Geometric series — convergence condition, proof: Question bank

4.3.4 · D5 · Maths › Calculus III — Sequences & Series › Geometric series — convergence condition, proof

Yeh ek conceptual workout hai Geometric series — convergence condition & proof ke liye. Yahaan koi bhaari arithmetic nahi hai — har item ek misconception ya ek boundary case ko pakadta hai. Har prompt ko padhkar apna jawab zor se bolkar confirm karo, PHIR reveal karo. Answer side mein hamesha reason diya hota hai, sirf verdict nahi.

Shuru karne se pehle, parent note ke teen anchors yaad rakho:

Series hai $\sum a r^{n} = a + a r + a r^{2} + \dots$ , jahan $== a ==$ woh actual first term hai jo tum likhte ho aur $== r ==$ hai common ratio (har term ÷ pehle wali term).
Closed form $\frac{a}{1 - r}$ sirf tab valid hai jab $== ∣ r ∣ < 1 ==$ , kyunki iske derivation mein secretly $r^{N} \to 0$ use kiya gaya tha.
"Converges" ka matlab hai ki partial sums $S_{N}$ ka sequence (dekho Partial sums and series convergence) ek single finite number ke paas jaata hai jab $N \to \infty$ .

Sahi ya galat — justify karo

Sahi ya galat: Agar kisi series ke terms chhote se chhhote hote jaayein, toh series zaroor converge karegi.

Galat. Shrinking terms necessary hain par sufficient nahi — harmonic series

1 + \frac{1}{2} + \frac{1}{3} + \dots

ke terms

\to 0

hain phir bhi diverge karti hai. Geometric series ke liye shrinking fast honi chahiye (constant ratio

∣ r ∣ < 1

), jo sirf

\to 0

se zyada strong condition hai.

Sahi ya galat:

\sum_{n = 0}^{\infty} a r^{n}

converge karta hai agar aur sirf agar

- 1 < r < 1

Sahi. Condition

∣ r ∣ < 1

exactly

- 1 < r < 1

mein khulti hai, dono endpoints exclude hain (

r = 1

deta hai

S_{N} = a N \to \infty

;

r = - 1

mein

r^{N}

oscillate karta hai

1, - 1, 1, \dots

aur kabhi settle nahi karta).

Sahi ya galat: Negative

r

wali geometric series ka sum hamesha negative hota hai.

Galat.

∣ r ∣ < 1

ke saath sum hai

\frac{a}{1 - r}

; negative

r

1 - r > 1

hota hai, toh sum

a

se neeche aata hai par

a

ka sign rakhta hai. Jaise

\sum 3 (- \frac{1}{3})^{n} = \frac{9}{4} > 0

Sahi ya galat: Agar

∣ r ∣ < 1

toh saare terms positive hone chahiye.

Galat.

r

ka sign convergence se independent hai.

r = - \frac{1}{2}

ke saath terms alternating sign hain, phir bhi

∣ r ∣ < 1

hai toh converge karta hai — alternation actually partial cancellation se help karta hai.

Sahi ya galat: Formula

S_{N} = \frac{a ( 1 - r ^{N} )}{1 - r}

har real

r

ke liye hold karta hai.

Galat. Yeh har

r

ke liye hold karta hai siwaaye

r = 1

ke, jahan

1 - r = 0

se divide karna illegal hai; wahan

S_{N} = a N

directly aata hai. (Yeh formula ek finite sum hai, toh isko koi convergence assumption nahi chahiye — sirf

r \neq = 1

Sahi ya galat: Pehle term

a

ko double karne se infinite sum bhi double ho jaata hai.

Sahi. Sum

\frac{a}{1 - r}

mein

a

linear hai;

r

(jo convergence control karta hai) untouched rehta hai, toh total usi factor se scale karta hai. Isliye

a

ek pure "volume knob" hai.

Sahi ya galat: Agar ek geometric series diverge karti hai, toh iska partial-sum formula

\frac{a ( 1 - r ^{N} )}{1 - r}

galat hai.

Galat. Partial-sum formula har finite

N

ke liye sahi rehta hai (jab tak

r \neq = 1

). Jo fail hota hai woh hai limit lena:

r^{N}

0

nahi jaata, toh

S_{N}

ka limit exist nahi karta. Formula galat nahi hai; limit step unavailable hai.

Error pakdo

Flaw dhundo: "

\sum_{n = 0}^{\infty} 2 \cdot 3^{n}

ke liye,

\frac{a}{1 - r} = \frac{2}{1 - 3} = - 1

milta hai, toh sum

- 1

hai."

