Ratio test — proof, limitations: Question bank — Hinglish deep dive

4.3.12 · D5 · Maths › Calculus III — Sequences & Series › Ratio test — proof, limitations

Questions se pehle, ek shared mental picture — aur dhyan rakho yeh letter $r$ introduce karta hai, jise humein earn karna hoga:

Recall Picture formal

$ε$ -proof kaise banti hai "Settles below the $r$ -line" exactly wahi $ε$ -statement hai: $ε = r - L > 0$ ke saath, limit ek $N$ deta hai jisse $n \geq N \Rightarrow ∣ a_{n + 1} / a_{n} ∣ < L + ε = r$ . Us inequality ko $k$ baar chain karne se milta hai $∣ a_{N + k} ∣ < r^{k} ∣ a_{N} ∣$ — yahi geometric wall hai. Phir $\sum_{k} r^{k} ∣ a_{N} ∣ = ∣ a_{N} ∣/ (1 - r) < \infty$ kyunki $r < 1$ , aur comparison test us finiteness ko $\sum ∣ a_{N + k} ∣$ pe transfer kar deta hai. Toh yeh motaa phrase "the wall shrinks" hi hai woh precise reason jis se $r^{k}$ se bound karna ek convergent comparison series deta hai.

True or false — justify

TF1. If

lim_{n \to \infty} ∣ a_{n + 1} / a_{n} ∣ = 1

then

\sum a_{n}

diverges.

False.

L = 1

inconclusive case hai;

\sum 1/ n^{2}

L = 1

hai aur converge karta hai jabki

\sum 1/ n

L = 1

hai aur diverge karta hai — same

L

, opposite fate.

TF2. If

\sum a_{n}

converges then its ratio limit

L

must be less than

1

False. Convergence

L < 1

force nahi karta;

\sum 1/ n^{2}

converge karta hai phir bhi uska

L = 1

hai. Sirf

L < 1

convergence imply karta hai, reverse nahi.

TF3. The ratio test can prove conditional convergence of the alternating harmonic series

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

False. Ratio absolute values use karta hai aur yahan

L = 1

hai, toh test khamosh hai; conditional convergence ke liye alternating series test chahiye.

TF4. If the ratio test gives

L > 1

, the root test could still return

L = 1

on the same series.

False. Jab dono limits exist karte hain toh agree karte hain; agar ratio limit

> 1

hai toh root limit same value hogi, toh root test bhi diverge bolega (kabhi

1

nahi).

TF5. A series of strictly positive terms with

L < 1

converges "absolutely".

True, aur trivially bhi — positive terms ke liye

∣ a_{n} ∣ = a_{n}

, toh absolute convergence aur plain convergence same statement hai.

TF6. If

∣ a_{n + 1} / a_{n} ∣ < 1

for every

n

, the series must converge.

False. Ratio ka

1

se kam hona kaafi nahi; use ek fixed

r < 1

ke neeche rehna chahiye. Harmonic

\sum 1/ n

n / (n + 1) < 1

hamesha hai phir bhi diverge karta hai kyunki ratios

1

ki taraf creep karte hain.

TF7. If the shrink-ratios oscillate and

lim_{n \to \infty} ∣ a_{n + 1} / a_{n} ∣

does not exist, the ratio test simply reports "inconclusive".

True in the basic (limit) form. Koi limit nahi matlab theorem ka hypothesis fail ho gaya — lekin sharper

lim sup

ratio test phir bhi kaam kar sakta hai: agar

lim sup ∣ a_{n + 1} / a_{n} ∣ < 1

toh converge karta hai, aur agar

lim inf ∣ a_{n + 1} / a_{n} ∣ > 1

toh diverge karta hai (EC7 dekho).

TF8.

L = \infty

and

L > 1

lead to the same conclusion.

True. Dono matlab terms eventually badhti hain, toh

a_{n} \neq \to 0

aur series term test se diverge karti hai.

Spot the error

SE1. "For

\sum (- 1)^{n} / 2^{n}

, the ratio is

a_{n + 1} / a_{n} = - 1/2

, and since

- 1/2 < 1

the series converges." — find the flaw.

Verdict sahi hai lekin reasoning broken hai: tumhe

∣ a_{n + 1} / a_{n} ∣ = 1/2

use karna chahiye. Raw ratio

- 1/2

woh nahi hai jo theorem compare karta hai, aur negative ratios geometric wall bound nahi kar sakte.

