Power series — centre, radius of convergence, interval of convergence: Worked examples

4.3.14 · D3 · Maths › Calculus III — Sequences & Series › Power series — centre, radius of convergence, interval of co

Yeh page parent topic ka exhaustive worked-example companion hai. Parent ne tumhe tools diye (Ratio Test, Root Test, master limsup formula). Yahan hum ensure karte hain ki koi bhi situation tumhe surprise na kare — hum har tarah ka case list karte hain jo yeh topic throw kar sakta hai, phir har ek ko ek fully worked example se hit karte hain.

Scenario matrix

Neeche har row ek case class hai. Sabse daayaan column us example ka naam batata hai jo us cell mein aata hai.

#	Case class	Kya khaas hai	Covered by
C1	$R = 0$ (degenerate)	sirf centre par converge karta hai	Ex 1
C2	$R = \infty$	har jagah converge karta hai, koi endpoints nahi	Ex 2
C3	Finite $R$ , dono endpoints diverge	open interval $(a - R, a + R)$	Ex 3
C4	Finite $R$ , ek endpoint hold karta hai, ek break	half-open interval	Ex 4
C5	Finite $R$ , dono endpoints hold karte hain	closed interval $[a - R, a + R]$	Ex 5
C6	Centre $a \neq = 0$ , coefficients mein negative sign	shifted, sign bookkeeping	Ex 4, Ex 6
C7	Ratio Test silent, limsup wins	oscillating coefficients	Ex 7
C8	Gap series ( $x^{2 n}$ , sirf even powers)	substitution trick	Ex 8
C9	Word problem (real-world quantity)	$R$ ko physically interpret karo	Ex 9
C10	Exam twist — nested/combined	ek series ka derivative	Ex 10

Neeche ki figure ek number line par exactly woh chaar shapes draw karti hai. Har row ek finite interval hai jo $a$ par centred hai aur same radius $R$ ke saath; rows ke beech sirf fark endpoints ka hai. Ek filled dot ka matlab hai "yeh endpoint included hai" (series wahan converge karti hai); ek open (hollow) dot ka matlab hai "excluded" (wahan diverge karti hai). Chaar coloured bars ko upar se neeche padho aur tumne har possible outcome dekh liya jo ek finite radius produce kar sakta hai — is page ke examples mein se har ek in rows mein se ek mein aata hai.

Figure — Power series — centre, radius of convergence, interval of convergence

Case C1 — radius zero hai

Worked example Ex 1 · cell C1 —

$n = 0 \sum \infty n^{n} x^{n}$ Forecast: coefficients $n^{n}$ bahut tezi se badhte hain. Kya tumhare hisaab se yeh $x$ ke bade set ke liye converge karega, chhote set ke liye, ya almost kuch nahi ke liye?

Step 1. Centre aur coefficients identify karo: $a = 0$ , $c_{n} = n^{n}$ . Yeh step kyun? Baaki sab kuch centre $a$ se distance measure karta hai; yahan series plain $x^{n}$ mein likhi gayi hai, isliye $a = 0$ .

Step 2. Root Test use karo kyunki $n n^{n}$ exactly simplify hota hai (ek coefficient jo khud ek $n$ -th power hai woh $n$ -th root ke liye tailor-made hai, jo outer power ko cancel karta hai — Ratio Test ek messy $(1 + \frac{1}{n})^{n}$ factor chhod deta). $n ∣ c_{n} ∣ ∣ x ∣^{n} = n n^{n} \cdot ∣ x ∣ = n ∣ x ∣.$ Yeh step kyun? $n n^{n} = n$ exactly — root $n$ -th power ko undo kar deta hai.

Step 3. Limit lo: kisi bhi fixed $x \neq = 0$ ke liye, $n ∣ x ∣ \to \infty > 1$ , isliye series har $x \neq = 0$ ke liye diverge karti hai. Yeh step kyun? Root Test kehta hai diverge karo jab limit $1$ se zyada ho; yahan yeh sab large $n$ ke liye $1$ se zyada hai.

Step 4. $x = 0$ par pahle ke baad har term $0$ hai, isliye yeh converge karta hai (convention se $0^{0} = 1$ maana jaata hai). Isliye $R = 0$ , interval $= {0}$ .

Verify: $1/ R = lim sup n n^{n} = lim sup n = \infty$ , aur $1/\infty$ convention se $R = 0$ milta hai. ✓ Match karta hai.