Yahaan

∣ r ∣ = 3 \geq 1

hai, toh series diverge karti hai — formula applicable nahi hai. Positive growing terms ka sum negative nahi ho sakta; "

- 1

" ek meaningless output hai uss assumption (

r^{N} \to 0

) ka jo false hai.

Flaw dhundo: "

\sum_{n = 1}^{\infty} (\frac{1}{2})^{n}

ke liye, coefficient

1

hai, toh

a = 1

aur

S = \frac{1}{1 - 1/2} = 2

Pehla actual term (

n = 1

plug karo) hai

(\frac{1}{2})^{1} = \frac{1}{2}

, toh

a = \frac{1}{2}

hai, jo deta hai

S = \frac{1/2}{1/2} = 1

. Bahar wala coefficient grab karna instead of pehla actual term compute karna — yeh classic

a

-misread hai.

Flaw dhundo: "

\sum_{n = 0}^{\infty} a r^{n}

aur

\sum_{n = 1}^{\infty} a r^{n}

mein same

a

aur

r

hain, toh dono ka sum same hai."

Dono exactly

n = 0

wale term se different hain, jo ki

a

hai. Toh

\sum_{n = 1}^{\infty} a r^{n} = \frac{a}{1 - r} - a = \frac{a r}{1 - r}

. Index start change hone se pehla term change hota hai, isliye sum bhi.

Flaw dhundo: "Series

1 - 1 + 1 - 1 + \dots

mein

a = 1, r = - 1

hai, toh sum

\frac{1}{1 - ( - 1 )} = \frac{1}{2}

hai."

r = - 1

ke saath

∣ r ∣ = 1

hai, not

< 1

, toh series diverge karti hai: iske partial sums

1, 0, 1, 0, \dots

oscillate karte hain aur kabhi

\frac{1}{2}

ke paas nahi jaate.

\frac{1}{2}

ek formal fiction hai (Cesàro/Abel average), sum nahi.

Flaw dhundo: "Jab

r \to 1^{-}

, sum

\frac{a}{1 - r}

ek large finite number ke paas jaata hai."

Jab

r \to 1^{-}

, denominator

1 - r \to 0^{+}

hota hai, toh

\frac{a}{1 - r} \to + \infty

(

a > 0

ke liye) — koi finite limit nahi hai. Terms barely shrink karte hain, toh pile unbounded grow karta hai.

Flaw dhundo: "Telescoping trick

S_{N} - r S_{N}

ke liye

∣ r ∣ < 1

chahiye."

Nahi — cancellation pure algebra hai finite sum par aur kisi bhi

r \neq = 1

ke liye kaam karta hai. Condition

∣ r ∣ < 1

sirf baad mein chahiye, jab hum

N \to \infty

lete hain aur

r^{N} \to 0

require karte hain.

Flaw dhundo: "

\sum_{n = 0}^{\infty} 0 \cdot r^{n}

diverge karta hai jab

r = 5

kyunki

∣ r ∣ \geq 1

a = 0

ke saath har term

0

hai, toh

S_{N} = 0

har

N

ke liye aur sum

0

hai — yeh trivially converge karta hai. "

∣ r ∣ \geq 1

hone par diverge karta hai" rule ka fine print hai

a \neq = 0

Why questions

\frac{a}{1 - r}

ka derivation exactly

∣ r ∣ = 1

par kyun break down hota hai?

Kyunki closed form aata hai

lim_{N \to \infty} (1 - r^{N})

se, jisko

r^{N} \to 0

chahiye.

r = 1

par,

r^{N} = 1

hamesha rehta hai (koi shrink nahi);

r = - 1

par,

r^{N}

oscillate karta hai. Sirf

∣ r ∣ < 1

r^{N} \to 0

force karta hai (dekho Sequences — limits and convergence).

Hum pehle

N

terms sum karke limit kyun lete hain, instead of directly infinitely many terms add karne ke?

Infinitely many additions perform nahi kar sakte. "Infinite sum" defined hai

lim_{N \to \infty} S_{N}

ke roop mein — woh value jiske paas finite partial sums jaate hain. Yahi series convergence ka poora matlab hai.

Negative ratio sum ko first term

a

se chhota kyun banata hai (jab

a > 0

Negative

r

terms ko alternating sign deta hai, toh partial cancellation hoti hai. Algebraically

1 - r > 1

hota hai, toh

\frac{a}{1 - r} < a

— bada denominator directly us cancellation ko reflect karta hai.

Hum sirf term-by-term

r^{N} \to 0

se convergence conclude kyun nahi kar sakte bina partial-sum formula ke?

Individual terms

\to 0

hona kabhi guarantee nahi karta ki sum converge karega (harmonic series counterexample hai). Humein explicit

S_{N} = \frac{a ( 1 - r ^{N} )}{1 - r}

chahiye taaki dekh sakein ki poora partial sum finite limit ke paas jaata hai.

