Geometric series — convergence condition, proof: Visual walkthrough

4.3.4 · D2 · Maths › Calculus III — Sequences & Series › Geometric series — convergence condition, proof

Is page mein poora result $n = 0 \sum \infty a r^{n} = \frac{a}{1 - r}$ sirf ek pile of shrinking (ya growing) rectangles se rebuild kiya gaya hai — kuch bhi nahi se shuru karke. Koi formula tab tak use nahi hoga jab tak hum dekh na lein ki woh kyun sach hona chahiye. Agar koi symbol aata hai, hum us picture ki taraf point karte hain jahan woh rehta hai.

Parent: 4.3.4 · Geometric series.

Step 1 — WHAT are we even adding up?

WHAT. Ek geometric series numbers ki ek list hoti hai jahan agla number paane ke liye aap usi ek fixed number se multiply karte ho — har baar. Pehle number ko $a$ kaho (the "starting height"), aur us fixed multiplier ko $r$ kaho (the "shrink/grow factor").

Har term ko exactly $1$ width ki ek bar ke roop mein draw karo, axis par khadi, jiska height us term ki value ke barabar ho. Kyunki har bar ki width $1$ hai, uska area uski height ke barabar hai — to shuru se hi, "ek term ki value" aur "uski bar ka area" ek hi number hain. List yeh hai: $term 0 a, term 1 a r, term 2 a r^{2}, term 3 a r^{3}, \dots$

Picture par har symbol ko dekho:

$a$ pehli bar ki height (= area) hai (woh tall burnt-orange wali bar jo left mein hai).
$r$ ratio hai "next bar $\div$ this bar". Har neighbouring pair ka ek hi ratio hota hai — yahi "geometric" word ka poora matlab hai.
$a r^{n}$ matlab hai " $a$ se shuru karo, phir $r$ se kul $n$ baar multiply karo". Chhota exponent $n$ sirf ek counter hai — kitni baar shrink hua.

WHY unit-width bars draw karo. Kyunki tab "numbers jodna" "areas stack karna" ban jaata hai: jab width $1$ ho har jagah, to total sum total shaded area hai. Yeh ek algebra ke sawaal ko picture ke sawaal mein badal deta hai, aur heights aur areas ko is poore page mein interchangeable rakhta hai.

Figure — Geometric series — convergence condition, proof

Step 2 — WHY shrinking is the whole game (and what a negative $r$ looks like)

WHAT. Teen piles of unit-width bars compare karo. Beech mein, $∣ r ∣ < 1$ aur $r > 0$ (mano $r = \frac{1}{2}$ ): har bar pichli bar ka ek shrunken copy hai, to heights zero ki taraf race karti hain. Right mein, $∣ r ∣ > 1$ (mano $r = 2$ ): har bar badi hai, to heights explode karti hain. Aur ek alag strip mein, ek negative ratio $r = - \frac{1}{2}$ : bars size mein shrink hoti hain, lekin ab woh flip ho jaati hain — ek bar upar point karti hai, agla neeche axis ke below, hamesha alternating.

Yahaan $∣ r ∣$ (padho " $r$ ki size", sign ignore karke) yeh honestly measure karta hai ki "terms kitni tezi se shrink hoti hain". Hume size chahiye, $r$ khud nahi, kyunki ek negative $r$ bhi bars ko shrink karta hai agar uski size $1$ se kam ho — woh bas unhe axis ke neeche flip karta hai, to up-bars aur down-bars aapas mein partly cancel ho jaati hain.

WHY yeh matter karta hai. $∣ r ∣ < 1$ ke liye total area finite hai (bars jaldi gayab ho jaati hain, bounded ink) chahe $r$ positive ho ya negative — negative $r$ ke saath below-axis bars area subtract karti hain, isliye running total apni limit ki taraf upar-neeche wobble karta hai. $∣ r ∣ > 1$ ke liye ink endless hai. To woh ek fact jo sab kuch decide karta hai yeh hai:

$Does ∣ r ∣ < 1 ? ⟺ do the bars die out fast enough?$

PICTURE. Left/middle: bars axis ki taraf nose-dive kar rahi hain, area capped. Ek alag panel mein $r = - \frac{1}{2}$ dikhta hai jisme bars axis ke upar aur neeche alternating hain (shrinking). Right: bars top se bahar ja rahi hain.

