Root test: Foundations — Hinglish deep dive

4.3.13 · D1 · Maths › Calculus III — Sequences & Series › Root test

Parent note ki ek bhi line padhne se pehle, tumhe har woh symbol chahiye jo woh tumhare samne throw karta hai. Neeche har ek notation piece hai, build-order mein. Koi bhi cheez use hone se pehle draw ki jaati hai.

1. Ek series: " $\sum a_{n}$ " ka matlab kya hai

Poori, precise notation yeh hai: $\sum_{n = 1}^{\infty} a_{n} = a_{1} + a_{2} + a_{3} + \dots$ Ise piece by piece padho:

$\sum$ — Greek capital sigma, matlab "inhe sab add karo".
$n = 1$ neeche — index position $1$ se shuru hota hai.
$\infty$ upar — yeh hamesha ke liye chalti rehti hai (koi last term nahi).
$a_{n}$ daayein taraf — general term, har position pe number dene wala rule.

Toh $n = 1 \sum \infty a_{n}$ exactly yeh kehta hai: " $n$ ko $1, 2, 3, \dots$ se endlessly chalao, aur har $a_{n}$ ko add karte jao." Jab range obvious ho toh hum ise sirf $\sum a_{n}$ likh dete hain.

$a_{n}$ = general term = position $n$ pe baitha number. Yeh ek rule hai, jaise $a_{n} = \frac{1}{n ^{2}}$ ka matlab hai position 1 pe $\frac{1}{1}$ hai, position 2 pe $\frac{1}{4}$ hai, position 3 pe $\frac{1}{9}$ hai, …
$n$ = index, ek whole-number counter $1, 2, 3, \dots$ jo positions ke through chalta hai.

Position n ::: woh counting index 1, 2, 3, … jo har term ko label karta hai

\sum_{n = 1}^{\infty} a_{n}

mein, upar ka

\infty

kya matlab rakhta hai

sum kabhi khatam nahi hoti — tum har

n = 1, 2, 3, \dots

ke liye ek term add karte ho

Symbol

\sum a_{n}

ka matlab hai

saare terms

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

ko hamesha ke liye add karo

2. "Converges" aur "diverges" — do fates

Upar ke figure mein, right panel ka flattening curve convergence hai; climbing curve divergence hai. Poora root test ek machine hai jo decide karta hai in do fates mein se kaun si series ko milti hai — bina tumhare poora add kiye.

3. Absolute value $∣ a_{n} ∣$ aur "absolute convergence"

Root test $a_{n}$ ki jagah $∣ a_{n} ∣$ kyun use karta hai? Kyunki terms negative ho sakti hain (jaise $(- 1)^{n} / 2^{n}$ ), aur root test measure karta hai terms kitni badi hain, unki direction nahi. Hum pehle sign strip karte hain.

Root ka conclusion " $L < 1 \Rightarrow$ converges absolutely" exactly yahi idea hai: humne sizes $∣ a_{n} ∣$ ko tame kiya, toh signs trouble cause nahi kar saktein.

4. Geometric series $\sum r^{n}$ — woh yardstick jiske against sab kuch measure hota hai

Yahan pehli baar symbol $r$ appear hota hai — isse pehle sab kuch sirf "ek fixed fraction" ki baat karta tha. Picture ek bouncing ball ki hai jo har bounce mein apni height ka fixed fraction $r$ rakhti hai.

Ise sirf state karne ki bajaye earn karte hain. Hum classic $S - r S$ trick se partial sum ki exact value derive karte hain.

Yeh yardstick kyun hai: geometric sabse saaf possible "shrinking" hai aur uska answer hume exactly pata hai. Root test ki poori strategy hai "apni series ko ek geometric jaisi banao, phir $r$ padho." Poori kahani ke liye Geometric series dekho.

