Root test — Hinglish notes

4.3.13 · Maths › Calculus III — Sequences & Series

YEH KEHTA KYA HAI

Hum $lim sup$ use karte hain taaki test ko hamesha ek value mile, lekin exam ke zyaadatar problems mein ordinary limit $lim n ∣ a_{n} ∣$ exist karti hai aur aap wohi use karte ho.

YEH KAAM KYUN KARTA HAI (derivation scratch se)

Hum ise poori tarah geometric series $\sum r^{n}$ se banate hain, jo converge karta hai tabhi jab $∣ r ∣ < 1$ .

Formula Case

$L < 1$ — comparison se convergence Maano $L = lim n ∣ a_{n} ∣ < 1$ . Koi bhi number $r$ chuno jo strictly $L$ aur $1$ ke beech ho: $L < r < 1.$ Yeh step kyun? Hume ek fixed ratio $r$ chahiye jisse hum terms ko dominate kar sakein, aur jo ek convergent geometric series bhi banaye — isliye $r$ ko $1$ se neeche aur $L$ se upar hona chahiye.

Kyunki $n ∣ a_{n} ∣ \to L < r$ , limit ki definition se ek $N$ exist karta hai aur sabhi $n \geq N$ ke liye: $n ∣ a_{n} ∣ < r .$ Yeh step kyun? Eventually $n$ -th roots $r$ se neeche squeeze ho jaati hain kyunki $r$ limit $L$ se bada hai.

Dono sides ko $n$ -th power tak raise karo (dono sides positive hain, isliye inequality preserve hoti hai): $∣ a_{n} ∣ < r^{n} (n \geq N) .$ Ab $\sum_{n \geq N} r^{n}$ ek convergent geometric series hai ( $0 \leq r < 1$ ). Comparison test se tail $\sum ∣ a_{n} ∣$ converge karti hai, isliye $\sum a_{n}$ absolutely converge karta hai. $■$

ISKA USE KAISE KAREIN (worked examples)

Steel-manned mistakes

Ise kab use karein (80/20)

Root test limit formula

L = lim sup_{n \to \infty} n ∣ a_{n} ∣

L < 1

hone par kya conclusion nikalte hain

Series converges absolutely.

L > 1

L = \infty

hone par kya conclusion nikalte hain

Series diverges (terms

0

tak nahi jaate).

L = 1

hone par kya conclusion nikalte hain

Inconclusive — koi doosra test use karo.

Root test

L < 1

ke liye kaam kyun karta hai?

Aisa

r

chuno jahan

L < r < 1

ho; eventually

∣ a_{n} ∣ < r^{n}

hota hai, aur

\sum r^{n}

converge karta hai, isliye comparison se convergence milti hai.

Root test

L > 1

ke liye kaam kyun karta hai?

∣ a_{n} ∣ > 1

infinitely often hota hai, isliye

a_{n} \neq \to 0

;

n

-th term test divergence force karta hai.

lim_{n \to \infty} n n

kya hai?

1

(kyunki

n^{1/ n} = e^{l n n / n} \to e^{0} = 1

Root test kab prefer karna chahiye?

Jab general term mein

n

-th power

(\dots)^{n}

ya base

^{n}

ho.

\sum (2 n + 1)^{n} / (3 n + 5)^{n}

ke liye

L

nikalo

L = 2/3 < 1

, converges.

Har

\sum 1/ n^{p}

ke liye

L = 1

kyun hota hai?

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

kyunki

ln n / n \to 0

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho series mein har term ek ball ke bounce ke baad girne ki height hai. Agar har bounce pichli height ka ek fixed fraction $r$ rakhta hai ( $r < 1$ ), toh total distance finite hoti hai — ball ruk jaati hai. Root test ek clever trick hai us fraction ko dhoondne ke liye: agar ek term $r$ ka khud se $n$ baar multiply kiya hua jaisa lagta hai ( $r^{n}$ ), toh aap us multiplying ko "undo" karte ho $n$ -th root lekar, aur $r$ nikalta hai. Agar $r < 1$ toh bounces khatam ho jaate hain (sum finite hai = converges). Agar $r > 1$ toh bounces hamesha ke liye badhte rehte hain (diverges). Agar $r$ exactly $1$ aaye, toh trick nahi bata sakti — doosra tool chahiye.

YEH KEHTA KYA HAI

YEH KAAM KYUN KARTA HAI (derivation scratch se)

ISKA USE KAISE KAREIN (worked examples)

Steel-manned mistakes

Ise kab use karein (80/20)

Concept Map