Exercises — Sequences — convergence, divergence, boundedness, monotonicity

Neeche ki figure in teen sequences ke pehle das terms ek axis pe plot karti hai — yahi picture hai jisse answer padha ja sakta hai.

Figure kaise padhen: cyan circles ($a_n=1/n$) neeche ki taraf march karte hain aur dotted floor $0$ se chipke rehte hain — decreasing aur kabhi $(0,1]$ se bahar nahi jaate. Amber squares ($b_n=(-1{)}^n$) hamesha $+1$ aur $-1$ ke beech hop karte rehte hain — do dotted lines ke beech trapped hain lekin zig-zag karte hain, isliye koi single direction nahi. White triangles ($d_n=2-1/n$) dotted ceiling $2$ ki taraf upar climb karte hain bina use touch kiye.

(a) $a_n=1,\frac{1}{2},\frac{1}{3},\dots$ — har term $(0,1]$ mein hai, isliye bounded. Har term pichle se chhota hai, isliye monotone (decreasing). (b) $b_n=-1,1,-1,1,\dots$ — $[-1,1]$ mein trapped hai isliye bounded, lekin upar-neeche jump karta hai (amber squares), isliye not monotone. (c) $c_n=1,2,3,\dots$ — har ceiling ko cross kar jaata hai, isliye unbounded (above). Hamesha increasing hai, isliye monotone (increasing). (Plot nahi kiya — yeh frame ke upar se nikal jaata, jo "unbounded" ka visual meaning hi hai.) (d) $d_n=1,\frac{3}{2},\frac{5}{3},\dots$ $2$ ki taraf climb karta hai lekin pahuncha nahi (white triangles) — $[1,2)$ mein trapped hai isliye bounded, aur har term pichle se zyada hai, isliye monotone (increasing).

(a) $\to 0$ — cyan circles floor pe settle ho jaate hain. (b) diverge karta hai — amber squares kabhi settle nahi karte. (c) diverge karta hai ($\infty$ ki taraf). (d) $\to 2$ — white triangles ceiling $2$ ke neeche pack ho jaate hain.

$n^2$ se divide kyun karein? Bade $n$ ke liye $n$ ki sabse badi power dominate karti hai; upar aur neeche se divide karne par pata chalta hai kaunse pieces survive karte hain aur kaunse vanish ho jaate hain. $\frac{3n^2+1}{n^2-5}=\frac{3+\frac{1}{n^2}}{1-\frac{5}{n^2}}.$ Algebra of limits se, $\frac{1}{n^2}\to 0$ aur $\frac{5}{n^2}\to 0$, isliye limit hai $\frac{3+0}{1-0}=\boxed{3}$.

Rationalize kyun karein? Jaisa likha hai yeh $\infty -\infty$ hai, ek indeterminate form. Conjugate se multiply karne par difference ek aisi fraction mein badal jaata hai jo hum padh sakte hain. $\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\frac{(n+1)-n}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}.$ Denominator bina bound ke badhta hai, isliye fraction $\to \boxed{0}$.

Squeeze kyun karein? Hum $\cos n$ ko term-by-term track nahi kar sakte (yeh wander karta hai), lekin hum use bound kar sakte hain: $\cos n$ kabhi $[-1,1]$ se bahar nahi jaata. Us band ko positive number $n$ se divide karne par milta hai $-\frac{1}{n}\le \frac{\cos n}{n}\le \frac{1}{n}.$ Squeeze ke peeche $\epsilon$–$N$ engine (clearly bataya gaya): koi bhi $\epsilon >0$ fix karo. Lower bound $-\frac{1}{n}\to 0$ ka matlab hai ek $N_1$ exist karta hai jahan $n>N_1\Rightarrow -\frac{1}{n}>-\epsilon$; upper bound $\frac{1}{n}\to 0$ ka matlab hai ek $N_2$ exist karta hai jahan $n>N_2\Rightarrow \frac{1}{n}<\epsilon$. $n>N=\max (N_1,N_2)$ ke liye, sandwich chain karo: $-\epsilon <-\frac{1}{n}\le \frac{\cos n}{n}\le \frac{1}{n}<\epsilon ⟹\left|\frac{\cos n}{n}-0\right|<\epsilon .$ Yahi $\epsilon$–$N$ condition hai, isliye trapped middle $\to \boxed{0}$. (Yeh exactly L5·Q1 ka general Squeeze proof hai $L=0$ ke saath apply kiya gaya.)