Common ratio

r

\sum r^{n}

har term apni previous term ka

r

times hai

r^{0}

barabar hai

1

(empty product convention)

1 + r + \dots + r^{N}

ki closed form

\frac{1 - r ^{N + 1}}{1 - r}

(

r \neq = 1

ke liye)

\sum r^{n}

exactly tab converge karta hai jab

∣ r ∣ < 1

, aur tab total hai

\frac{1}{1 - r}

5. Powers aur $n$ -th root $n \cdot$ — show ka star

Yeh inverse operations hain, jaise ek lock aur uski chaabi: $n x^{n} = x (x \geq 0) .$

n y

is sawaal ka jawaab deta hai

kaun sa number power

n

pe raise hone par

y

ke barabar hoga

n 2^{n}

barabar hai

2

(root

n

-th power cancel kar deta hai)

6. Limits: arrow $\to$ aur $lim_{n \to \infty}$

Ek dotted horizontal line $L$ height pe imagine karo; points $b_{n}$ us line se ever tighter hug karte hain.

Hume yeh isliye chahiye kyunki $n ∣ a_{n} ∣$ usually $n$ ke saath change karta rehta hai — yeh ek fixed ratio nahi hai, yeh drift karta hai. Limit $L$ hai settling value, effective geometric ratio jis pe term approach karti hai.

b_{n} \to L

ka matlab hai

terms

b_{n}

L

ke arbitrarily close ho jaate hain aur rehte hain jab

n \to \infty

n n

tend karta hai

1

ki taraf

7. $e$ aur $ln$ — us limit ke andar ki machinery

Hum inse sirf ek fact use karte hain: yeh hume ek awkward "variable exponent" jaise $n^{1/ n}$ ko $e^{(l n n) / n}$ mein rewrite karne dete hain, ek mushkil root ko ek easy exponent-limit mein badal dete hain. Yahi exact trick hai $n n \to 1$ ke peeche aur parent note ke $n 1/ n^{p} \to 1$ ke peeche.

ln x

poochta hai

e

ko kis power pe raise karne se

x

milega

n^{1/ n}

e^{(l n n) / n}

kyun likhte hain

yeh ek variable-power root ko ek easy exponent limit mein convert karta hai

8. $sup$ aur $lim sup$ — "hamesha-jawaab-dene-wala" limit

Ek ceiling imagine karo jo jitni neeche ho sake ho, phir bhi $S$ ke har point pe ya uske upar baith sake.

sup S

hai

least upper bound — sabse chhota number jo

S

ke har element se

\geq

T_{N} = sup {b_{n} : n \geq N}

jab

N

badhta hai

kabhi nahi badhta (har tail ek subset hai), toh settle ho jaata hai

lim sup

hamesha exist karta hai kyunki

tail-ceilings

T_{N}

sirf decrease karte hain, toh hamesha converge karte hain

9. Symbol $L$ ko saath mein banana — aur test prove karna

Ab $L = n \to \infty lim sup n ∣ a_{n} ∣$ ka har piece earn ho gaya hai:

$∣ a_{n} ∣$ — terms ki sizes, signs remove karke (§3),
$n$ — geometric power ke liye un-do button, ratio extract karta hai (§5),
$lim sup / lim$ — un extracted ratios ki settling value (§6, §8).

Toh $L$ hi hai woh effective geometric ratio jaise tumhari series behave karti hai. Ise yardstick $∣ r ∣ < 1$ se compare karo (§4).

Formula

$L < 1$ case ka proof (bounding argument, poora) Maan lo $L = lim sup n ∣ a_{n} ∣ < 1$ . Step 1 — ek buffer $r$ chuno. Kyunki $L < 1$ hai toh $L$ aur $1$ ke beech jagah hai; koi bhi fixed $r$ chuno jahan $L < r < 1.$ Kyun: hume ek geometric ratio chahiye jo (a) settling value se upar ho taaki eventually terms ko dominate kare, aur (b) $1$ se neeche ho taaki uski geometric series converge kare. $L$ aur $1$ ke beech ka gap guarantee karta hai ki aisa $r$ exist karta hai. Step 2 — tail $r$ ke neeche aa jaati hai. $lim sup$ ke tail-ceiling picture se (§8), ceilings $T_{N} = sup {m ∣ a_{m} ∣ : m \geq N}$ decrease hokar $L < r$ tak aate hain. Toh kisi cut-off $N$ ke liye, tail ka har term $r$ se neeche hai: $n ∣ a_{n} ∣ < r for all n \geq N .$ Step 3 — $n$ -th power pe raise karo. Dono sides $\geq 0$ hain, toh inequality survive karti hai: $∣ a_{n} ∣ < r^{n} (n \geq N) .$ Step 4 — geometric yardstick se compare karo. Tail $n \geq N \sum r^{n}$ ek convergent geometric series hai ( $0 \leq r < 1$ , §4). Kyunki har $∣ a_{n} ∣$ chhota hai, Comparison test ke hisaab se tail $\sum ∣ a_{n} ∣$ converge karta hai. Finite front chunk wapas add karne se woh finite rehta hai, toh $\sum ∣ a_{n} ∣$ converge karta hai — yaani $\sum a_{n}$ absolutely converge karta hai. $■$

