Question bank — Sequences — convergence, divergence, boundedness, monotonicity

True. Agar $a_n\to L$, toh $\epsilon =1$ lo: kuch $N$ ke baad saari terms $(L-1,L+1)$ mein hoti hain, aur finitely many early terms ka ek max/min hota hai, toh ek global bound $K$ exist karta hai.

False. Boundedness terms ko infinity tak escape karne se rokti hai lekin unhe wobble karne se nahi — $a_n=(-1{)}^n$ $1$ se bounded hai phir bhi kabhi settle nahi karta. Tumhe monotonicity bhi chahiye, ya Bolzano–Weierstrass jo sirf ek convergent subsequence deta hai.

False. Monotone akela kaafi nahi hai: $a_n=n$ hamesha badhta hai aur $\infty$ tak diverge karta hai. Tumhe boundedness bhi add karni hogi — yahi exactly Monotone Convergence Theorem hai.

False. Yeh sequence of terms aur series of partial sums ko confuse karta hai. $a_n=1/n\to 0$ phir bhi $\sum 1/n$ diverge karta hai. Term $\to 0$ series convergence ke liye necessary hai, kabhi sufficient nahi.

False. Convergence tail ke baare mein ek statement hai ($n>N$ arbitrarily large $N$ ke liye). Koi bhi finite block of terms change karne se tail — aur isliye limit — unchanged rehti hai.

True. Kyunki $||a_n|-0|=|a_n-0|$, wahi $N$ dono ke liye kaam karta hai, toh $\epsilon$–$N$ condition directly transfer ho jaati hai.

False. Sirf eventually: kuch $N$ ke baad terms $\epsilon =L/2$ ke andar $L$ ke paas hoti hain toh positive hain, lekin early terms kuch bhi ho sakti hain (jaise $a_1=-100$).

True. Khaas taur par poori sequence khud apni hi ek subsequence hai. (Zyada strong fact: agar $(a_n)$ diverge kare, toh koi do subsequences alag values ki taraf jaati hain.)

False. Limits unique hote hain: agar $a_n\to L$ aur $a_n\to M$ jahan $L\ne M$, toh $\epsilon =|L-M|/2$ choose karo aur tail ek saath do disjoint bands mein nahi baith sakti.

$\mathbb{R}$ mein True hai, exactly completeness ki wajah se. $\mathbb{Q}$ mein yeh fail ho sakta hai — jaise $\sqrt{2}$ ke rational approximations Cauchy hain lekin unka limit rational nahi hai.

==True sirf tab jab $B\ne 0$==. Algebra-of-limits quotient rule ko non-zero denominator limit chahiye; agar $B=0$ toh ratio diverge kar sakta hai ya kuch bhi ho sakta hai.

Error doosre clause mein hai: bounded hona converge karna imply nahi karta. $(-1{)}^n$ standard counterexample hai — yeh $1$ aur $-1$ ke beech hamesha oscillate karta hai.

Convergence ke liye distance ka monotonically shrink karna zaroori nahi hai. Terms overshoot kar ke aage-peeche ja sakti hain (jaise $a_n=(-1{)}^n/n\to 0$); bas yeh zaroori hai ki eventually har term har band ke andar ho.

Galat: Squeeze Theorem lagao. Kyunki $-\frac{1}{n}\le \frac{\sin n}{n}\le \frac{1}{n}$ aur dono bounds $\to 0$, sequence $\sin n$ ke wobble se befikar $0$ par squeeze ho jaati hai.

Fixed-point equation solve karna tab hi valid hai jab pehle yeh prove karo ki sequence converge karti hai (yahan monotone + bounded se). Fixed-point step batata hai limit kya hai, kabhi nahi ki limit exist karti hai.

"$\infty /\infty$" koi jawab nahi hai — upar aur neeche ko $n^2$ se divide karo toh $\frac{3+1/n^2}{1-5/n^2}\to 3$ milta hai. Limit ko dominant-power ratio govern karta hai.

Infinity tak diverge karna divergence ka ek form hai, convergence nahi. Convergence ke liye ek finite limit $L$ chahiye; $\infty$ koi real number nahi hai aur $\epsilon$–$N$ definition satisfy nahi ho sakti.