SE2. "

\sum 1/ n^{3}

: since

a_{n + 1} / a_{n} = n^{3} / (n + 1)^{3} \to 1

, the ratio test says it diverges." — find the flaw.

L = 1

matlab koi information nahi, divergence nahi. Actually

\sum 1/ n^{3}

converge karta hai (ek p-series jisme

p = 3 > 1

); ratio test yahaan galat tool hai.

SE3. "The terms of

\sum 1/ n

go to zero, so the ratio test's

L < 1

branch applies and it converges." — find the flaw.

Do errors hain:

a_{n} \to 0

ratio-test condition nahi hai, aur yahan

L = 1

hai (

< 1

nahi). Terms ka zero jaana convergence ke liye necessary hai lekin kabhi sufficient nahi.

SE4. "In the

L < 1

proof we pick any

r

with

L < r < 1

; if

L = 0

there is no such

r

, so the proof fails for

L = 0

." — find the flaw.

Jab

L = 0

ho tab bhi aisa

r

milta hai —

r = 1/2

lo,

0

aur

1

ke beech strictly koi bhi number kaam karega. Gap

(L, 1)

nonempty hai jab bhi

L < 1

ho.

SE5. "Because

∣ a_{N + k} ∣ < r^{k} ∣ a_{N} ∣

, the whole series equals

∣ a_{N} ∣/ (1 - r)

." — find the flaw.

Woh value tail ka upper bound hai, sum nahi. Real sum chhota hai, aur

N

se pehle bhi finitely many terms hain; comparison sirf finiteness guarantee karta hai, koi specific value nahi.

SE6. "

\sum n! /1 0^{n}

diverges because

n!

is huge, so its ratio limit is some big constant like

10

." — find the flaw.

Ratio

(n + 1) /10

hai, jo bina bound ke badhta hai, toh

L = \infty

hai, koi fixed constant nahi. Factorials eventually kisi bhi fixed exponential ko beat kar dete hain — ratio level off nahi hota.

SE7. "Applying the ratio test to

\sum 2^{n}

gives

a_{n + 1} / a_{n} = 2 > 1

, so it diverges — but only because it's not a geometric series." — find the flaw.

\sum 2^{n}

ratio

2

ke saath geometric hai; ratio test sirf geometric rule recover kar raha hai. Parenthetical claim galat hai, koi caveat nahi.

Why questions

WHY1. Why does the proof compare against a geometric series specifically, not a p-series?

Kyunki ek fixed shrink-factor

r

exactly wahi define karta hai jo geometric series hai, aur hum exactly jaante hain kab yeh converge karta hai (

∣ r ∣ < 1

). Ratio test wahi factor measure karta hai, toh geometric natural yardstick hai.

WHY2. Why must the trapping

r

be a fixed number rather than "the ratio at each step"?

Per-step ratio

1

tak drift kar sakta hai (jaise harmonic ka

n / (n + 1)

) aur geometric wall ko kabhi nahi shrinkne deta. Sirf ek fixed

r < 1

force karta hai

r^{k} \to 0

aur isliye finite total.

WHY3. Why does

L > 1

guarantee divergence but

L < 1

does not guarantee the terms are monotonically shrinking?

L > 1

eventually force karta hai

∣ a_{n + 1} ∣ > ∣ a_{n} ∣

, toh

a_{n} \neq \to 0

— ek hard failure. Lekin

L < 1

sirf kehta hai ki long-run ratio chhota hai; early terms

N

ke baad geometric domination kick karne se pehle bounce kar sakte hain.

WHY4. Why is the ratio test "blind" to every power-law series

\sum 1/ n^{p}

Kisi bhi

p

ke liye,

a_{n + 1} / a_{n} = (n / (n + 1))^{p} \to 1

, exactly

L = 1

dead zone mein jaake baith jaata hai. Test exponent

p

nahi dekh sakta, jo exactly wahan convergence decide karta hai.

WHY5. Why does the ratio test shine on factorials and exponentials?

\frac{a _{n + 1}}{a _{n}}

mein factorials telescope ho jaate hain (

n! / (n + 1)! = 1/ (n + 1)

) aur exponentials ek constant pe collapse ho jaate hain (

r^{n + 1} / r^{n} = r

), ek stubborn

1

ki jagah clean limit milta hai.

