Direct comparison test: Question bank — Hinglish deep dive

4.3.8 · D5 · Maths › Calculus III — Sequences & Series › Direct comparison test

Almost har answer ke peeche do ideas hain, toh pehle inhe simple words mein pin karte hain.

Recall Do arrows (poora game)

Convergence neeche flow karta hai: kisi series ko ek finite roof ke andar trap karne ke liye, tumhe ek badi series chahiye jo converge karti ho. Agar tumhari series kisi finite cheez ke neeche baithe, toh woh bhi finite hai. Divergence upar flow karta hai: kisi series ko ek infinite floor ke upar push karne ke liye, tumhe ek choti series chahiye jo diverge karti ho. Agar tumhari series kisi infinite cheez ke upar baithe, toh woh bhi infinite hai. Kuch bhi aur — ek chota convergent partner, ya ek bada divergent partner — tumhe kuch nahi batata.

Neeche ki picture is page par har answer ke peeche ki mental image hai — jab bhi koi arrow ambiguous lage, isme wapas dekho.

Recall Roof/floor actually conclusion kyun force karta hai (engine)

Do arrows magic nahi hain — yeh Monotone convergence theorem par chalte hain. Upar ki figure dekho: partial sums $S_{n} = a_{1} + \dots + a_{n}$ banao. Kyunki har term $a_{n} \geq 0$ hai, har naya term sirf height add karta hai, toh $S_{n}$ kabhi neeche step nahi kar sakta — yeh ek increasing staircase hai (blue). Agar iske upar ek convergent $\sum b_{n}$ baitha ho, toh woh staircase bhi finite total $B$ (yellow dashed roof) se capped ho jaata hai. Ek increasing staircase jo kabhi ek fixed ceiling se upar nahi ja sakti uske paas settle hone ke alawa koi option nahi — yahi hai Monotone Convergence, aur yahi wajah hai ki "finite roof ke neeche" ka matlab "converges" hai.

True or false — justify

"Eventually" ka matlab hai "kisi fixed starting index $N$ ke liye sabhi $n \geq N$ ke liye" — shuru ke finitely many terms kabhi convergence ko affect nahi karte.

Agar

0 \leq a_{n} \leq b_{n}

sabhi

n

ke liye aur

\sum b_{n}

converge karta hai, toh

\sum a_{n}

converge karta hai.

True — DCT ka convergence wala half.

a_{n}

ke partial sums

S_{n}

increasing hain (har term

\geq 0

height add karta hai, kabhi subtract nahi karta) aur

b_{n}

ke finite total se capped hain; ek increasing-aur-capped staircase ko Monotone convergence theorem ki wajah se settle karna hi padega, isliye

\sum a_{n}

converge karta hai.

Agar

0 \leq a_{n} \leq b_{n}

aur

\sum a_{n}

converge karta hai, toh

\sum b_{n}

converge karta hai.

False — choti series ka finite hona badi ke baare mein kuch nahi kehta. Badi series phir bhi blow up kar sakti hai;

a_{n} = 1/ n^{2}

converge karta hai lekin

b_{n} = 1/ n

diverge karta hai, jabki

a_{n} \leq b_{n}

hai.

Agar

0 \leq a_{n} \leq b_{n}

aur

\sum a_{n}

diverge karta hai, toh

\sum b_{n}

diverge karta hai.

True — divergence wala half.

b_{n}

ke partial sums term-by-term

a_{n}

ke partial sums ko dominate karte hain, aur

a_{n}

ke partial sums pehle se infinity ki taraf ja rahe hain, toh badi staircase bhi infinity tak drag ho jaati hai (koi finite cap exist nahi kar sakti).

Agar

\sum a_{n}

converge karta hai aur

\sum b_{n}

converge karta hai, toh DCT woh tool tha jisne yeh prove kiya.

False — DCT ek possible tool hai, akela nahi. Convergence ratio, root, integral, ya alternating-series tests se bhi aa sakti hai; DCT ko specifically ek jaane-maane benchmark ke against term-by-term inequality chahiye.

DCT ko

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

par directly apply kiya ja sakta hai.

False — terms ka sign badalta hai, toh partial-sum staircase neeche step kar sakti hai, jo Monotone convergence theorem ko chahiye wali monotonicity ko tod deta hai. Iske bajaaye absolute values compare karo (dekho Absolute convergence) ya Alternating Series Test use karo.

a_{n} \leq b_{n}

sabhi

n

ke liye" DCT se actually require hone se stronger hai.

True — DCT ko inequality sirf eventually chahiye (i.e.,

n \geq N

ke liye). Finitely many terms ek finite sum hain aur yeh nahi badal sakte ki tail converge karta hai ya nahi.

Agar

lim_{n \to \infty} a_{n} = 0

, toh

\sum a_{n}

comparison se converge karta hai.

