Absolute vs conditional convergence: Question bank

4.3.11 · D5 · Maths › Calculus III — Sequences & Series › Absolute vs conditional convergence

Shuru karne se pehle, har question ke liye ek anchor. Hum hamesha ek saath do cheezein dekh rahe hote hain:

True or false — justify karo

Har absolutely convergent series convergent hoti hai.

True. Yeh Absolute Convergence Theorem hai:

\sum ∣ a_{n} ∣

ka converge karna

\sum a_{n}

ko converge hone pe majboor karta hai, kyunki cancellation sirf us sum ki help kar sakti hai jo already all-positive terms ke saath converge karta ho.

Har convergent series absolutely convergent hoti hai.

False. Alternating harmonic series

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

converge karti hai (to

ln 2

) lekin

\sum 1/ n

diverge karta hai, isliye yeh absolutely converge kiye bina converge karti hai.

Agar

\sum ∣ a_{n} ∣

diverge kare, toh

\sum a_{n}

zaroor diverge karega.

False. Absolute series ka diverge karna sirf absolute convergence ko rule out karta hai; signed series phir bhi conditionally converge kar sakti hai, exactly jaisa ki

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

karta hai.

Ek conditionally convergent series woh hai jo "barely" ya "slowly" converge karti hai.

False. Yeh poori tarah se ek genuine finite real number tak converge karti hai; "conditional" ka matlab hai ki absolute series diverge karti hai, na ki signed limit shaky ya partial hai.

Agar

\sum a_{n}

converge kare aur har term positive ho, toh yeh absolutely convergent hai.

True. Jab saare

a_{n} \geq 0

hon toh

∣ a_{n} ∣ = a_{n}

hota hai, isliye absolute series hi original series hai aur woh bhi converge karti hai — ek positive-term convergent series automatically absolute hoti hai.

Agar

\sum a_{n}

absolutely converge kare, toh

\sum a_{n}^{2}

converge karta hai.

True. Absolute convergence force karti hai ki

∣ a_{n} ∣ \to 0

, isliye eventually

∣ a_{n} ∣ < 1

aur

a_{n}^{2} \leq ∣ a_{n} ∣

; convergent

\sum ∣ a_{n} ∣

ke saath comparison ise finish kar deta hai (dekho Comparison Test).

Ek conditionally convergent series ke liye, terms ko rearrange karne se sum kabhi nahi badalta.

False. Riemann's Rearrangement Theorem kehta hai ki tum ek conditionally convergent series ko reorder karke koi bhi real number ya

\pm \infty

tak pahunch sakte ho — value order pe depend karti hai (dekho Riemann Rearrangement Theorem).

Ek absolutely convergent series ke liye, koi bhi rearrangement same sum deta hai.

True. Absolute convergence series ko ek finite sum ki tarah behave karata hai, isliye order irrelevant hai aur har rearrangement same limit pe land karta hai.

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

conditionally convergent hai kyunki yeh alternate karta hai.

False. Alternating signs type decide nahi karte; yahan

\sum 1/ n^{2}

(ek p-series with

p = 2 > 1

) converge karta hai, isliye yeh absolutely convergent hai.

Agar ek power series apne interval of convergence ke strictly andar kisi point pe converge kare, toh wahan absolutely converge karta hai.

True. Radius of convergence ke andar convergence hamesha absolute hoti hai; sirf endpoints pe convergence merely conditional ho sakti hai.

Error dhundo

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

Alternating Series Test pass karta hai, isliye yeh absolutely converge karta hai."

Alternating Series Test sirf signed series ko certify karta hai. Absolute convergence test karne ke liye tumhe

\sum 1/ n

, yaani Harmonic series, examine karni hogi, jo diverge karti hai — isliye yeh series sirf conditional hai.

"Terms

1/ n \to 0

hain, isliye

\sum 1/ n

converge karta hai."

a_{n} \to 0

necessary hai lekin sufficient nahi. Harmonic series ke terms zero ki taraf jaate hain phir bhi diverge karti hai, isliye "zero ki taraf jaana" akele kabhi prove nahi karta ki sum converge karta hai.

\sum a_{n}

aur

\sum ∣ a_{n} ∣

dono diverge karte hain, isliye

\sum a_{n}

conditionally convergent hai."

Conditional convergence ke liye signed series ka converge karna zaroori hai. Agar

\sum a_{n}

diverge kare toh koi convergence hi nahi hai — series simply divergent hai, conditional nahi.

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

ko classify karne ke liye maine Alternating Series Test run kiya."

Pehle

∣ a_{n} ∣

pe Ratio Test lagao:

∣ a_{n + 1} ∣/∣ a_{n} ∣ = 2/ (n + 1) \to 0 < 1

, jo seedha absolute convergence deta hai, isliye koi alternating machinery ki zaroorat nahi.

"Maine Ratio Test apply ki aur

L = 1

mila, isliye series diverge karti hai."

L = 1

Ratio Test ka inconclusive case hai — yeh kuch decide nahi karta. Tumhe tools switch karne honge (jaise

p

-series ya Comparison) series ko classify karne ke liye.

"Kyunki

\sum a_{n}

conditionally converge karta hai, uske partial sums bounded aur monotone hone chahiye."

Woh bounded hote hain lekin monotone nahi — ek conditionally convergent series is baat pe rely karti hai ki partial sums signs alternate hone se upar aur neeche oscillate karein; monotone bounded sums absolutely converge karte.