Step 3 — WHAT is a partial sum $S_{N}$ ?

WHAT. Hum kabhi bhi physically infinitely many bars nahi jod sakte, isliye pehli $N$ bars jodte hain aur us running total ko $S_{N}$ kehte hain: $S_{N} = bar 0 a + bar 1 a r + bar 2 a r^{2} + \dots + bar N - 1 a r^{N - 1} .$

Last symbol dhyan se padho: sabse bada exponent $N - 1$ hai, nahi $N$ . Kyun? Kyunki humne bar $0$ se count karna shuru kiya, isliye "pehli $N$ bars" bars $0, 1, 2, \dots, N - 1$ hain — kul $N$ bars, $N$ se ek kadam neeche khatam hoti hain.

WHY yeh karo. Ek infinite sum define hota hai as $S_{N}$ kya approach karta hai jab $N$ badhta hai. Isliye pehle $S_{N}$ ke liye ek clean closed formula chahiye — ek expression, koi " $\dots$ " nahi. Steps 4–5 us formula ki talaash karte hain.

PICTURE. Pehli $N$ bars solid shaded hain; bar $N - 1$ ke aage ki ghostly grey bars woh tail hai jo humne abhi nahi jodi. $S_{N}$ sirf solid area hai.

Step 4 — WHY poore sum ko $r$ se multiply karo (the telescoping trick)

WHAT. Usi pile $S_{N}$ ko lo aur shift karo: har term ko $r$ se multiply karo. Yeh poori staircase ko ek bar aage khiskaata hai: $r S_{N} = was bar 0 a r + was bar 1 a r^{2} + \dots + was bar N - 2 a r^{N - 1} + new tail a r^{N} .$

$S_{N}$ aur $r S_{N}$ ko line up karo taaki equal-height bars ek doosre ke seedha upar baith jayen.

WHY. Kyunki ab $r S_{N}$ ki almost har bar ka $S_{N}$ mein ek identical twin hai. Agar hum subtract karte hain, to woh saare matched twins zero ho jaate hain. Sirf woh bars bachti hain jinका koi partner nahi:

$S_{N}$ ke left mein ek unmatched bar hai: original $a$ (bar $0$ — $r S_{N}$ ne use kabhi produce nahi kiya).
$r S_{N}$ ke right mein ek unmatched bar hai: brand-new $a r^{N}$ (woh $S_{N}$ ke end se ek aage nikal gaya).

Yeh deliberate line-up-then-cancel hi "telescoping" ka matlab hai — ek lambi chain sirf apne do ends par collapse ho jaati hai. (Same trick jaise Telescoping series mein.)

PICTURE. Do staircases ek doosre ke upar draw ki hain; matched bars ko faint teal ties se link kiya gaya hai, do akele survivors ( $a$ far left par, $a r^{N}$ far right par) plum mein circle kiye hue hain.

Step 5 — WHAT algebra wapas deta hai: the closed formula

WHAT. $S_{N} - r S_{N} = a - a r^{N}$ ke dono sides ko factor karo.

Left side: dono terms $S_{N}$ share karte hain, isliye $S_{N} - r S_{N} = S_{N} (1 - r)$ . Right side: dono terms $a$ share karte hain, isliye $a - a r^{N} = a (1 - r^{N})$ .

Setting them equal aur common factor $(1 - r)$ se divide karte hue: $running total S_{N} = \frac{a first bar ( 1 - r ^{N} tail size )}{cannot be 0 1 - r} .$

Har piece padho:

$a$ sab kuch scale karta hai (pehli bar double karo, total double — obvious).
$r^{N}$ us pehli bar ki height hai jo humne nahi jodi; yeh "kitna bacha hua hai" wala knob hai.
$1 - r$ neeche woh jagah hai jahan shrinking magic bake in hoti hai.

WHY $r = 1$ exclude karo. Humne $(1 - r)$ se divide kiya. Agar $r = 1$ ho to yeh zero se divide karna hai — illegal. Aur yeh hona hi chahiye ek alag case: $r = 1$ par har bar ki height $a$ hi hoti hai, isliye $N$ equal bars dete hain $S_{N} = a N$ , jo bas hamesha badhta rehta hai (jab tak $a = 0$ na ho).

