Divergence test (necessary but not sufficient): Question bank

4.3.6 · D5 · Maths › Calculus III — Sequences & Series › Divergence test (necessary but not sufficient)

Shuru karne se pehle, vocabulary ka ek reminder, taaki koi symbol unexplained na lage:

True ya false — justify karo

Terms → 0 hona guarantee karta hai ki series converge karti hai.

False. Ye test ka converse hai aur ye fail hota hai:

\sum \frac{1}{n}

ke terms

\to 0

hain phir bhi diverge karta hai (dekho Harmonic Series). Terms → 0 necessary hai, sufficient nahi.

Agar

lim a_{n} \neq = 0

toh

\sum a_{n}

diverge karta hai.

True. Ye exactly test hai; ek non-vanishing term matlab running total hamesha roughly same-size amount se kick hota rehta hai, isliye settle nahi ho sakta.

Divergence Test prove kar sakta hai ki ek series converge karti hai.

False. Ye ek one-way street hai: iske sirf do outputs hain — "diverges" ya "inconclusive". Ye kabhi "converges" nahi bol sakta.

Agar

lim a_{n} = 0

toh series definitely converge karti hai.

False. Zero limit sirf necessary condition pass karti hai; series phir bhi diverge kar sakti hai (harmonic) ya converge kar sakti hai ($p$-series

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

). Aapko koi stronger test use karna hoga.

"Necessary but not sufficient" ka matlab hai ki har convergent series ke terms → 0 hote hain, lekin har terms-→-0 series converge nahi karti.

True. Convergence terms → 0 ko force karti hai (necessary), jabki terms → 0 convergence ko force nahi karti (not sufficient).

Agar

lim a_{n}

exist nahi karta, toh series diverge karti hai.

True. "DNE"

0

ke barabar nahi hai, isliye ye test trigger karta hai bilkul nonzero limit ki tarah. Example:

\sum (- 1)^{n}

diverge karta hai kyunki

(- 1)^{n}

kabhi settle nahi hota.

Divergence Test "converges

\Rightarrow

terms

\to 0

" ka contrapositive hai.

True.

P \Rightarrow Q

(converges

\Rightarrow

limit is 0) se hum not-

Q \Rightarrow

not-

P

(limit not 0

\Rightarrow

diverges) banate hain, jo logically identical hai.

Agar do series ke terms ka limit zero same ho, toh unka fate bhi same hona chahiye.

False.

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

converge karta hai aur

\sum \frac{1}{n}

diverge karta hai, phir bhi dono ke terms

\to 0

hain. Same limit, opposite fates — proof ki test inhe distinguish nahi kar sakta.

Ek series converge kar sakti hai even though uske infinitely many terms large hoon.

False. Agar infinitely many terms

0

se door bounded rahein, toh limit

0

nahi hai (ya DNE), isliye test ke hisaab se ye diverge karega; convergence isko allow nahi karta.

Error dhundo

lim \frac{1}{n} = 0

, isliye Divergence Test se

\sum \frac{1}{n}

converge karta hai."

Error ye hai ki "converges" ka verdict expect kiya ja raha hai. Jab limit

0

ho tab test silent hota hai; ye kabhi convergence output nahi karta. Actually

\sum \frac{1}{n}

diverge karta hai — iske liye aapko Integral Test ya doubling-bracket argument chahiye.

lim (- 1)^{n} = 0

hai kyunki values average out ho jaati hain, isliye series converge karti hai."

Do errors hain. Limit

0

nahi hai — ye exist hi nahi karta, kyunki terms

- 1, + 1

pe hamesha alternate karte rehte hain. Aur kyunki limit

0

nahi hai, test conclude karta hai ki series diverge karti hai, converge nahi.

\sum \frac{3 n ^{2} + 1}{5 n ^{2} + n}

ke terms

0

ki taraf jaate hain, isliye test inconclusive hai."

Limit

\frac{3}{5}

hai,

0

nahi (upar aur neeche

n^{2}

se divide karo). Kyunki

\frac{3}{5} \neq = 0

, test decisive hai: series diverge karti hai. Ye claim ek nonzero limit ko galat se zero bol raha hai.

"Maine

lim a_{n} = 0

dikhaya, isliye main done hoon — series classify ho gayi."

Zero limit kuch bhi finish nahi karta. Test "inconclusive" return karta hai, isliye classification abhi bhi open hai. Aapko koi aur tool use karna hoga (comparison,

p

-series, integral, ratio, alternating).

S_{N} \to 0

, isliye series

0

ki sum pe converge karti hai."

Term aur running total ko confuse kiya ja raha hai.

S_{N}

partial sum hai;

a_{n}

individual term hai. Convergence ka matlab hai

S_{N}

kisi finite

S

ke paas jaaye (jo aksar nonzero hota hai), na ki

S_{N}

0

ke paas jaaye.

\sum \frac{1}{l n n}

ke terms chhote hain jo vanish ho jaate hain, isliye ye converge karta hai."

Vanishing terms kabhi convergence prove nahi karte.

\frac{1}{l n n} \to 0

lekin ye

\frac{1}{n}

se slower shrink karta hai, isliye Harmonic Series se comparison karke ye diverge karta hai. Itna slow shrinking phir bhi infinitely many pieces ko infinity tak pile karne deta hai.

