Limit comparison test: Question bank — Hinglish deep dive

4.3.9 · D5 · Maths › Calculus III — Sequences & Series › Limit comparison test

Deep dive child of Limit Comparison Test (4.3.9) · saath mein padhein Hinglish version.

Ye conceptual traps hain — koi bhaari algebra nahi. Har ek us jagah ko target karta hai jahan students sochte hain ki unhe Limit Comparison Test (LCT) samajh aa gayi, lekin koi boundary case ya hidden assumption unhe kaata hai. Answer cover karo, ek sentence mein sochke reason karo, phir reveal karo.

True or false — justify

True or false: Agar

lim_{n \to \infty} a_{n} = 0

hai, toh

\sum a_{n}

converge karta hai.

False.

a_{n} \to 0

sirf necessary hai, sufficient nahi — harmonic series

\sum 1/ n

ke terms

0

ki taraf jaate hain phir bhi diverge karta hai kyunki woh bahut dheere shrink karte hain.

True or false: LCT mein terms

a_{n}

aur

b_{n}

positive hone chahiye.

True. Proof ek ratio inequality ko

b_{n}

se multiply karta hai bina usse flip kiye (iske liye

b_{n} > 0

chahiye) aur Direct Comparison Test par rely karta hai, jo khud positive terms maangta hai.

True or false: Agar

0 < L < \infty

hai toh

\sum a_{n}

hamesha converge karta hai.

False.

0 < L < \infty

ka sirf matlab hai ki do series ek hi fate share karti hain — ya dono converge karti hain ya dono diverge. Kuch bhi conclude karne ke liye tumhe abhi bhi pata hona chahiye ki

\sum b_{n}

converge karta hai ya nahi.

True or false: Agar

L = 0

hai, toh

\sum a_{n}

converge karta hai.

False.

L = 0

sirf tab help karta hai jab

\sum b_{n}

converge kare (tab

a_{n}

eventually chota ho jaata hai aur neeche khinch jaata hai). Agar

\sum b_{n}

diverge kare, toh

L = 0

kuch nahi batata.

True or false: Agar

L = \infty

hai, toh

\sum a_{n}

diverge karta hai.

False.

L = \infty

sirf tab help karta hai jab

\sum b_{n}

diverge kare (tab

a_{n}

eventually bada hota hai aur upar push hota hai). Agar

\sum b_{n}

converge kare, toh

L = \infty

inconclusive hai.

True or false: Agar

L = lim a_{n} / b_{n}

exist nahi karta, toh LCT apni main form mein apply nahi hoti.

True. Main test ko ek genuine limit

L

chahiye; agar ratio hamesha oscillate kare toh koi single

L

nahi hai, isliye tumhe Direct Comparison Test ya Ratio Test jaisa koi alag tool use karna hoga.

True or false:

a_{n}

aur

b_{n}

ke roles swap karna (matlab

b_{n} / a_{n}

compute karna) change kar sakta hai ki tum kaun se conclusions draw kar sakte ho.

True. Main case mein (

0 < L < \infty

) kuch nahi badalta, kyunki

1/ L

bhi finite aur positive hota hai. Lekin one-sided cases mein flip ho jaata hai:

lim a_{n} / b_{n} = 0

ban jaata hai

lim b_{n} / a_{n} = \infty

, toh jo card tumhe convergence inherit karne deta tha woh divergence inherit karne wala ban jaata hai — usable direction swap ho jaata hai.

True or false: LCT ko

\sum (- 1)^{n} / n

par use kiya ja sakta hai.

Directly false, kyunki terms sab positive nahi hain. Tum instead

\sum ∣ a_{n} ∣ = \sum 1/ n

test karoge (jo diverge karta hai) ya original par Alternating Series Test use karoge.

Spot the error

Ek student likhta hai: "

\sum \frac{n + 1}{n ^{3} + n}

ka behaviour

\frac{1}{n ^{3}}

jaisa hai, isliye

b_{n} = 1/ n^{3}

." Galti kahan hai?

Unhone numerator ki growth drop kar di. Strongest power alag-alag top aur bottom mein rakho: top

\sim n

, bottom

\sim n^{3}

, isliye

b_{n} \sim n / n^{3} = 1/ n^{2}

Ek student

\sum \frac{l n n}{n ^{2}}

ke liye

b_{n} = 1/ n

choose karta hai aur

L

compute karta hai. Actually kya hota hai, aur

1/ n

poor choice kyun hai?