Prerequisite map

Equipment checklist

Khud test karo — har ek tab reveal karo jab tum use zor se bol sako.

Main

\sum_{n = 1}^{\infty} a_{n}

ka har part padh sakta/sakti hoon

sigma = add up;

n = 1

start index;

\infty

= hamesha ke liye;

a_{n}

= term rule.

Main explain kar sakta/sakti hoon

\sum a_{n}

ka matlab sigma symbol ke bina

Yeh endless sum hai

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

un terms ka jo rule

a_{n}

deta hai.

Mujhe pata hai converge aur diverge mein fark

Converge = partial sums

S_{N}

ek finite number pe settle hote hain; diverge = nahi hote.

Mujhe pata hai

∣ a_{n} ∣

kyun use karte hain

Test term size measure karta hai, sign ignore karke, aur absolute convergence deta hai.

Main geometric partial sum derive kar sakta/sakti hoon

S_{N}

r

se multiply karo, subtract karo:

(1 - r) S_{N} = 1 - r^{N + 1}

, toh

S_{N} = \frac{1 - r ^{N + 1}}{1 - r}

Main exactly bata sakta/sakti hoon

\sum r^{n}

kab converge karta hai aur uski value

Jab

∣ r ∣ < 1

; tab total hai

\frac{1}{1 - r}

(kyunki

r^{N + 1} \to 0

Mujhe pata hai geometric sum

n = 0

n = 1

se kyun shuru ho sakta hai

Single term

r^{0} = 1

add/drop karna total ko finite amount se change karta hai, convergence kabhi nahi.

Main

n 2^{n}

aur

n n^{2}

compute kar sakta/sakti hoon

n 2^{n} = 2

;

n n^{2} = (n n)^{2} \to 1

Mujhe pata hai

n

-th root sahi tool kyun hai

Yeh ek

n

-th power cancel karta hai, toh

n r^{n} = r

hidden ratio bahar nikalta hai.

Mujhe

lim n n

ki value pata hai aur kyun

1

, kyunki

n^{1/ n} = e^{(l n n) / n}

aur

\frac{l n n}{n} \to 0

Mujhe pata hai supremum kya hai aur

lim sup

hamesha kyun exist karta hai

Least upper bound; tail-ceilings

T_{N}

sirf decrease karte hain, toh hamesha settle karte hain.

Main

L < 1

proof outline kar sakta/sakti hoon

r

chuno jahan

L < r < 1

; eventually

∣ a_{n} ∣ < r^{n}

; convergent

\sum r^{n}

se compare karo.

Main do series de sakta/sakti hoon jahan

L = 1

ho aur fates opposite hon

\sum 1/ n

diverge karta hai,

\sum 1/ n^{2}

converge karta hai — dono ka

L = 1

hai.

Mujhe pata hai

L < 1

L > 1

L = 1

ka matlab

Converge, diverge, inconclusive respectively.

Recall Ek-line recap

Root test mein har symbol ek hi sawaal ka jawaab dene ke liye exist karta hai: kya $a_{n}$ eventually $r^{n}$ jaisa behave karta hai kisi fixed $r < 1$ ke liye? $n$ -th root woh $r$ extract karta hai, limit uski settling value $L$ padhta hai, aur geometric yardstick $∣ r ∣ < 1$ ( $S - r S$ se prove hua) fate decide karta hai.

Parent pe wapas jao: 4.3.13 Root test · Hinglish: 4.3.13 Root test (Hinglish)