Hum kya chahte hain: koi bhi $\epsilon >0$ diya ho, ek cutoff $N$ produce karo taaki $n>N\Rightarrow |a_n-2|<\epsilon$. Pehle gap simplify karo: $|a_n-2|=\left|\frac{2n}{n+1}-2\right|=\left|\frac{2n-2(n+1)}{n+1}\right|=\left|\frac{-2}{n+1}\right|=\frac{2}{n+1}.$ Kyun: gap ek clean formula hai jise hum $\epsilon$ se neeche force kar sakte hain. $\frac{2}{n+1}<\epsilon ⟺n+1>\frac{2}{\epsilon }⟺n>\frac{2}{\epsilon }-1.$ Choose $N=\lceil \frac{2}{\epsilon }\rceil$. Tab $n>N\Rightarrow n>\frac{2}{\epsilon }-1\Rightarrow \frac{2}{n+1}<\epsilon$. ✓ Isliye $a_n\to 2$.

Method (i) — Squeeze (fully self-contained, koi external constant use nahi). Term ko $n$ factors ke product ke roop mein likho: $a_n=\frac{n!}{n^n}=\frac{1\cdot 2\cdot 3\cdots n}{n\cdot n\cdot n\cdots n}=\underset{\text{pehla factor}}{\underbrace{\frac{1}{n}}}\cdot \underset{\text{har ek }\le 1}{\underbrace{\frac{2}{n}\cdot \frac{3}{n}\cdots \frac{n}{n}}}.$ Baaki $n-1$ factors $\frac{k}{n}$ (jahan $k=2,\dots ,n$) mein se har ek $0<\frac{k}{n}\le 1$ satisfy karta hai, isliye unka product $\le 1$ hai. Isliye $0<a_n\le \frac{1}{n}.$ Dono outer bounds ($0$ aur $\frac{1}{n}$) $0$ ki taraf jaate hain, isliye Squeeze Theorem se $a_n\to \boxed{0}$. Yeh method apne aap complete hai aur koi nayi constant introduce nahi karta.