PICTURE. Finished formula ko is tarah dikhaya gaya hai: (poora rectangle of area $\frac{a}{1 - r}$ ) minus (area $\frac{a r ^{N}}{1 - r}$ ka ek scaled copy jo un-added tail ko represent karta hai).

Recall

$(1 - r)$ neeche kyun aata hai Yeh woh leftover factor hai cancel karne ke baad. Geometrically, $\frac{1}{1 - r}$ woh "amplification" hai jo height $a$ ki ek bar ko poora infinite stack mein badal deta hai jab tail gayab ho jaati hai.

Step 6 — WHAT tail $r^{N}$ ka hota hai jab hum hamesha ke liye jodte hain

WHAT. Infinite sum define hota hai as $N$ ke bina bound ke badhne par $S_{N}$ ki limit (yeh "list ki limit" wala idea Sequences — limits and convergence mein rehta hai). Formula mein sirf ek piece $N$ par depend karta hai: tail $r^{N}$ . To use dekho.

Paanch cases — sab ke sab, koi gap nahi (note karo ki $∣ r ∣ > 1$ sign ke hisaab se split hota hai):

case	$r^{N}$ kya karta hai	picture
$∣ r ∣ < 1$	$r^{N} \to 0$ : leftover bar kuch bhi nahi ho jaata	tail dot axis par slide ho jaata hai
$r > 1$	$r^{N} \to + \infty$ : leftover bar upar explode karta hai	dots top se ud jaate hain
$r < - 1$	$∣ r^{N} ∣ \to \infty$ aur sign flip: bars bhi badhti hain aur up/down alternate bhi karti hain	dots alag-alag sign ke saath door phink jaate hain
$r = 1$	$r^{N} = 1$ hamesha; $S_{N} = a N$	equal bars ki staircase, koi ceiling nahi
$r = - 1$	$r^{N}$ flip karta hai $1, - 1, 1, - 1, \dots$	dot ping-pong karta hai, kabhi settle nahi hota

WHY. $1$ se kam size ke number se baar baar multiply karne se guarantee hai decay zero ki taraf — har multiply ek fixed fraction remove karta hai. $1$ se badi size se multiply karna guarantee karta hai blow-up; agar woh bada multiplier negative bhi ho ( $r < - 1$ ), to har step dono magnify aur sign flip karta hai, to tail ever-larger positive aur negative values ki taraf jump karti hai — doubly hopeless. Do boundary values $r = \pm 1$ par kuch bhi shrink nahi hota, isliye sum kabhi settle nahi ho sakta.

PICTURE. Ek axis har case ke liye $r^{N}$ sequence ko dots ki tarah dikhata hai: teal dots $0$ ki taraf aa rahe hain; orange dots upar escape kar rahe hain ( $r > 1$ ); $r < - 1$ ka trace kabhi dur upar kabhi dur neeche alternately phink raha hai; plum dots height $1$ par stuck hain; plum dots $\pm 1$ ke beech oscillate kar rahe hain.

Step 7 — WHAT hum par land karte hain: the convergence condition

WHAT. Sirf ek surviving good case $∣ r ∣ < 1$ mein, tail $r^{N} \to 0$ , isliye $lim_{N \to \infty} S_{N} = \frac{a ( 1 - 0 )}{1 - r} = \frac{a}{1 - r} .$

WHY yeh iff hai. Agar $∣ r ∣ \geq 1$ (aur $a \neq = 0$ ) to tail vanish nahi hoti, limit exist nahi karti, aur koi finite sum nahi hota. Isliye formula valid hai exactly jab $∣ r ∣ < 1$ — ek bhi value zyada nahi: $\sum_{n = 0}^{\infty} a r^{n} = \frac{a}{1 - r} ⟺ ∣ r ∣ < 1.$

Picture par sanity checks (sign ke sab cases):

$r$ bahut chhota positive ⇒ $1 - r \approx 1$ ⇒ sum $\approx a$ : sirf pehli bar aur crumbs. ✓
$r \to 1^{-}$ ⇒ $1 - r \to 0^{+}$ ⇒ sum $\to + \infty$ : bars barely shrink hoti hain, ink pile hoti hai. ✓
$r < 0$ ⇒ $1 - r > 1$ ⇒ sum $< a$ : bars axis ke upar aur neeche alternate karti hain aur partly cancel hoti hain. ✓

Step 8 — WHAT agar pehli bar ki height zero ho? ( $a = 0$ , the trivial edge)

WHAT. Mano pehli bar ki height $a = 0$ hai. Tab $a r^{n} = 0$ har $n$ ke liye: koi bars hi nahi hain, sirf axis par ek flat line hai. Sum hai $0 + 0 + 0 + \dots = 0$ , chahe $r$ kuch bhi ho — ek bada growing $r$ bhi zero ko multiply karta hai aur zero hi rehta hai.