"Kyunki

a_{N} = S_{N} - S_{N - 1}

hai aur dono

S

ki taraf jaate hain, toh difference

S

hai, isliye terms

\to S

Arithmetic ki galti:

S - S = 0

hota hai,

S

nahi. Dono partial sums same limit

S

ki taraf jaate hain, isliye unka difference

0

ki taraf jaata hai — aur exactly yahi reason hai ki convergent series ke terms → 0 hote hain.

Why questions

lim a_{n} = 0

sirf necessary kyun hai, convergence ke liye sufficient kyun nahi?

Kyunki terms bahut slowly shrink kar sakte hain: infinitely many tiny pieces phir bhi infinity tak add ho sakte hain (harmonic series). Shrinking hona ek hope ke liye zaruri hai, lekin ye rate control nahi karta, jo actually sum decide karta hai.

Identity

a_{N} = S_{N} - S_{N - 1}

derivation ke heart mein kyun hai?

Ye term aur running total ke beech bridge banata hai: "naya term = abhi ka total minus ek step pehle ka total". Limits lene par phir ye force karta hai ki

a_{N} \to S - S = 0

jab bhi series converge karti hai. Yahi telescoping identity Telescoping Series ko bhi power karta hai.

"Limit exist nahi karta" same kyun count hota hai jaise "limit nonzero hai"?

Test ko silent rehne ke liye limit ka

0

ke barabar hona zaroori hai. Iska koi bhi failure — nonzero value ya koi value hi nahi — matlab hai ki terms zero ki taraf shrink nahi kar rahe, isliye necessary condition violate hoti hai aur series diverge karti hai.

Divergence Test ko sabse sasta pehla check kyun kehte hain?

Isme sirf

a_{n}

ki ek limit lagti hai (Limits of Sequences ke through) aur ye instantly kaafi series ko kisi bhi harder machinery se pehle kill kar sakta hai. Agar limit clearly nonzero hai, toh aap ek hi line mein done ho.

Hum harmonic series ko

\geq \frac{1}{2}

wale brackets mein group karke convergence conclude kyun nahi kar sakte?

Woh grouping ulta dikhata hai — har doubling block kam se kam

\frac{1}{2}

sum karta hai, isliye infinitely many blocks add karne par total infinity ki taraf push ho jaata hai. Ye ek divergence argument hai, jo demonstrate karta hai ki terms → 0 kaafi nahi hai.

Test

p = 1

wale

p

-series par versus

p = 2

par kyun nahi fire karta?

Dono ke terms

\frac{1}{n ^{p}} \to 0

hain, isliye limit condition dono cases mein pass hoti hai aur test silent rehta hai.

p = 1

(diverges) aur

p = 2

(converges) mein distinguish karne ke liye p-Series Test ya Integral Test chahiye.

Edge cases

\sum_{n = 1}^{\infty} (1 + \frac{1}{n})^{n}

ke baare mein test kya kehta hai?

Terms

e \approx 2.718 \neq = 0

ki taraf jaate hain, isliye necessary condition fail hoti hai aur series diverge karti hai. Ek limit jo lagta hai

1

ke paas jaayegi woh actually

e

ke paas jaati hai.

Ek constant nonzero series

\sum 5

ka kya?

lim 5 = 5 \neq = 0

, isliye ye turant diverge karta hai — aap hamesha ke liye

5

add karte ja rahe ho aur running total bina bound ke badhta jaata hai.

\sum 0

, yaani all-zeros series ka kya?

lim a_{n} = 0

, isliye rule ke hisaab se test inconclusive hai; lekin yahan partial sums sab

0

hain, isliye series trivially

0

pe converge karti hai. Ek reminder ki "inconclusive" ka matlab "diverges" nahi hota.

Agar sirf terms ka tail matter karta hai — kya koi early spike verdict change karta hai?

Nahi. Convergence aur

lim a_{n}

dono sirf long-run behaviour par depend karte hain, isliye finitely many early terms change karna test ke conclusion ko alter nahi kar sakta.

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

ke baare mein test kya kehta hai?

Terms

1 \neq = 0

ki taraf jaate hain, isliye ye diverge karta hai. Chahe har term

1

se kam ho, ye kabhi

0

ki taraf shrink nahi karte.

\sum sin n

(angle radians mein) ka kya?

lim sin n

exist nahi karta — values

[- 1, 1]

par densely wander karti hain aur kabhi settle nahi hoti. Kyunki limit

0

nahi hai, test divergence declare karta hai.

Agar terms oscillate karein lekin shrinking amplitude ke saath, jaise

a_{n} = \frac{( - 1 ) ^{n}}{n}

, toh test kya kehta hai?

lim a_{n} = 0

, isliye test silent hai. Alternating structure kahin aur decide hoti hai (actually ye converge karta hai) — Divergence Test yahan koi information nahi deta.

Connections

Parent: Divergence Test — wo poori derivation jinhe ye traps target karti hain.
Partial Sums and Series Convergence — "spot the error" set ke peeche $S_{N}$ vs $a_{n}$ ka distinction.
Harmonic Series — canonical "terms → 0 yet diverges" counterexample.
p-Series Test, Integral Test, Comparison Test — woh tools jo inconclusive ( $lim a_{n} = 0$ ) cases resolve karte hain.
Telescoping Series — $a_{N} = S_{N} - S_{N - 1}$ identity ko directly use karta hai.
Limits of Sequences — $lim a_{n}$ compute karne ka engine.