False — terms ka zero ki taraf jaana necessary hai lekin sufficient nahi. Harmonic series

\sum 1/ n

ke terms zero ki taraf jaate hain phir bhi diverge karta hai; DCT ko phir bhi ek genuine convergent upper bound chahiye.

Yeh dikhana ki

a_{n} \leq \frac{1}{2 ^{n}}

(aur

a_{n} \geq 0

) proves karta hai ki

\sum a_{n}

converge karta hai.

True —

\sum 1/ 2^{n}

ek convergent Geometric series hai (ratio

1/2 < 1

), ek badi series jo

a_{n}

ke increasing partial sums ko ek finite roof ke neeche cap karti hai, toh Monotone Convergence settling force karta hai. Sahi direction.

Yeh dikhana ki

a_{n} \geq \frac{1}{2 ^{n}}

proves karta hai ki

\sum a_{n}

converge karta hai.

False — yeh ek chota convergent partner hai, useless direction. Kisi finite cheez ke upar baithne se staircase par koi ceiling nahi aata;

a_{n}

phir bhi diverge kar sakta hai.

Spot the error

Har argument ek real inequality use karke ek conclusion tak pahuncha hai. Dhundho ki logic kahan galat hai, algebra nahi. Neeche ki figure ek nazar mein do useless directions dikhati hai.

"Yeh dikhane ke liye ki

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

converge karta hai, main note karta hun ki

\frac{1}{n ^{2} + 1} \geq \frac{1}{2 n ^{2}}

, aur

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

converge karta hai, toh kaam ho gaya."

Wrong direction — partner chota aur convergent hai, jo kuch prove nahi karta. Iske bajaaye

\frac{1}{n ^{2} + 1} \leq \frac{1}{n ^{2}}

use karo (ek bada convergent series); yeh ise upar se trap karta hai.

"Yeh dikhane ke liye ki

\sum \frac{1}{n}

diverge karta hai, main note karta hun ki

\frac{1}{n} \leq \frac{1}{1} = 1

, aur divergent constant series

\sum 1

se compare karne se kaam khatam ho jaata hai."

Wrong direction —

\frac{1}{n}

\sum 1

se neeche hai, aur kisi divergent series se chota hona kuch prove nahi karta. Iske bajaaye

\frac{1}{n} \geq \frac{1}{n}

use karo (

p = 1/2 \leq 1

vs harmonic), ek chota divergent partner jo upar push karta hai.

"Kyunki

\frac{2 + s i n n}{n ^{2}}

ka numerator kabhi kabhi negative-ish hai, DCT apply nahi ho sakta."

Premise galat hai:

- 1 \leq sin n \leq 1

1 \leq 2 + sin n \leq 3

milta hai, jo hamesha positive hai. Terms non-negative hain, toh DCT bound

\leq \frac{3}{n ^{2}}

ke saath bilkul apply hota hai.

\frac{1}{n ^{2} - 10} \leq \frac{1}{n ^{2}}

sabhi

n

ke liye, toh

\sum \frac{1}{n ^{2} - 10}

converge karta hai."

Jo inequality batayi gayi hai woh ulti hai: har

n

ke liye jahan

n^{2} > 10

(i.e.

n \geq 4

) denominator

n^{2} - 10

n^{2}

se chota hai, toh

\frac{1}{n ^{2} - 10} > \frac{1}{n ^{2}}

— ek badi series, convergence ke liye useless direction. (

n \leq 3

ke liye term negative bhi hai.) Fix karne ke liye, thoda loose convergent roof se bound karo jaise

\frac{1}{n ^{2} - 10} \leq \frac{2}{n ^{2}}

sufficiently large

n

ke liye (yahan

n^{2} - 10 \geq n^{2} /2 ⟺ n^{2} \geq 20 ⟺ n \geq 5

), phir

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

converge karta hai.

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

converge karta hai kyunki

\frac{l n n}{n} \leq \frac{n}{n} = 1

Term-by-term ko constant

1

se bound karna useless hai —

\sum 1

diverge karta hai, aur kisi divergent series se neeche hona kuch nahi kehta. Sahi approach:

ln n \geq 1 ⟺ n \geq e \approx 2.718

, toh eventually (

n \geq 3

ke liye)

\frac{l n n}{n} \geq \frac{1}{n}

, ek chota divergent partner — isliye yeh diverge karta hai.

\sum a_{n}

jahan

a_{n} = \frac{c o s ^{2} n}{n ^{2}}

— kyunki

cos^{2} n

oscillate karta hai, koi clean comparison exist nahi karta, toh test inconclusive hai."

