Geometric series — convergence condition, proof

4.3.4 · Maths › Calculus III — Sequences & Series

Ek geometric series woh sum hoti hai jisme har term apne pehle wali term ka ek fixed multiple hoti hai. "Infinite sums jo sense karte hain" ki poori theory yahaan se shuru hoti hai — yeh sabse simple infinite series hai jise hum poori tarah samajh sakte hain.

The Setup — HUM KYA SUM KAR RAHE HAIN?

Partial Sum ko Scratch se Derive karna

Hum seedha infinitely many cheezein kabhi sum nahi karte. Hum pehle $N$ terms sum karte hain, ek formula lete hain, phir limit lete hain. Pehle $N$ terms (indices $0$ se $N - 1$ tak): $S_{N} = a + a r + a r^{2} + \dots + a r^{N - 1}$

$r = 1$ ko exclude kyun karte hain? Tab har term sirf $a$ hoti hai, toh $S_{N} = a N$ — ek alag, trivial case hai (yeh diverge karta hai jab tak $a = 0$ na ho).

Limit Lena — Convergence Condition

Infinite sum ko define kiya jaata hai $n = 0 \sum \infty a r^{n} = N \to \infty lim S_{N}$ ke roop mein.

$lim_{N \to \infty} S_{N} = \frac{a}{1 - r} lim_{N \to \infty} (1 - r^{N})$

Sab kuch $N \to \infty lim r^{N}$ par depend karta hai. $r^{N}$ kaisa behave karta hai?

Agar $∣ r ∣ < 1$ : size mein $1$ se chhoti cheez se baar baar multiply karne se $r^{N} \to 0$ ho jaata hai.
Agar $∣ r ∣ > 1$ : $∣ r^{N} ∣ = ∣ r ∣^{N} \to \infty$ , koi limit nahi.
Agar $r = 1$ : $r^{N} = 1$ hamesha, lekin yeh excluded case hai ( $S_{N} = a N \to \pm \infty$ ).
Agar $r = - 1$ : $r^{N}$ , $1, - 1, 1, - 1, \dots$ flip karta hai — kabhi settle nahi hota.

Figure — Geometric series — convergence condition, proof

Worked Examples

Common Mistakes (Steel-manned)

Recall Feynman: ek 12-saal ke bachche ko explain karo

Socho tum ek deewar ki taraf chal rahe ho. Pehle aadha raasta chalte ho, phir baaki ka aadha, phir us ka aadha, hamesha ke liye. Har kadam pehle se chhota hota hai. Infinitely many kadam hone ke bawajood, tum poori deewar tak ki distance cover karte ho — ek finite number! Yeh $r = \frac{1}{2}$ wali geometric series hai. Lekin agar har kadam bada hota (jaise double), toh tum hamesha chalte rehte aur kabhi nahi rukte — woh series "diverge" karti hai. Magic rule: total settle hone ke liye kadam chhote hote rehne chahiye ( $r$ ki size $1$ se kam).

Active Recall — Flashcards

\sum a r^{n}

ke liye convergence condition kya hai?

∣ r ∣ < 1

(jab

a \neq = 0

); warna diverge karta hai.

Ek convergent geometric series ka sum kya hota hai?

\frac{a}{1 - r}

jahan

a

pehla term hai,

r

common ratio hai.

Partial sum

S_{N}

derive karo: kaun sa trick use hota hai?

S_{N} - r S_{N}

compute karo; beech ke terms cancel ho jaate hain (telescope), deta hai

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

∣ r ∣ < 1

convergence guarantee kyun karta hai?

Kyunki tab

r^{N} \to 0

hota hai, toh

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

ho jaata hai.

r = 1

par kya hota hai?

Formula invalid hai; har term

a

hai toh

S_{N} = a N \to \infty

(diverges unless

a = 0

r = - 1

par kya hota hai?

r^{N}

1, - 1, 1, \dots

oscillate karta hai; partial sums settle nahi hote — diverges.

\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots

ka sum?

1

(yahaan

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

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

Geometric series ke through

0. \overline{7}

ko fraction likho.

\frac{7/10}{1 - 1/10} = \frac{7}{9}

∣ r ∣ > 1

hone par

\frac{a}{1 - r}

plug karna galat kyun hai?

Derivation ne

r^{N} \to 0

assume kiya tha, jo fail hota hai; limit exist nahi karti toh value meaningless hai.

Connections

Sequences — limits and convergence (woh $lim r^{N}$ define karta hai jis par hum depend kiye)
Partial sums and series convergence (series = partial sums ki limit)
Ratio Test ("constant ratio" idea ko varying ratios tak generalize karta hai)
Power series and radius of convergence (geometric series model hai: $\sum x^{n} = \frac{1}{1 - x}$ )
Repeating decimals as fractions
Telescoping series (proof mein same cancellation idea use hoti hai)