WHY mention karo. Taaki koi reader wonder na kare ki jab $a$ vanish ho to rules break hote hain kya. Nahi hote — woh bas trivial ho jaate hain:

Formula abhi bhi agree karta hai: $\frac{a}{1 - r} = \frac{0}{1 - r} = 0$ kisi bhi $r \neq = 1$ ke liye. ✓
Convergence condition " $∣ r ∣ < 1$ " $a \neq = 0$ ke liye stated hai; jab $a = 0$ hai to series sab $r$ ke liye (including $∣ r ∣ \geq 1$ ) $0$ par converge karti hai, kyunki literally kuch bhi blow up karne ke liye nahi hai.
Ek genuinely-excluded spot $r = 1$ abhi bhi $S_{N} = a N = 0$ deta hai jab $a = 0$ — abhi bhi $0$ , abhi bhi theek.

PICTURE. Ek flat axis jisme sab bars ki height $0$ hai; running total $r$ se regardless $0$ par pinned rehta hai.

The one-picture summary

Sab kuch ek saath: shrinking bars, shifted copy jo telescope karti hai, leftover tail $\frac{a r ^{N}}{1 - r}$ jo pighal rahi hai, aur poora rectangle $\frac{a}{1 - r}$ jo khada reh jaata hai — sirf band $∣ r ∣ < 1$ ke andar valid (trivial $a = 0$ case hamesha $0$ sum karta hai).

Recall Feynman retelling — walkthrough plain words mein

Socho bars ek row mein khadi hain, har ek unit wide, isliye ek bar ki height hi uska area hai. Pehli bar ki height $a$ hai, aur har agla bar pichle se $r$ times jitna tall hai. Numbers jodna total shaded area measure karne ke barabar hai. Agar bars shrink hoti rehti hain ( $r$ ki size ek se kam), to ink ki ek limited amount hai — ek finite total. Agar $r$ negative hai lekin size mein chhota hai, to bars axis ke upar aur neeche flip karti hain aur partly cancel hoti hain, lekin phir bhi ek finite total tak shrink hoti hain. Agar woh same rehti hain ya badhti hain, to ink kabhi khatam nahi hoti — aur agar woh growing $r$ negative bhi hai, to bars dono blow up aur flip karti hain, doubly hopeless. Pehli kuch bars ka total nikalne ke liye, main poori staircase ki ek doosri copy banata hoon lekin ek bar nudge karke (yahi hai " $r$ se multiply karna"). Jab main shifted copy ko original se subtract karta hoon, har middle bar ka ek twin hai aur cancel ho jaata hai — sirf pehli bar $a$ aur ek leftover tail bar $a r^{N}$ bachti hai. Theek karne par milta hai $S_{N} = \frac{a ( 1 - r ^{N} )}{1 - r}$ . Ab socho bars hamesha ke liye jodte hain. Sirf ek piece badalta hai: leftover tail $r^{N}$ . Agar bars shrink hoti hain, woh tail zero ho jaati hai aur total $\frac{a}{1 - r}$ par settle ho jaata hai. Agar woh shrink nahi hoti, tail kabhi nahi marti aur koi total nahi hota. Aur agar shuruaat mein hi pehli bar ki height zero thi, to poori cheez sirf zero hai. Isliye woh ek rule jo poora show chalata hai yeh hai: multiplier ki size ek se kam honi chahiye.

Connections

Geometric series — convergence condition, proof (parent — poora result aur examples)
Sequences — limits and convergence ( $lim r^{N}$ ka behaviour Step 6 mein)
Partial sums and series convergence (ek series apne $S_{N}$ ki limit hai)
Telescoping series (Step 4 ka shift-and-cancel move)
Ratio Test (constant-ratio idea ko generalise karta hai)
Power series and radius of convergence (model case $\sum x^{n} = \frac{1}{1 - x}$ )
Repeating decimals as fractions (ek favourite application)