Case C2 — radius infinite hai

Worked example Ex 2 · cell C2 —

$n = 0 \sum \infty \frac{x ^{n}}{( 2 n )!}$ Forecast: denominator mein factorials terms ko crush kar dete hain. Guess: finite $R$ ya infinite $R$ ?

Step 1. $a = 0$ , $c_{n} = 1/ (2 n)!$ . Ratio Test use karo — factorials ratios mein beautifully cancel ho jaate hain. Yeh step kyun? $\frac{( 2 n )!}{( 2 n + 2 )!}$ ek chhota finite product hai, ratio ke liye ideal.

Step 2. $\frac{a _{n + 1}}{a _{n}} = \frac{∣ x ∣ ^{n + 1} / ( 2 n + 2 )!}{∣ x ∣ ^{n} / ( 2 n )!} = \frac{∣ x ∣}{( 2 n + 1 ) ( 2 n + 2 )} .$ Yeh step kyun? $(2 n + 2)! = (2 n + 2) (2 n + 1) (2 n)!$ , isliye factorials ka ratio $\frac{1}{( 2 n + 1 ) ( 2 n + 2 )}$ hai.

Step 3. Jaise $n \to \infty$ yeh har $x$ ke liye $\to 0 < 1$ ho jaata hai. Isliye yeh har jagah converge karta hai: $R = \infty$ . Yeh step kyun? Limit $0$ hai, hamesha $1$ se neeche, $x$ chahe kuch bhi ho — koi finite endpoints nahi hain toh endpoint check exist nahi karta.

Verify: $x = 1$ par number series $\sum 1/ (2 n)! = cosh (1) \approx 1.5431$ , ek finite value — convergence ke saath consistent. ✓ ( $x = t^{2}$ likhne par pata chalta hai yeh $cosh (t) = cosh (x)$ ki Maclaurin series hai; note karo argument $x$ hai, $x$ nahi.)

Case C3 — finite $R$ , dono endpoints diverge (open interval)

Worked example Ex 3 · cell C3 —

$n = 0 \sum \infty \frac{( x - 5 ) ^{n}}{3 ^{n}}$ Forecast: yeh disguise mein ek Geometric Series hai, ratio $\frac{x - 5}{3}$ . $R$ kya hai, aur dono ends par kya hoga?

Step 1. $a = 5$ , $c_{n} = 1/ 3^{n}$ . Rewrite karo: $\sum (\frac{x - 5}{3})^{n}$ — ratio $r = \frac{x - 5}{3}$ ke saath ek geometric series. Yeh step kyun? Geometric Series form ko pehchanna turant answer deta hai: yeh converge karta hai iff $∣ r ∣ < 1$ .

Step 2. $∣ r ∣ < 1 ⟺ ∣ x - 5∣ < 3 \Rightarrow R = 3$ , open interval $(2, 8)$ . Yeh step kyun? Geometric convergence exactly $∣ r ∣ < 1$ hai; distance inequality solve karne par radius milti hai.

Step 3. Endpoint $x = 8$ : $r = 1$ , series hai $\sum 1^{n} = 1 + 1 + 1 + \dots$ — diverges (terms $\to 0$ nahi jaate). Endpoint $x = 2$ : $r = - 1$ , series hai $\sum (- 1)^{n}$ — diverges (terms $\to 0$ nahi jaate). Yeh step kyun? $∣ x - a ∣ = R$ par Ratio Test silent hai, isliye hum substitute karte hain; dono mein terms divergence test fail kar dete hain.

Step 4. Dono ends break karte hain $\Rightarrow$ open interval $(2, 8)$ .

Verify: midpoint $x = 5$ (centre) par sum $= \sum 0^{n} = 1$ , finite ✓. $x = 6$ par, $r = 1/3$ , sum $= \frac{1}{1 - 1/3} = \frac{3}{2}$ , finite ✓.

Case C4 & C6 — ek endpoint hold karta hai, ek break karta hai; shifted centre

Worked example Ex 4 · cells C4, C6 —

$n = 1 \sum \infty \frac{( - 1 ) ^{n} ( x + 4 ) ^{n}}{n 5 ^{n}}$ Forecast: centre negative hai, aur ek alternating sign hai. Shape predict karo: $[)$ , $(]$ , $[]$ ya $()$ ?

Step 1. $a = - 4$ (kyunki $x + 4 = x - (- 4)$ ), $c_{n} = \frac{( - 1 ) ^{n}}{n 5 ^{n}}$ . Yeh step kyun? Centre hamesha us $x$ ki value ke roop mein padho jo base ko zero banata hai: $x + 4 = 0 \Rightarrow x = - 4$ .