Ratio hai

\frac{( ln n ) / n ^{2}}{1/ n} = \frac{ln n}{n} \to 0

, isliye

L = 0

hai (na ki

\infty

— denominator mein extra

n

slow

ln n

ko beat karta hai). Lekin

\sum 1/ n

diverge karta hai, aur divergent

b_{n}

ke saath

L = 0

inconclusive hai.

b_{n} = 1/ n^{1.5}

par switch karo, phir bhi

L = 0

milega lekin ab ek convergent series ke against, isliye tum convergence conclude kar sakte ho.

Ek student kehta hai: "

a_{n} < b_{n}

eventually aur

\sum b_{n}

diverge karta hai, isliye LCT se

\sum a_{n}

diverge karta hai." Isko theek karo.

Ek divergent series se chota hona kuch prove nahi karta (chota wala converge bhi ho sakta hai). Divergence sirf ek chote divergent series se bade wale mein transfer hoti hai, matlab tumhe

a_{n} > b_{n}

chahiye jab

\sum b_{n}

divergent ho.

Ek student

\sum \frac{1}{n ^{2} + 1}

b_{n} = 1/ n^{2}

se compare karta hai, limit nikalte hain, phir kehta hai "converges." Galti pakdo.

Ratio hai

\frac{1/ n ^{2} + 1}{1/ n ^{2}} = \frac{n ^{2}}{n ^{2} + 1} \approx n \to \infty

, isliye

L = \infty

hai —

0

nahi. Aur convergent

b_{n} = 1/ n^{2}

ke against

L = \infty

inconclusive hai, isliye koi verdict nahi niklega. Sahi skeleton hai

b_{n} = 1/ n

⇒

L = 1

⇒ diverges.

Ek student likhta hai

L = lim \frac{a _{n}}{b _{n}} = \frac{l i m a _{n}}{l i m b _{n}} = \frac{0}{0}

aur ruk jaata hai. Galti kya hai?

Tum ek ratio ki limit ko limits ke ratio mein split nahi kar sakte jab dono

0

ki taraf jaayein — yahi indeterminate form

0/0

hai. Limit lene se pehle tumhe ratio ko khud simplify karna hoga (factor karo, top power se divide karo).

Ek student claim karta hai ki

\sum \frac{1}{2 ^{n} + n}

ke liye

2^{n}

ki wajah se Ratio Test chahiye. Kuch galat hai kya?

Galat nahi, lekin LCT zyada clean hai: bade

n

ke liye

2^{n}

dominate karta hai, isliye

b_{n} = 1/ 2^{n}

(ek geometric series

r = \frac{1}{2}

ke saath)

L = 1

aur immediately convergence deta hai.

Why questions

Proof mein

ε = L /2

kyun choose karte hain, koi bhi chota

ε

kyun nahi?

Limit definition kehti hai: kisi bhi

ε > 0

ke liye ek

N

hai jisme

∣ a_{n} / b_{n} - L ∣ < ε

hoga

n > N

ke liye, matlab

L - ε < a_{n} / b_{n} < L + ε

ε = L /2

choose karne se lower end

L - ε = L /2

ban jaata hai, jo strictly positive hai kyunki

L > 0

; koi bada

ε

lower bound ko

0

ya usse neeche gira sakta hai aur useful inequality

\frac{L}{2} b_{n} < a_{n}

khatam ho jaati.

LCT sirf "large

n

" ki parwaah kyun karta hai aur pehle few terms ignore kyun karta hai?

Convergence ek tail property hai — finitely many terms add ya change karne se sum ek finite amount shift hoti hai lekin ek convergent series ko divergent ya ulta nahi kar sakti. Limit

L

sirf tail behaviour padhta hai.

b_{n}

banane ke liye mnemonic "strongest rakho, baaki chodo" kyun kaam karta hai?

Bahut bade

n

ke liye highest-power term saare kamzor waalon ko dabaata hai, isliye leading terms ka ratio growth/decay ki sahi rate capture karta hai — yahi akela cheez decide karti hai convergence.

LCT aksar Direct Comparison Test se zyada preferable kyun hota hai?

Direct comparison ek exact inequality

a_{n} \leq b_{n}

(ya

\geq

) sahi direction mein maangta hai, jo painful ho sakta hai ya galat direction point kar sakta hai; LCT sirf do terms ko same rate se shrink karne ki zaroorat hoti hai, inequality se completely bachke.

\sum 1/ n^{2} + 1

LCT ke liye ek textbook case kyun hai na ki direct comparison ke liye?

