: Agar koi bhi box unticked hai, toh answer unverified hai.