Step 2. $∣ a_{n} ∣ = \frac{∣ x + 4 ∣ ^{n}}{n 5 ^{n}}$ par Ratio Test: $\frac{a _{n + 1}}{a _{n}} = \frac{∣ x + 4∣}{5} \cdot \frac{n}{n + 1} \to \frac{∣ x + 4∣}{5} .$ Yeh step kyun? $∣ x + 4 ∣^{n}$ ek factor mein collapse ho jaata hai, $5^{n}$ ek $\frac{1}{5}$ mein, aur $\frac{n}{n + 1} \to 1$ .

Step 3. Converge karo jab $\frac{∣ x + 4∣}{5} < 1 \Rightarrow ∣ x + 4∣ < 5 \Rightarrow R = 5$ , interval $(- 9, 1)$ . Yeh step kyun? Ratio Test (absolute) convergence exactly tab guarantee karta hai jab ratio limit $1$ se neeche ho; $\frac{∣ x + 4∣}{5} < 1$ set karke solve karne par distance $∣ x + 4∣ < 5$ milti hai, jiska bound hi radius $R = 5$ hai aur jiska solution set interval $(a - R, a + R) = (- 9, 1)$ hai.

Step 4. Endpoint $x = 1$ : $∣ x + 4∣ = 5$ , term $= \frac{( - 1 ) ^{n} 5 ^{n}}{n 5 ^{n}} = \frac{( - 1 ) ^{n}}{n}$ — alternating harmonic, Alternating Series Test se converges (terms $↓ 0$ ). Yeh conditional convergence hai. Endpoint $x = - 9$ : $x + 4 = - 5$ , term $= \frac{( - 1 ) ^{n} ( - 5 ) ^{n}}{n 5 ^{n}} = \frac{( - 1 ) ^{n} ( - 1 ) ^{n}}{n} = \frac{1}{n}$ — $p = 1$ ke saath harmonic p-series, diverges. Yeh step kyun? Dono endpoints genuinely alag series hain; tumhe har ek ko substitute karke sahi test choose karna hoga.

Step 5. Left break karta hai, right hold karta hai $\Rightarrow (- 9, 1]$ .

Verify: $x = - 9$ par dono signs $(- 1)^{n} (- 1)^{n} = 1$ collapse ho jaate hain, divergent harmonic series milti hai ✓. $x = 1$ par $\sum (- 1)^{n} / n = - ln 2 \approx - 0.693$ , finite ✓.

Case C5 — dono endpoints hold karte hain (closed interval)

Worked example Ex 5 · cell C5 —

$n = 1 \sum \infty \frac{( x - 1 ) ^{n}}{n ^{2}}$ Forecast: denominator mein $n^{2}$ strong hai. Guess karo ki dono endpoints survive karte hain ya nahi.

Step 1. $a = 1$ , $c_{n} = 1/ n^{2}$ . Yahan Ratio Test use karo: coefficient ek simple power $n^{- 2}$ hai, isliye ratio $\frac{c _{n + 1}}{c _{n}} = \frac{n ^{2}}{( n + 1 ) ^{2}}$ ek clean rational function hai jiska obvious limit hai — $n 1/ n^{2}$ lene se kahin zyada aasaan, jisme $n^{1/ n} \to 1$ fact chahiye. $\frac{a _{n + 1}}{a _{n}} = ∣ x - 1∣ \cdot \frac{n ^{2}}{( n + 1 ) ^{2}} \to ∣ x - 1∣.$ Yeh step kyun? $\frac{n ^{2}}{( n + 1 ) ^{2}} \to 1$ , isliye poori limit sirf $∣ x - 1∣$ hai.

Step 2. Converge karo jab $∣ x - 1∣ < 1 \Rightarrow R = 1$ , interval $(0, 2)$ .

Step 3. Endpoint $x = 2$ : $\sum 1/ n^{2}$ — $p = 2 > 1$ ke saath p-series, converges. Endpoint $x = 0$ : $\sum (- 1)^{n} / n^{2}$ — absolutely converge karta hai (iska absolute version same $p = 2$ series hai) absolute convergence se. Yeh step kyun? Jab $p > 1$ toh p-series hold karta hai; aage ka sign already-absolutely-convergent series ko break nahi kar sakta.

Step 4. Dono hold karte hain $\Rightarrow [0, 2]$ .