Kyunki

\frac{1}{n ^{2} + 1} < \frac{1}{n}

hai, jo divergence prove karne ke liye direct comparison mein galat direction hai (divergent series se chota hona kuch prove nahi karta). LCT

1/ n

ke against

L = 1

deta hai aur settle kar deta hai.

L = \infty

ke saath ek divergent

b_{n}

actually

\sum a_{n}

ko diverge hone par kyun force karta hai?

L = \infty

ka matlab hai ki kisi bhi bade

M

ke liye ek

N

hai jisme

a_{n} / b_{n} > M

hoga

n > N

ke liye, matlab

a_{n} > M b_{n}

. Kyunki

\sum b_{n}

diverge karta hai,

\sum M b_{n}

bhi diverge karta hai, aur

a_{n}

term-by-term uske upar baitha hai — Direct Comparison Test se

\sum a_{n}

bhi diverge karta hai.

[[p-Series|

p

-series]] aur Geometric Series ka fate jaanna LCT ke liye itna important kyun hai?

LCT khud kabhi decide nahi karta — yeh ek known verdict

b_{n}

a_{n}

mein transfer karta hai. Isliye tumhare paas un series ka stockpile hona chahiye jinka fate tumhe pehle se pata ho, aur

p

-series aur geometric series do workhorses hain.

Edge cases

Agar

a_{n} = b_{n}

sab

n

ke liye ho, toh

L

kya hai aur test kya kehta hai?

L = 1

hai, safely

0 < L < \infty

ke andar, isliye kehta hai "same fate" — trivially sach hai kyunki yeh same series hai. Harmless lekin content-free.

Maano

a_{n} / b_{n}

hamesha

1

aur

2

ke beech oscillate karta rahta hai (koi limit nahi). Kya tum LCT bachaa sakte ho?

Main limit form nahi, lekin ratio

[1, 2]

mein trapped rehta hai, isliye tum

b_{n} \leq a_{n} \leq 2 b_{n}

bound kar sakte ho aur Direct Comparison Test ya ratio par Squeeze Theorem pe fall back kar sakte ho.

Kya hota hai agar

\sum b_{n}

khud ek borderline series ho jaise harmonic

\sum 1/ n

Kuch special nahi toot ta — LCT

\sum 1/ n

ka (divergent) verdict

\sum a_{n}

mein tab bhi transfer karta hai jab

0 < L < \infty

ho.

b_{n}

ka borderline nature exactly wahi reason hai ki tumhe

b_{n}

ka exponent carefully choose karna padta hai.

Agar dono

\sum a_{n}

aur

\sum b_{n}

converge karte hain, toh kya

0 < L < \infty

hona zaruri hai?

Nahi. Do convergent series wildly different rates se decay kar sakti hain,

L = 0

deke (jaise

a_{n} = 1/ n^{3}

b_{n} = 1/ n^{2}

) ya

L = \infty

. Finite positive

L

shared fate ke liye sufficient hai lekin necessary nahi.

Verdict kya hai agar

L = 0

aur

\sum b_{n}

diverge kare?

Inconclusive.

a_{n}

eventually

b_{n}

se chota hai, lekin ek divergent series se chota hona kuch nahi kehta —

\sum a_{n}

dono taraf ja sakta hai. Koi alag

b_{n}

choose karo.

Verdict kya hai agar

L = \infty

aur

\sum b_{n}

converge kare?

Inconclusive.

a_{n}

eventually

b_{n}

se bada hai, lekin ek convergent series se bada hona

\sum a_{n}

ke baare mein kuch nahi kehta. Koi aisa

b_{n}

chuno jo

0 < L < \infty

de.

Kya LCT kabhi ek convergent

b_{n}

use karke series ka diverge hona prove kar sakta hai?

Sirf

L = \infty

mismatch mein, jahan woh kar nahi sakta — toh effectively nahi. Divergence prove karne ke liye ya toh

0 < L < \infty

chahiye

\sum b_{n}

divergent ke saath, ya

L = \infty

chahiye

\sum b_{n}

divergent ke saath.

Recall Jaane se pehle quick self-test

$L = 0$ sirf tab help karta hai jab $\sum b_{n}$ ::: converge kare. $L = \infty$ sirf tab help karta hai jab $\sum b_{n}$ ::: diverge kare. $0 < L < \infty$ ka matlab hai ki do series ::: same fate share karti hain (dono converge ya dono diverge). Terms par single required condition hai ::: dono $a_{n} > 0$ aur $b_{n} > 0$ sab bade $n$ ke liye.