Yeh jhuta despair hai —

0 \leq cos^{2} n \leq 1

0 \leq a_{n} \leq \frac{1}{n ^{2}}

milta hai, ek clean bada convergent bound. Bounded numerator mein oscillation kabhi obstacle nahi hoti; ise uske maximum se bound karo.

Why questions

DCT terms ke non-negative hone par kyun insist karta hai?

Tabhi partial sums monotone increasing hote hain, jo ki Monotone convergence theorem ko exact hypothesis chahiye ki ek bounded increasing sequence converge karti hai. Sign changes monotonicity aur poora proof tod dete hain.

Choti series ka converge karna badi series ke baare mein koi information kyun nahi deta?

Ek finite lower structure uske upar ki cheez ko cap nahi kar sakta; badi series ke paas grow karne ka extra "room" hai. Sirf ek upper bound (ek badi convergent series) finiteness force kar sakta hai.

p = 2

wali

p

-series go-to convergent benchmark kyun hai?

p-series test ke mutabik,

\sum 1/ n^{p}

exactly tabhi converge karta hai jab

p > 1

, aur

1/ n^{2}

simple, positive, aur kaafi rational-term series ke tails ko dominate karta hai — ek aasaan roof jise dhundhna aasaan hai.

Harmonic series canonical divergent benchmark kyun hai?

Yeh borderline

p = 1

case hai jo bahut mushkil se diverge karta hai, toh koi bhi series jinke terms eventually

\geq 1/ n

hain woh ek infinite floor ke upar push ho jaate hain — sabse clean divergence lever.

DCT humein sum ki value kabhi kyun nahi batata?

Yeh sirf behaviour compare karta hai (finite vs infinite) term-by-term; actual total exactly sabhi terms par depend karta hai, jo ek inequality deliberately chhod deti hai. Yeh "kya yeh converge karta hai?" ka jawab deta hai na ki "kis value par?".

Hum finitely many misbehaving early terms ko kyun ignore kar sakte hain?

Ek finite sum hamesha finite hota hai, toh yeh nahi badal sakta ki infinite tail converge karta hai ya nahi. Convergence tail ki property hai, isliye "eventually

n \geq N

" kaafi hai.

Jab inequality awkward ho toh Limit comparison test kyun prefer karte hain?

DCT ko chahiye ki tum ek clean term-by-term inequality prove karo; agar

a_{n}

aur

b_{n}

sirf alike behave karte hain lekin tum direction pin nahi kar sakte, toh limit-comparison ratio inequality ko sidestep kar deta hai jabki same conclusion deta hai.

Integral test kabhi kabhi DCT ke saath kyun use hota hai?

Yeh benchmark series (

\sum 1/ n^{p}

jaise) ko convergent ya divergent classify karne ka alternate tarika hai, jo tumhe trusted partner deta hai jise DCT phir compare karta hai.

Edge cases

DCT kya conclude karta hai agar

a_{n} = 0

infinitely many

n

ke liye?

Kuch nahi toot-ta — zeros non-negative hain aur

0 \leq a_{n} \leq b_{n}

trivially satisfy karte hain. Zero terms sirf help karte hain; comparison aur monotonicity arguments unaffected hain.

Agar

a_{n} \leq b_{n}

lekin

\sum b_{n}

diverge karta hai, toh

\sum a_{n}

ke baare mein DCT kya keh sakta hai?

Kuch nahi — badi series ka diverge karna choti series ko converge ya diverge karne ki freedom deta hai. Tumhare paas divergence ke liye wrong-direction pairing hai aur convergence ke liye koi upper bound nahi.

Series ke terms

\geq 0

hain lekin natural bound sirf even

n

ke liye hold karta hai, odd

n

ke liye fail karta hai — kya DCT usable hai?

Directly nahi, kyunki DCT ko inequality sabhi

n \geq N

ke liye chahiye, kisi subsequence ke liye nahi. Tumhe ek single bound dhundhna hoga jo poori tail ke liye valid ho (ya split karke har piece handle karo).

Dono

a_{n} \geq 0

aur

a_{n} \leq b_{n}

hold karte hain, aur

\sum a_{n}

converge karta hai — kya yeh

b_{n} \to 0

force karta hai?

Nahi — badi series

b_{n}

choti series ke converge karne se bilkul unconstrained hai.

b_{n}

ko zero ki taraf jaana bhi zaruri nahi; DCT koi downward information nahi deta.

Ek series hai jisme

a_{n} \to 0

aur yeh

\frac{1}{n ^{2}} \leq a_{n} \leq \frac{1}{n}

ke beech squeezed hai — convergent hai ya divergent?

Sirf in bounds se inconclusive: lower (

1/ n^{2}

) converge karta hai aur upper (

1/ n

) diverge karta hai, toh DCT koi verdict nahi deta. Koi bhi trap fire nahi karta;

a_{n}

ke true dominant term ke saath Limit comparison test try karo.