Verify: $x = 2$ par $\sum 1/ n^{2} = π^{2} /6 \approx 1.6449$ , finite ✓. $x = 0$ par $\sum (- 1)^{n} / n^{2} = - π^{2} /12 \approx - 0.8225$ , finite ✓.

Case C6 (revisited) — base ke andar sign, careful bookkeeping

Worked example Ex 6 · cell C6 —

$n = 0 \sum \infty \frac{( 3 - x ) ^{n}}{2 ^{n} ( n + 1 )}$ Forecast: base $3 - x$ hai, $x - 3$ nahi. Kya isse kuch flip hota hai? Centre kahan hai?

Step 1. Note karo $3 - x = - (x - 3)$ , isliye $(3 - x)^{n} = (- 1)^{n} (x - 3)^{n}$ . Centre $a = 3$ . Yeh step kyun? $(- 1)^{n}$ factor out karne par true centre dikhta hai aur ek hidden alternating sign ek endpoint par reveal hoti hai — ek classic trap.

Step 2. $∣ a_{n} ∣ = \frac{∣ x - 3 ∣ ^{n}}{2 ^{n} ( n + 1 )}$ par Ratio Test: $\frac{a _{n + 1}}{a _{n}} = \frac{∣ x - 3∣}{2} \cdot \frac{n + 1}{n + 2} \to \frac{∣ x - 3∣}{2} .$ Step 3. $\frac{∣ x - 3∣}{2} < 1 \Rightarrow R = 2$ , interval $(1, 5)$ . Yeh step kyun? Ratio Test convergence precisely tab deliver karta hai jab uski limit $1$ se kam ho; inequality $\frac{∣ x - 3∣}{2} < 1$ rearrange karke distance condition $∣ x - 3∣ < 2$ milti hai, jiska bound radius $R = 2$ hai aur jiska solution interval $(a - R, a + R) = (1, 5)$ hai.

Step 4. Endpoint $x = 5$ : $3 - x = - 2$ , term $= \frac{( - 2 ) ^{n}}{2 ^{n} ( n + 1 )} = \frac{( - 1 ) ^{n}}{n + 1}$ — alternating, converges (Alternating Series Test). Endpoint $x = 1$ : $3 - x = 2$ , term $= \frac{2 ^{n}}{2 ^{n} ( n + 1 )} = \frac{1}{n + 1}$ — harmonic (shifted), diverges.

Step 5. $(1, 5]$ .

Verify: $x = 1$ par $\sum \frac{1}{n + 1}$ = harmonic tail, diverges ✓. $x = 5$ par $\sum \frac{( - 1 ) ^{n}}{n + 1} = ln 2 \approx 0.693$ , finite ✓.

Case C7 — Ratio Test silent hai, limsup wins karta hai

Worked example Ex 7 · cell C7 — coefficients

$c_{n} = {3^{n} 1 n even n odd$ , series $\sum c_{n} x^{n}$ Forecast: ratio $∣ c_{n + 1} / c_{n} ∣$ wildly swing karega. Kya Ratio Test yahan $R$ de sakta hai?

Step 1. Ratio compute karo: even→odd se yeh $\frac{1}{3 ^{n}}$ hai (tiny), odd→even se yeh $3^{n + 1}$ hai (huge). Koi limit exist nahi karta. Yeh step kyun? Yeh Ratio formula ka failure mode directly demonstrate karta hai.

Step 2. Master root/limsup tool (upar Cauchy–Hadamard formula) par switch karo, jo hamesha exist karta hai: $n ∣ c_{n} ∣ = {n 3^{n} = 3 n 1 = 1 n even n odd$ Yeh step kyun? $n$ -th root $3^{n}$ ko constant $3$ mein convert kar deta hai, oscillation ko tame karta hai.

Step 3. Values ${1, 3}$ par cluster karti hain; sabse bada cluster point $lim sup n ∣ c_{n} ∣ = 3$ hai. Yeh step kyun? Limsup kisi bhi subsequence ki sabse badi limit hai — yahan $3$ .

Step 4. Cauchy–Hadamard: $R = \frac{1}{lim sup n ∣ c _{n} ∣} = \frac{1}{3}$ , abhi-tak-open interval $(- \frac{1}{3}, \frac{1}{3})$ .

