Limit comparison test — Hinglish notes

4.3.9 · Maths › Calculus III — Sequences & Series

Tags: #calculus #series #convergence Subject: Maths — Calculus III: Sequences & Series

The big picture

Hum decide karna chahte hain ki $\sum a_{n}$ converge karega ya diverge, jahan $a_{n} > 0$ hai (positive-term series). Kisi known series se directly compare karna (Direct Comparison Test) thoda fragile hota hai: tumhe exact inequality $a_{n} \leq b_{n}$ chahiye hoti hai jo prove karna mushkil ho sakta hai. Limit Comparison Test (LCT) zyada smooth tool hai — isse sirf ye matter karta hai ki $a_{n}$ aur $b_{n}$ large $n$ ke liye same tarah behave karein ya nahi.

YE KEHTA KYA HAI

"Main" case $0 < L < \infty$ hai. Baaki do edge cases hain jo phir bhi one-directional information dete hain.

YE SACH KYU HAI (derivation from scratch)

Maano main case: $L = lim a_{n} / b_{n}$ jahan $0 < L < \infty$ hai.

Step 1 — Limit ko ek sandwich mein translate karo. Ye step kyun? "Limit equals $L$ " ka matlab hai ki ratio eventually $L$ ke around kisi bhi chhoti window mein rehta hai. Hum ek aisi window choose karte hain jo sab kuch positive rakhe.

$ε = L /2$ choose karo. Limit ki definition se, ek $N$ exist karta hai aise ki sabhi $n > N$ ke liye: $\frac{a _{n}}{b _{n}} - L < \frac{L}{2} .$ Absolute value ko unpack karo: $L - \frac{L}{2} < \frac{a _{n}}{b _{n}} < L + \frac{L}{2} ⟹ \frac{L}{2} < \frac{a _{n}}{b _{n}} < \frac{3 L}{2} .$

Step 2 — Fraction clear karo (legal hai kyunki $b_{n} > 0$ hai). Ye step kyun? Positive $b_{n}$ se multiply karne par ratio statement ek direct comparison statement ban jaati hai. $\frac{L}{2} b_{n} < a_{n} < \frac{3 L}{2} b_{n} (n > N) .$

Step 3 — Har inequality ko Direct Comparison Test mein daalo.

Agar $\sum b_{n}$ converges: toh $\sum \frac{3 L}{2} b_{n}$ converge karti hai (constant multiple). Kyunki $a_{n} < \frac{3 L}{2} b_{n}$ hai, right inequality se $\sum a_{n}$ converge karti hai.
Agar $\sum b_{n}$ diverges: toh $\sum \frac{L}{2} b_{n}$ diverge karti hai. Kyunki $a_{n} > \frac{L}{2} b_{n}$ hai, left inequality $\sum a_{n}$ ko bhi diverge hone par majboor karti hai.

Toh $\sum a_{n}$ aur $\sum b_{n}$ saath jeete hain ya saath marte hain. $■$

ISEY KAISE USE KAREIN (recipe)

Large $n$ ke liye $a_{n}$ dekho. Numerator aur denominator mein sirf dominant powers rakho.
Unhi dominant terms se $b_{n}$ banao — usually ek $p$ -series $\sum 1/ n^{p}$ ya geometric series, jinki fate tumhe pehle se pata hai.
$L = lim a_{n} / b_{n}$ compute karo.
Conclusion nikalo. (Agar $b_{n}$ achhi tarah choose ki, toh $L$ ek nice finite positive number hogi.)

Reference facts jinse compare karte ho: $\sum \frac{1}{n ^{p}} converges ⟺ p > 1, \sum r^{n} converges ⟺ ∣ r ∣ < 1.$

Worked examples

Common mistakes

Forecast-then-Verify drill

Compute karne se pehle convergence predict karo, phir LCT se confirm karo.

$\sum \frac{2 n}{n ^{3} + 1}$ → forecast: $1/ n^{2}$ jaisa behave karta hai → converges. Verify: $L = 2$ . ✔
$\sum \frac{n + 1}{n + 4}$ → $n / n = 1/ n^{1/2}$ jaisa behave karta hai, $p = \frac{1}{2} \leq 1$ → diverges. Verify: $L = 1$ . ✔
$\sum \frac{1}{2 ^{n} + n}$ → $1/ 2^{n}$ geometric jaisa behave karta hai → converges. Verify: $L = 1$ . ✔

Flashcards

LCT mein terms

a_{n}, b_{n}

ke baare mein kya require hota hai?

Ye positive honi chahiye (sabhi large

n

ke liye).

LCT mein kya compute kiya jaata hai?

L = lim_{n \to \infty} a_{n} / b_{n}

LCT mein agar

0 < L < \infty

ho, toh kya conclude kar sakte ho?

\sum a_{n}

aur

\sum b_{n}

dono converge ya dono diverge karti hain.

Agar

L = 0

ho, toh

\sum a_{n}

ki convergence ke liye kaunsi extra condition chahiye?

\sum b_{n}

converge karni chahiye.

Agar

L = \infty

ho, toh

\sum a_{n}

ki divergence ke liye kaunsi extra condition chahiye?

\sum b_{n}

diverge karni chahiye.

Ratio inequality ko

b_{n}

se multiply karna valid kyun rehta hai?

Kyunki

b_{n} > 0

hai, isliye inequality ka direction preserve hota hai.

Proof mein kaun sa

ε

choose kiya jaata hai aur kyun?

ε = L /2

, taaki lower bound

L /2

strictly positive rahe.

\sum \frac{3 n ^{2} + 5}{n ^{4} - n + 2}

ke liye sahi

b_{n}

kya hai?

1/ n^{2}

(leading powers ka ratio).

a_{n} \to 0

convergence ke liye sufficient kyun nahi hai?

Ye necessary hai sufficient nahi; jaise

\sum 1/ n

diverge karta hai jabki

1/ n \to 0

LCT inconclusive kab hota hai?

Jab

L = 0

ho aur

\sum b_{n}

divergent ho, ya

L = \infty

ho aur

\sum b_{n}

convergent ho.

p

-series

\sum 1/ n^{p}

exactly kab converge karti hai?

p > 1

par.

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

Socho do runners hain. Hum akele measure nahi kar sakte ki har ek kitna jaata hai, lekin unme se ek ko hum perfectly jaante hain. Agar, kaafi time baad, unknown runner hamesha roughly same speed se ja raha ho jis runner ko hum samajhte hain (unka speed ratio ek steady positive number par settle ho jaata hai), toh dono ya toh finish line tak pahunchenge ya dono saath mein forever dauraate rahenge. Series convergence bas ye hai ki "kya running total kahin ruk jaata hai?" — aur matching speeds ka matlab hai matching destinies.

Connections

Direct Comparison Test — LCT iske upar build kiya gaya hai sandwich step ke zariye.
p-Series — favourite $b_{n}$ benchmark ( $\sum 1/ n^{p}$ ).
Geometric Series — benchmark jab $a_{n}$ mein $r^{n}$ ho.
Ratio Test — better hota hai jab factorials/exponentials dominate karein.
Integral Test — monotone positive terms ke liye alternative.
Harmonic Series — classic " $a_{n} \to 0$ phir bhi diverges" counterexample.
Squeeze Theorem — same "quantity ko ek window mein trap karo" technique jaise $ε = L /2$ step mein.