Sirf tab jab yeh above bounded ho. Ek unbounded increasing sequence ka $\sup =+\infty$ hota hai aur yeh diverge karta hai; sup-is-the-limit fact Monotone Convergence Theorem ka bounded waala half hai.

Kyunki "kareebi aana" har tolerance ke liye sach hona chahiye, chahe kitni bhi choti ho. Challenger ko sabse mushkil $\epsilon$ choose karne dena terms ko genuinely close hone par majboor karta hai, sirf vaguely near nahi.

Condition "$n>N\Rightarrow |a_n-L|<\epsilon$" sirf $N$ ke baad ki terms par constraint lagaati hai, jise arbitrarily far tak push kiya ja sakta hai — toh pehle $N$ terms kabhi test nahi hote.

Theorem ke proof ko ek bounded increasing sequence ke liye least upper bound $\sup \{a_n\}$ ka exist karna zaroori hai — aur completeness exactly yahi guarantee hai ki reals ke har bounded-above set ka supremum hota hai.

Yeh ek sasta necessary filter hai (convergent $\Rightarrow$ bounded), toh ek unbounded sequence turant divergent jaani jaati hai; aur monotonicity ke saath mil ke yeh MCT se sufficient ban jaata hai.

Consecutive terms $+1$ aur $-1$ hain, hamesha $2$ door. Convergence poori tail ko $2\epsilon$ width ke band mein force kar deti; $\epsilon <1$ choose karne se band itna narrow ho jaata hai ki dono ko hold nahi kar sakta.

Products/factorials ke liye terms multiply karte hain, toh ratio cleanly simplify ho jaata hai (yahan $(\frac{n}{n+1}{)}^n$ tak) jabki difference ek messy subtraction rehta hai. Ratio ko $1$ se compare karna increase/decrease seedha reveal karta hai.

$L$ ek actual real number hai jiske around terms cluster karte hain; "$\to \infty$" shorthand hai terms eventually har bound $M$ ko exceed kar deti hain — koi number nahi hai jise yeh approach karte hain, toh yeh unbounded growth ka description hai.

Agar $a_n$ do sequences ke beech trapped hai jo dono same $L$ par funnel kar rahi hain, toh outer bands close hote hain aur $a_n$ ko bhi $L$ ke around har $\epsilon$-band ke andar kheench laate hain — uske paas kahin aur jaane ki jagah hi nahi hoti.

Haan, $5$ par. Kisi bhi $\epsilon$ ke liye, $|5-5|=0<\epsilon$ sab $n$ ke liye hold karta hai, toh $N=1$ kaam karta hai. Constant sequences sabse simple convergent (aur monotone-both-ways, bounded) case hain.

Sirf tab agar yeh constant ho. $a_{n+1}\ge a_n$ aur $a_{n+1}\le a_n$ saath mein har $n$ ke liye $a_{n+1}=a_n$ force karte hain, jo ek constant sequence deta hai.

Haan, $0$ par. Rationalise karo: $\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\to 0$. Do diverging pieces ka difference phir bhi converge kar sakta hai.

Nahi. $(-1{)}^n$ sirf do values leta hai aur diverge karta hai. Yeh converge karta hai sirf tabhi jab jo values yeh repeat karta hai eventually ek single value par settle ho jaayein.

Iska koi limit nahi hai. $\cos (n\pi )=(-1{)}^n$, jo $1$ aur $-1$ ke beech oscillate karta hai — wahi divergence jo alternating sign sequence mein hoti hai.

Nahi. Yeh seedha converge bhi kar sakti hai (jaise $\frac{(-1{)}^n}{n}\to 0$), aur diverge bhi kare toh Bolzano–Weierstrass ek convergent subsequence guarantee karta hai.

Haan. $a_n=1-\frac{1}{n}$ $1$ ki taraf badhti hai lekin koi bhi term $1$ ke equal nahi; limit $\sup$ hai, jo un-attained ho sakta hai.

Nahi. Har term strictly $e$ se neeche hai (sequence usi ki taraf increase karti hai); $e$ woh supremum hai jise approach kiya jaata hai lekin kabhi reach nahi kiya jaata.