WHY6. Why is the ratio test the tool of choice for finding a power series' radius of convergence?

L = lim_{n \to \infty} ∣ a_{n + 1} / a_{n} ∣ \cdot ∣ x ∣ < 1

set karne se directly

∣ x ∣

ki range solve ho jaati hai; boundary

L = 1

radius mark karta hai, exactly wahan jahan geometric domination break hoti hai.

WHY7. Why can Raabe's test sometimes decide cases where the ratio test returns

L = 1

Raabe's test us rate mein zoom in karta hai jis par ratio

1

approach karta hai (via

n (1 - a_{n + 1} / a_{n})

), ek finer measurement jo

1/ n

aur

1/ n^{2}

mein farq karta hai jahan coarse ratio limit nahi kar sakta.

WHY8. Why does the term-by-term comparison step actually finish the convergence proof?

Jab ek baar

∣ a_{N + k} ∣ \leq r^{k} ∣ a_{N} ∣

aur

\sum r^{k}

converge karta hai, comparison kehta hai ki ek convergent positive series se dominated series khud bhi (absolutely) convergent hai — woh theorem closing karta hai.

Edge cases

EC1. What does the ratio test say about a series with a zero term in the middle, like

a_{5} = 0

Basic ratio

a_{n + 1} / a_{n}

us step pe undefined hota hai (zero se division). Tumhe sirf itna chahiye ki terms eventually nonzero hon; ek stray zero early on theek hai kyunki limit sirf tail ki parwah karta hai.

EC2. What if

a_{n} = 0

for all

n

beyond some point?

Toh series finite sum hai — trivially convergent — aur ratio test ki zaroorat nahi (actually

a_{n + 1} / a_{n}

0/0

ban jaata hai). Test force karne ki bajaye degenerate case ko pehchano.

EC3. Ratio test verdict when

L = 0

Absolutely converge karta hai, aur bahut strongly —

L = 0

matlab terms kisi bhi geometric series se tez crush ho jaate hain, toh yeh "most convergent" outcome hai, koi special exception nahi.

EC4. Two series both hit

L = 1

\sum 1/ n

and

\sum 1/ n^{2}

. What single sharper fact separates them?

p-series rule mein exponent

p

p = 1

diverge karta hai,

p = 2 > 1

converge karta hai. Ratio test

p

nahi padh sakta; integral/p-test padh sakta hai.

EC5. At the endpoints

∣ x ∣ = 1

of a power series where

L = 1

, what must you do?

Har endpoint ko apni alag series ki tarah haath se test karo; ratio test wahan already surrender kar chuka hai. Jaise

\sum x^{n} / n

ke liye,

x = 1

diverge karta hai (harmonic) lekin

x = - 1

converge karta hai (alternating harmonic).

EC6. If a series has

L < 1

but individual ratios exceed

1

for the first hundred terms, does it still converge?

Haan. Convergence sirf kisi

N

ke baad tail par depend karta hai; growing terms ka ek finite block finite amount add karta hai aur convergence kharab nahi kar sakta jab ratio ek fixed

r < 1

ke neeche settle ho jaaye.

EC7. The plain limit

lim_{n \to \infty} ∣ a_{n + 1} / a_{n} ∣

doesn't exist — is the ratio test truly useless?

Hamesha nahi.

lim sup

variant phir bhi convergent case decide kar sakta hai: agar

lim sup ∣ a_{n + 1} / a_{n} ∣ < 1

toh tum phir bhi tail ko kisi

N

ke baad fixed

r < 1

ke neeche trap kar sakte ho, toh converge karta hai; aur agar

lim inf ∣ a_{n + 1} / a_{n} ∣ > 1

toh terms eventually badhti hain, toh diverge karta hai. Sirf tab genuinely inconclusive hai jab limsup

1

ke aas-paas straddle kare.

Connections

Ratio test — proof, limitations — woh parent jise yeh bank stress-test karta hai.
Geometric Series — woh yardstick jo har $L$ comparison use karta hai.
Comparison Test — $L < 1$ convergence case close karta hai.
Term Test (nth-term divergence) — $L > 1$ case khatam karta hai.
Root Test — woh sibling jo agree karta hai jab dono limits exist karein.
p-Series and Integral Test · Raabe's Test — $L = 1$ dead zone ke liye sharper rulers.
Radius of Convergence — ratio test ka headline application.