Geometric series ko power series ka "model" kyun kehte hain?

a = 1, r = x

rakhne par

\sum x^{n} = \frac{1}{1 - x}

milta hai

∣ x ∣ < 1

ke liye. Convergence boundary

∣ x ∣ < 1

radius of convergence ban jaata hai; dekho Power series and radius of convergence.

Telescoping cancellation mein sirf do surviving terms kyun bachte hain?

S_{N} - r S_{N}

mein, har internal

a r^{k}

ek baar

S_{N}

mein aur ek baar

r S_{N}

mein aata hai, cancel hoke. Sirf

S_{N}

ka sabse pehla term (

a

) aur

r S_{N}

ka sabse aakhri (

a r^{N}

) ka koi partner nahi hota — same idea jaise ek Telescoping series mein.

Ratio Test geometric series ke liye same threshold

∣ r ∣ < 1

kyun deta hai?

Consecutive terms ka ratio exactly

r

hai (constant), toh Ratio Test ki limit

L = ∣ r ∣

hai; yeh

L < 1

ke liye convergence declare karta hai — identical boundary. Geometric series special case hai jahan ratio kabhi vary nahi karta.

Edge cases

Edge case:

r = 0

hone par sum kya hoga?

Pehle ke baad har term vanish ho jaata hai (

0^{n} = 0

for

n \geq 1

), toh series sirf

a

hai. Formula agree karta hai:

\frac{a}{1 - 0} = a

. (Convention se

0^{0} = 1

n = 0

term ke liye.)

Edge case: Agar

a = 0

aur

∣ r ∣ \geq 1

ho, maan lo

r = 10

Series hai

0 + 0 + 0 + \dots = 0

, jo

0

pe converge karta hai. Divergence rule "

∣ r ∣ \geq 1 \Rightarrow

diverges" assume karta hai

a \neq = 0

; zero first term har term ko kill kar deta hai

r

ki parwah kiye bina.

Edge case: Kya

\sum_{n = 0}^{\infty} r^{n}

endpoint

r = 1

par converge karta hai?

Nahi. Yeh

1 + 1 + 1 + \dots

ban jaata hai, toh

S_{N} = N \to \infty

(- 1, 1)

ke dono endpoints fail karte hain, toh convergence ka interval open hai.

Edge case:

\sum 7 (\frac{1}{10})^{n}

ke liye

n = 1

se shuru karte hue, kya boundary case

∣ r ∣ = 1

kabhi arise hota hai?

Nahi — yahaan

r = \frac{1}{10}

hai, comfortably

∣ r ∣ < 1

ke andar, toh yeh

\frac{7/10}{9/10} = \frac{7}{9}

mein converge karta hai. Isliye har repeating decimal ek fraction hoti hai (dekho Repeating decimals as fractions); iska ratio hamesha

1 0^{- k}

hota hai jisme

∣ r ∣ < 1

Edge case: Kya ek geometric series

a

ke opposite sign wali value mein converge kar sakti hai?

Nahi. Sum

\frac{a}{1 - r}

mein

1 - r > 0

hota hai jab bhi

∣ r ∣ < 1

, toh sum ka sign hamesha

a

ke sign se match karta hai. Koi bhi "positive terms ka negative sum" signal karta hai ki tumne

∣ r ∣ < 1

check ignore kar diya.

Edge case:

r = - 1

par

S_{N}

N

badhne ke saath kya hota hai?

a = 1

ke saath, partial sums cycle karte hain

1, 0, 1, 0, \dots

— yeh bounded hain par kabhi ek value par settle nahi karte, toh series oscillation se diverge karti hai. Boundedness akele convergence nahi hai.

Recall Upar ke har trap ke liye ek-line litmus test

$\frac{a}{1 - r}$ likhne se pehle: (1) start index plug karke true first term $a$ identify karo, (2) confirm karo $== ∣ r ∣ < 1 ==$ , (3) tabhi formula apply karo. Step 2 skip karo aur tum "prove" kar doge nonsense jaise $2 + 6 + 18 + \dots = - 1$ .

Connections

Parent: convergence condition & proof
Sequences — limits and convergence — woh $lim r^{N}$ jis par har trap lean karta hai
Partial sums and series convergence — "converges" defined via $S_{N}$
Ratio Test — same $∣ r ∣ < 1$ threshold, generalised
Power series and radius of convergence — geometric series as the model $\sum x^{n} = \frac{1}{1 - x}$
Repeating decimals as fractions — edge case with $r = 1 0^{- k}$
Telescoping series — proof ke peeche cancellation