Root test: Question bank — Hinglish deep dive

4.3.13 · D5 · Maths › Calculus III — Sequences & Series › Root test

Setup: ye page jo kuch bhi assume karti hai, zero se

Hum ek infinite sum $\sum a_{n} = a_{1} + a_{2} + a_{3} + \dots$ study karte hain. "Converges" ka matlab hai running totals ek single finite number pe settle ho jaayein; "diverges" ka matlab hai nahi hote.

Recall Is page par naam liye gaye baaki tests ki quick reminders

Comparison test: agar $0 \leq ∣ a_{n} ∣ \leq b_{n}$ eventually aur $\sum b_{n}$ converge karta hai, toh $\sum ∣ a_{n} ∣$ converge karta hai. ( $L < 1$ case ka engine.) n-th term divergence test: agar $a_{n} \neq \to 0$ , toh series diverge karti hai. ( $L > 1$ case ka engine.) Ratio test: roots ki jagah $lim ∣ a_{n + 1} / a_{n} ∣$ dekhta hai; jab wo limit exist kare toh same $L$ deta hai. p-series rule: $\sum 1/ n^{p}$ converge karta hai tab hi jab $p > 1$ . (Standard $L = 1$ examples.)

Neeche ki picture poori logic dikhati hai: $r$ ko rooting se nikalo, phir $1$ ke against judge karo.

Neeche har trap ek specific weak spot pe attack karta hai: $L = 1$ ka matlab kya hai, proof ko $L < r < 1$ kyun chahiye, $n$ -th roots polynomials vs exponentials ke saath kaise behave karte hain, aur root test ka Ratio test se kya relation hai.

True or false — justify

Agar

L = 1

toh series diverge karti hai.

False.

L = 1

ka matlab hai test koi information nahi deta —

\sum 1/ n

diverge karta hai lekin

\sum 1/ n^{2}

converge karta hai, aur dono ka

L = 1

hai. Aapko p-series rule jaisa koi aur tool switch karna hoga.

Agar

L < 1

toh series absolutely converge karti hai, sirf conditionally nahi.

True. Proof mein

∣ a_{n} ∣ < r^{n}

bound milti hai aur

\sum r^{n}

converge karta hai, isliye

\sum ∣ a_{n} ∣

converge karta hai — ye exactly absolute convergence hai.

Root test conditional convergence prove kar sakta hai (converges but not absolutely).

False. Ye sirf

∣ a_{n} ∣

ko bound karta hai, isliye "converges" ka verdict hamesha absolute convergence hota hai. Conditional cases ke liye (jaise alternating harmonic series) aapko alternating series test chahiye.

Agar

L > 1

, toh series diverge karti hai kyunki uske partial sums oscillate karte hain.

Reasoning mein False. Ye diverge karta hai kyunki

∣ a_{n} ∣ > 1

infinitely often hota hai, isliye

a_{n} \neq \to 0

; n-th term divergence test oscillation se independent divergence force karta hai.

Agar

n ∣ a_{n} ∣ \to 0

, toh series phir bhi diverge ho sakti hai.

False.

L = 0 < 1

, aur

0

bilkul valid case hai

L < 1

ka, isliye ye absolutely converge karta hai.

L = 0

par kuch special nahi hota.

Root test aur Ratio test hamesha same series par agree karte hain.

False. Bahut si series par same

L

dete hain, lekin root test kuch aisi series par succeed karta hai jahan ratio limit exist hi nahi karti — root test strictly stronger hai.

Agar ratio test inconclusive hai (

L = 1

), toh root test bhi inconclusive hoga.

True (dono same $L$ share karte hain). Jab bhi ratio limit exist kare wo root limit ke barabar hoti hai, isliye ek se

1

aaya toh doosre se bhi

1

aayega. Root test tabhi jeet ta hai jab ratio limit bilkul exist na kare.

L = \infty

wali series ko phir bhi alag divergence argument chahiye.

False.

L = \infty

sirf

L > 1

ka extreme case hai; same

n

-th term argument apply hota hai aur test khud divergence conclude karta hai.

Spot the error

a_{n} = n^{2} / 2^{n}

ke liye,

n a_{n}

ek tiny number ka root hai, isliye

L \approx 0

Error. Polynomial factors root ho kar

1

dete hain:

n n^{2} = n^{2/ n} \to 1

. Sirf exponential survive karta hai,

L = 1/2

milta hai,

0

nahi.

n n \to 0

kyunki aap ek bade number ka bada root le rahe ho."

Error.

n n = e^{l n n / n}

aur

ln n / n \to 0

, isliye limit

e^{0} = 1

hai,

0

nahi. Root

n

se bahut dheere badhta hai —

1

ki taraf chadta hai aur wahi rukta hai, kabhi blow up nahi karta.

L > 1

ka matlab hai terms eventually

1

se zyada ho jaate hain sabhi bade

n

ke liye."

Error.

lim sup

ke saath,

∣ a_{n} ∣ > 1

sirf infinitely often hona chahiye, har bade

n

ke liye nahi. Ye akela

a_{n} \to 0

todne ke liye kaafi hai.

"Kyunki koi bhi

r

L < r < 1

satisfy kare kaam karta hai,

r

ki choice proof ko sloppy bana deti hai."

Error. Sirf ek aisa

r

chahiye jo tail ko dominate kare — ye ek strength hai; existence guaranteed hai kyunki

L < 1

ek gap chhodta hai. Choice ki freedom ek logically valid step ko weak nahi karti.

n ∣ a_{n} ∣ < r

sabhi

n

ke liye hold karta hai, isliye

∣ a_{n} ∣ < r^{n}

everywhere."