Step 5. Endpoints $x = \pm \frac{1}{3}$ : sirf even terms dekho, $c_{2 k} x^{2 k} = 3^{2 k} x^{2 k} = (3 x)^{2 k}$ . $x = \pm \frac{1}{3}$ par yeh har $k$ ke liye $(\pm 1)^{2 k} = 1$ hai, isliye infinitely many terms $1$ ke barabar hain aur $0$ nahi jaate. Divergence test fail ⇒ series dono endpoints par diverge karti hai. Yeh step kyun? $∣ x ∣ = R$ par koi bhi test formula apply nahi hota, isliye hum terms ko directly inspect karte hain; terms ki ek subsequence $1$ par tikी rehti hai jo already convergence ko khatam kar deti hai.

Step 6. Dono ends break karte hain $\Rightarrow$ interval $(- \frac{1}{3}, \frac{1}{3})$ .

Verify: $1/ R = 3 \Rightarrow R = 1/3$ . Sanity: even terms $3^{n} x^{n} = (3 x)^{n}$ ek geometric part form karte hain jise $∣3 x ∣ < 1 ⟺ ∣ x ∣ < 1/3$ chahiye — exactly $R = 1/3$ , aur $∣ x ∣ = 1/3$ par woh geometric part ka ratio $∣3 x ∣ = 1$ hai, diverge karta hai. ✓

Case C8 — gap series (sirf even powers)

Worked example Ex 8 · cell C8 —

$n = 1 \sum \infty \frac{x ^{2 n}}{n 9 ^{n}}$ Forecast: koi $x^{2 n + 1}$ term nahi hai. Kya $R$ is tarah aata hai $∣ x ∣ < kuch$ ya $x^{2} < kuch$ ?

Step 1. $u = x^{2}$ substitute karo. Tab series hai $\sum \frac{u ^{n}}{n 9 ^{n}}$ , $u$ mein ek ordinary power series. Yeh step kyun? $x^{2 n}$ wali gap series standard series ban jaati hain jab $x^{2}$ ko ek single variable ka naam do.

Step 2. $u$ mein Ratio Test: $\frac{a _{n + 1}}{a _{n}} = \frac{∣ u ∣}{9} \cdot \frac{n}{n + 1} \to \frac{∣ u ∣}{9}$ . Converge karo jab $∣ u ∣ < 9$ . Yeh step kyun? $u^{n}$ ek factor mein cancel ho jaata hai, $9^{n}$ $\frac{1}{9}$ mein, aur $\frac{n}{n + 1} \to 1$ ; Ratio Test tab converge karta hai jab yeh limit $1$ se neeche ho, jo exactly $∣ u ∣ < 9$ hai.

Step 3. Wapas translate karo: $∣ u ∣ = ∣ x^{2} ∣ = x^{2} < 9 \Rightarrow ∣ x ∣ < 3$ . Isliye $R = 3$ , interval $(- 3, 3)$ $x$ mein. Yeh step kyun? $x^{2} < 9 ⟺ ∣ x ∣ < 3$ ; $x$ mein radius $u$ mein radius ka square root hai.

Step 4. Endpoints $x = \pm 3$ : tab $x^{2} = 9$ , term $= \frac{9 ^{n}}{n 9 ^{n}} = \frac{1}{n}$ — harmonic, dono ends par diverges. Yeh step kyun? Even powers dono endpoints ko same value $x^{2} = 9$ dete hain, isliye woh identically behave karte hain.

Step 5. $(- 3, 3)$ .

Verify: $x^{2} = 9$ par $\sum 1/ n$ , diverges ✓. $x = 1$ par: $\sum \frac{1}{n 9 ^{n}} = - ln (1 - 1/9) = ln (9/8) \approx 0.1178$ , finite ✓.

Case C9 — ek real-world word problem

Worked example Ex 9 · cell C9 — ek settling deposit

Ek financial model year $n$ mein $\frac{r ^{n}}{n}$ ke proportional amount pay karta hai jahan $r = \frac{x - 100}{20}$ measure karta hai ki interest offset $x$ (same units mein) baseline $100$ se kitna door hai. Total value hai $V (x) = n = 1 \sum \infty \frac{( x - 100 ) ^{n}}{n 2 0 ^{n}}$ . Kin $x$ ke liye total finite rehta hai (model blow up nahi karta)? Forecast: "safe operating range" $x = 100$ ke around ek interval hai. Uski width guess karo.

Step 1. $a = 100$ , $c_{n} = 1/ (n 2 0^{n})$ . Ratio Test: $\frac{∣ x - 100∣}{20} \cdot \frac{n}{n + 1} \to \frac{∣ x - 100∣}{20}$ . Yeh step kyun? Same mechanics jaise Ex 4/6; offset centre $100$ hai.