\sum sin (n) / n^{2}

alternate nahi karta, isliye main convergence ke baare mein kuch nahi keh sakta."

Tumhe alternation ki zaroorat nahi:

∣ sin n / n^{2} ∣ \leq 1/ n^{2}

, isliye convergent

\sum 1/ n^{2}

ke against Comparison Test seedha absolute convergence deta hai.

Why questions

Absolute convergence intuitively ordinary convergence ko kyun imply karta hai?

Signs hatana har partial sum ko as large as possible banata hai; agar woh all-positive sum bhi settle ho jaata hai, toh signed sum — jahan cancellation sirf terms ko aur kareeb laati hai — zaroor settle hoga.

Decision recipe mein hum pehle

\sum ∣ a_{n} ∣

kyun test karte hain?

Agar yeh converge kare toh hum seedha sabse strong verdict (absolute) ke saath done hain aur saare sign-based tests skip kar dete hain; sirf agar yeh diverge kare tab hum zyada delicate Alternating Series Test ki taraf jaate hain.

Comparison Test ko directly signed series jaise

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

pe kyun nahi lagaya ja sakta?

Comparison ke liye non-negative terms chahiye taaki inequalities consistently ek direction mein point karein; exactly isliye Absolute Convergence Theorem ka proof pehle non-negative quantity

a_{n} + ∣ a_{n} ∣

banata hai.

Ratio Test automatically absolute convergence kyun test karta hai?

Yeh

∣ a_{n + 1} ∣/∣ a_{n} ∣

, yaani magnitudes ke ratios ke saath kaam karta hai, isliye jab

L < 1

hota hai toh yeh

\sum ∣ a_{n} ∣

hai jo converge hota dikhta hai — isliye verdict absolute hai, merely conditional nahi.

Mathematicians absolute ko conditional convergence se zyada kyun value karte hain?

Sirf absolute convergence ke liye "the sum" ek well-defined number hai jo term order se independent hai; conditional sums ordering ke artifacts hain, jaise Riemann's theorem dramatically dikhata hai.

Ek conditionally convergent series mein infinitely many positive aur infinitely many negative terms kyun hone chahiye?

Agar sirf finitely many terms ek sign ke hote, toh tail sab ek sign ki hoti, jisse

\sum a_{n}

aur

\sum ∣ a_{n} ∣

tail mein agree karte — isliye ek ka converge karna doosre ko force karta, jo "conditional" ko contradict karta.

Edge cases

Kya ek finite sum (sirf finitely many nonzero terms) absolutely convergent hai?

Haan. Finitely many terms hamesha signs strip karke bhi ek finite number mein sum hote hain, isliye

\sum a_{n}

aur

\sum ∣ a_{n} ∣

dono trivially converge karte hain — yeh absolutely convergent hai.

Kya

a_{n} \geq 0

wali series kabhi conditionally convergent ho sakti hai?

Nahi. Koi negative terms nahi hone se

\sum a_{n} = \sum ∣ a_{n} ∣

, isliye agar series converge kare toh absolutely converge karti hai; conditional convergence ko sign cancellation chahiye jo positive terms provide nahi kar sakte.

Kya

\sum (- 1)^{n}

(terms

- 1, 1, - 1, \dots

) conditionally convergent hai?

Nahi. Terms zero ki taraf nahi jaate, isliye signed series outright diverge karti hai — yeh necessary condition fail karti hai aur na absolutely na conditionally convergent hai.

\sum 0

(har term zero) ka classification kya hai?

Absolutely convergent (to

0

). Absolute series bhi saari zeros hai aur converge karti hai, isliye yeh firmly absolute box mein hai — ek trivial lekin valid case.

Power series ke interval of convergence ke endpoints pe, kaun se convergence types possible hain?

Teeno mein se koi bhi: yeh absolutely converge kar sakti hai, sirf conditionally converge kar sakti hai, ya diverge kar sakti hai — har endpoint ko alag se test karna padta hai (dekho Power series & radius of convergence).

Kya

p

-series

\sum 1/ n^{p}

kabhi conditionally converge karti hai?

Nahi. Iske terms already positive hain, isliye convergence (jo exactly

p > 1

par hoti hai) hamesha absolute hoti hai; cancel karne ke liye koi signs hi nahi hain, isliye conditional convergence yahan impossible hai.

Recall Ek-line self-check

Agar tum kisi bhi series ke liye — jo bhi tumhare saamne aaye — yeh state kar sako ki kaun si do series mein se (signed ya absolute) har test actually examine kar raha hai, tum kabhi misclassify nahi karoge. Signed series ko test karta hai ::: the Alternating Series Test (aur convergence ki definition khud). Absolute series ko test karta hai ::: the Ratio Test, Comparison Test, aur $p$ -series jo $∣ a_{n} ∣$ pe apply hoti hai.

Connections

Alternating Series Test — wo ek tool jo legitimately conditional case certify karta hai
Comparison Test — zyaadatar absolute-convergence checks ko power karta hai
Ratio Test — $∣ a_{n} ∣$ test karta hai, isliye iska verdict hamesha absolute convergence ke baare mein hota hai
p-series · Harmonic series — benchmark examples ( $p > 1$ vs $p = 1$ )
Riemann Rearrangement Theorem — kyun conditional sums order pe depend karte hain
Power series & radius of convergence — jahan endpoint conditional convergence appear hoti hai