Error. Limit sirf

n \geq N

ke liye bound guarantee karta hai (eventually). Shuru ke finitely many terms convergence ke liye matter nahi karte, isliye hum sirf tail ko compare karte hain.

\sum \frac{3 ^{n}}{n ^{3}}

par root test use karne ke liye, main

n 3^{n} / n^{3} = 3/ n^{3}

leta hoon."

Error.

n n^{3} = (n n)^{3} \to 1

n^{3}

nahi. Correct root hai

3/ (n n)^{3} \to 3

, jo

L = 3

deta hai (diverges).

L = 1

borderline hai, toh main kahunga ye 'weakly converge' karta hai."

Error. Aisa koi verdict exist nahi karta.

L = 1

total silence hai — test kuch contribute nahi karta aur aapko ek alag test use karna hoga.

Why questions

Root test saari cheez ko sirf Geometric series se kyun compare karta hai aur kisi se nahi?

Kyunki geometric series wo sabse simple family hai jiska convergence hum exactly jaante hain (

∣ r ∣ < 1

), aur

n a_{n} \approx r

literally kisi bhi term se hidden geometric ratio nikaal leta hai.

Hum specifically

n

-th root kyun lete hain, koi aur operation kyun nahi?

Agar ek term

r^{n}

jaisi behave kare, toh

n

-th root "

n

-baar multiply" ko undo karta hai aur

r

seedha bahar aa jaata hai. Ye

n

-th power raise karne ka exact inverse hai, jo ki geometric growth hai.

Definition mein ordinary limit ki jagah

lim sup

kyun use hota hai?

Ordinary limit exist nahi kar sakta (roots forever wobble kar sakte hain), lekin

lim sup

— wo sabse badi value jis par ye baar baar aate hain — hamesha exist karta hai (finite ya

\infty

), isliye test hamesha defined hai. Exam problems mein plain limit usually exist karta hai aur uske barabar hota hai.

\sum 1/ n^{p}

mein har

p

ke liye

L = 1

kyun hota hai?

n 1/ n^{p} = e^{- p l n n / n}

aur

ln n / n \to 0

, isliye kisi bhi fixed

p

ke liye exponent vanish ho jaata hai. Root test literally wo exponent

p

nahi dekh sakta jo convergence decide karta hai.

(stuff)^{n}

jaisi terms ke liye root test Ratio test se zyada preferred kyun hai?

n \cdot

outer power ko ek clean step mein cancel kar deta hai, jabki ratio test aapko ek messy

(\dots)^{n + 1} / (\dots)^{n}

quotient simplify karne par force karta hai.

L > 1

case mein comparison ki jagah n-th term divergence test kyun use hota hai?

Comparison ko ek convergent dominating series chahiye, jo terms ke badhne par fail hoti hai. Isliye hum seedha note karte hain ki terms

0

par nahi jaate, jo ek direct divergence criterion hai.

Hum sirf

a_{n} \to 0

se convergence conclude kyun nahi kar sakte?

a_{n} \to 0

necessary hai lekin sufficient nahi —

\sum 1/ n

mein terms

0

par jaate hain phir bhi ye diverge karta hai. Root test demand karta hai ki shrinking ki rate geometric ho, jo zyada strong condition hai.

Edge cases

r = 1

ke saath

\sum r^{n}

ke baare mein root test kya kehta hai?

n 1 = 1

sabhi

n

ke liye, isliye

L = 1

— inconclusive. Aur sach mein

\sum 1

diverge karta hai, lekin test nahi bata sakta; ye

n

-th term test se seedha dikhtaa hai.

0 < r < 1

ke saath

\sum r^{n}

(model case) mein kya hota hai?

n r^{n} = r

, isliye

L = r < 1

exactly, aur test convergence confirm karta hai. Root test us family ke liye exact hone ke liye calibrated hai jis se ye compare karta hai.

Agar kuch terms mein

a_{n} = 0

ho — kya

n 0

test tod deta hai?

Nahi.

n 0 = 0

, jo sirf values neeche kheenchta hai;

lim sup

zero terms ko fine handle karta hai aur ye kabhi divergence cause nahi karte.

\sum \frac{1}{( l n n ) ^{n}}

ke liye

L

kya hai aur ye clean win kyun hai?

n 1/ (ln n)^{n} = 1/ ln n \to 0

, isliye

L = 0 < 1

, converges. Perfect

n

-th power root ke neeche instantly collapse ho jaata hai — is test ke liye ideal trigger.

Ek power series

\sum c_{n} x^{n}

ke liye root test kya ban jaata hai?

Ye Cauchy–Hadamard formula

R = 1/ lim sup n ∣ c_{n} ∣

ban jaata hai — root test

x

fold in karke apply karne se seedha Radius of convergence milta hai.

Agar

n ∣ a_{n} ∣

do values ke beech oscillate kare, maano

0.4

aur

0.9

lim sup

sabse badi recurring value leta hai, yahan

0.9 < 1

, isliye series phir bhi converge karti hai. Ordinary limit use karna fail ho jaata kyunki wo exist nahi karta, aur exactly isliye

lim sup

use hota hai.

Recall Har trap ki ek-line summary

$L < 1$ ⇒ absolute convergence; $L > 1$ ⇒ divergence (terms $0$ par nahi); $L = 1$ ⇒ silence. Polynomials root ho kar $1$ dete hain, sirf exponentials survive karte hain, aur $lim sup$ test ko hamesha defined rakhta hai.