Step 2. Converge karo jab $∣ x - 100∣ < 20 \Rightarrow R = 20$ , interval $(80, 120)$ . Yeh step kyun? $R = 20$ baseline $x = 100$ se maximum offset hai total diverge hone se pehle — model ki tolerance.

Step 3. Endpoint $x = 120$ : $\sum \frac{2 0 ^{n}}{n 2 0 ^{n}} = \sum \frac{1}{n}$ — diverges (model exactly top edge par blow up karta hai). Endpoint $x = 80$ : $\sum \frac{( - 20 ) ^{n}}{n 2 0 ^{n}} = \sum \frac{( - 1 ) ^{n}}{n}$ — alternating, converges (bottom edge safe hai).

Step 4. Safe range $[80, 120)$ : offset neeche 20 ja sakta hai (inclusive) lekin strictly +20 se neeche rehna chahiye.

Verify: $x = 80$ par $\sum (- 1)^{n} / n = - ln 2 \approx - 0.693$ , finite ✓. $x = 120$ par harmonic, diverges ✓. Radius $= 20$ , $∣ c_{n} / c_{n + 1} ∣ = \frac{( n + 1 ) 2 0 ^{n + 1}}{n 2 0 ^{n}} \to 20$ se match karta hai ✓.

Case C10 — exam twist: ek series differentiate karo

Worked example Ex 10 · cell C10 —

$f (x) = n = 1 \sum \infty \frac{x ^{n}}{n}$ ; $f^{'} (x)$ ka $R$ nikalo Forecast: Term-by-term differentiation coefficients change karta hai. Kya yeh radius change karta hai?

Step 1. $f$ mein $a = 0$ , $c_{n} = 1/ n$ hai; iska Ratio Test radius $R = lim ∣ c_{n} / c_{n + 1} ∣ = lim \frac{n + 1}{n} = 1$ hai. Yeh step kyun? Original radius establish karo taaki hum compare kar sakein.

Step 2. Term by term differentiate karo: $f^{'} (x) = \sum_{n = 1}^{\infty} \frac{n x ^{n - 1}}{n} = \sum_{n = 1}^{\infty} x^{n - 1} = \sum_{m = 0}^{\infty} x^{m}$ . Yeh step kyun? $\frac{d}{d x} \frac{x ^{n}}{n} = x^{n - 1}$ ; new series padhne ke liye $m = n - 1$ se reindex karo — ek plain Geometric Series.

Step 3. Derivative geometric hai ratio $x$ ke saath, isliye uska radius bhi $R = 1$ hai. Yeh step kyun? Theorem guarantee karta hai ki radius differentiation se unchanged rehta hai — endpoints change ho sakte hain par $R$ nahi.

Step 4. Endpoint check (woh differ karte HAIN): $f$ khud interval $[- 1, 1)$ mein hai ( $x = - 1$ converge karta hai, $x = 1$ harmonic diverge karta hai). Lekin $f^{'} = \sum x^{m}$ dono $x = \pm 1$ par diverge karta hai, interval $(- 1, 1)$ . Yeh step kyun? Same radius, narrower interval — ek favourite exam gotcha.

Verify: $f^{'} (x) = \frac{1}{1 - x}$ (geometric sum) for $∣ x ∣ < 1$ , aur indeed $\frac{d}{d x} (- ln (1 - x)) = \frac{1}{1 - x}$ — consistent hai, kyunki $\sum x^{n} / n = - ln (1 - x)$ . ✓ Dono ke liye radius $1$ ✓.

Recall

Recall Kaunsa shape? Series ko interval se match karo

$\sum \frac{( x - 2 ) ^{n}}{n ^{2}}$ → interval shape kya hai? ::: Closed $[1, 3]$ — $p = 2$ dono ends par hold karta hai. $\sum \frac{( x - 2 ) ^{n}}{3 ^{n}}$ → interval shape kya hai? ::: Open $(- 1, 5)$ — geometric, dono ends par $\sum (\pm 1)^{n}$ milta hai, diverge karta hai. Ratio $∣ c_{n + 1} / c_{n} ∣$ oscillate karta hai — kaun sa tool use karein? ::: root/limsup Cauchy–Hadamard formula $1/ R = lim sup n ∣ c_{n} ∣$ . Ek gap series $\sum a_{n} x^{2 n}$ jisme $u$ -radius $9$ hai — $x$ mein radius? ::: $9 = 3$ . Kya term-by-term differentiation $R$ change karta hai? ::: Nahi — $R$ preserve hota hai; sirf endpoints change ho sakte hain.