Method (ii) — ratio, ek honest geometric-decay proof ke saath. Factorials aur powers ke saath, consecutive terms compare karna aasaan hai: $\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{(n+1{)}^{n+1}}\cdot \frac{n^n}{n!}=\frac{(n+1)n^n}{(n+1{)}^{n+1}}=\frac{n^n}{(n+1{)}^n}={\left(\frac{n}{n+1}\right)}^n.$ Ab hume "ratio chhota hai" ko decay ka actual proof banana hai. Sabse clean self-contained bound: har $n\ge 1$ ke liye, $\frac{a_{n+1}}{a_n}={\left(\frac{n}{n+1}\right)}^n\le \frac{n}{n+1}<1,$ kyunki $(0,1)$ mein kisi number ko power $n\ge 1$ tak raise karne par woh $\le$ apne aap ho jaata hai (aur factors each $<1$ multiply karne par aur ghatta hai). Lekin hum ek fixed ceiling $1$ se neeche pin karna chahte hain: har $n\ge 1$ ke liye, $\frac{n}{n+1}\le \frac{1}{2}$ galat hai ($n=2$ se aage yeh $\frac{1}{2}$ se zyada hai), isliye hum pehle index pe evaluate karke ek uniform bound laate hain jo sab $n\ge 1$ ke liye valid ho: $\frac{a_{n+1}}{a_n}={\left(\frac{n}{n+1}\right)}^n\le {\left(\frac{n}{n+1}\right)}^1\le \frac{1}{2}\text{bade }n\text{ ke liye invalid hai};$ isliye hum genuinely uniform ceiling lete hain: doosre step se aage constant $q$ choose karo. $n\ge 2$ ke liye, $\frac{a_{n+1}}{a_n}={\left(\frac{n}{n+1}\right)}^n\le {\left(\frac{n}{n+1}\right)}^2\le {\left(\frac{2}{3}\right)}^2=\frac{4}{9}=:q<1.$ Yahan humne do monotone facts use kiye: base $<1$ ko badi power ($n\ge 2$) tak raise karne par woh chhota ho jaata hai, aur $\frac{n}{n+1}$ $n$ mein increase karta hai isliye $n=2$ par yeh sabse bada hai... nahi — $\frac{n}{n+1}$ $1$ ki taraf increase karta hai, isliye hum ise $n=2$ ki value se upar bound nahi kar sakte. Sahi uniform ceiling ke liye hum sirf exponent ko $2$ pe fixed rakhte hain aur ek factor ki drop bound karte hain: kyunki ${\left(\frac{n}{n+1}\right)}^n\le {\left(\frac{n}{n+1}\right)}^2$ aur yeh last quantity har $n\ge 2$ ke liye $<1$ hai lekin $1$ ki taraf creep karti hai, koi single $q<1$ fixed base se uniformly kaam nahi karta. Isliye hum ceiling ko globalize karne ki jagah localize karte hain, jo exactly wahi kaam karta hai jo ek limit karta hai — aur hum ab woh limit $e$ invoke kiye bina supply karte hain: ${\left(\frac{n}{n+1}\right)}^n=\frac{1}{{\left(1+\frac{1}{n}\right)}^n}.$ Denominator ${\left(1+\frac{1}{n}\right)}^n$ ek increasing sequence hai (yeh L4-style monotonicity parent note mein already use hui hai) aur $3$ se upar bounded hai (parent note, binomial bound). Isliye ${\left(1+\frac{1}{n}\right)}^n\ge {\left(1+\frac{1}{1}\right)}^1=2$ har $n\ge 1$ ke liye, isliye $\frac{a_{n+1}}{a_n}=\frac{1}{{\left(1+\frac{1}{n}\right)}^n}\le \frac{1}{2}=:q<1\text{sab }n\ge 1\text{ ke liye}.$ Yahi honest uniform ceiling hai. Ab geometric decay conclude karo: har $n\ge 1$ ke liye, $a_{n+1}\le q{a}_n⟹a_n\le q{a}_{n-1}\le q^2{a}_{n-2}\le \cdots \le q^{n-1}{a}_1={\left(\frac{1}{2}\right)}^{n-1}\cdot \frac{1!}{1^1}={\left(\frac{1}{2}\right)}^{n-1}.$ Kyunki $0<q<1$, $q^{n-1}\to 0$, isliye $0<a_n\le q^{n-1}\to 0$ aur Squeeze se phir $a_n\to 0$. Koi undefined constant kahin use nahi hua: ceiling $q=\frac{1}{2}$ sirf ${\left(1+\frac{1}{n}\right)}^n\ge 2$ se aayi, jo ek increasing sequence ke pehle term ke baare mein ek basic inequality hai. (Readers jo jaanna chahte hain ki is sequence ka limit kahan hai, woh The number e dekh sakte hain — lekin upar ka proof kabhi uski value use nahi karta.)

Parent ka recipe follow karo $L=2,\epsilon =1$ ke saath. L3·Q1 se, $|a_n-2|=\frac{2}{n+1}\le 1$ sab $n\ge 1$ ke liye, isliye har term $[1,3)$ mein hai: tail $|L|+1=3$ se bounded hai. (Single) head term $a_1=1$ hai. $K=\max (3,|a_1|)=\boxed{3}$ lo: $|a_n|\le 3$ sab $n$ ke liye.

Hum boundedness + monotonicity ⟹ MCT combine karte hain, phir fixed point solve karte hain. Staircase figure plan dikhata hai; Steps 1–4 padhne ke baad ise dekho.

Step 0 — positivity (baad mein zaroori). Har term positive hai: $a_1=1>0$, aur agar $a_n>0$ toh $a_{n+1}=\sqrt{2+a_n}>0$ (positive number ka square root). Induction se $a_n>0$ sab $n$ ke liye. Kyun bother karein: Step 2 mein $a_{n+1}$ aur $a_n$ ko unke squares ke through compare karenge, aur squaring order tabhi preserve karta hai jab dono nonnegative hon — isliye pehle jaanna zaroori hai ki dono positive hain. Step 1 — 2 se upar bounded (induction). $a_1=1\le 2$. Agar $a_n\le 2$ toh $a_{n+1}=\sqrt{2+a_n}\le \sqrt{2+2}=2$. Isliye $a_n\le 2$ sab $n$ ke liye. 2 kyun? Yeh suspected limit hai; jo ceiling tum expect karte ho usse test karna standard move hai. Step 2 — increasing. Squares ka difference compute karo: $a_{n+1}^2-a_n^2=(2+a_n)-a_n^2=-(a_n^2-a_n-2)=-(a_n-2)(a_n+1).$ Kyunki $a_n\le 2$ hai isliye $(a_n-2)\le 0$, aur kyunki $a_n>0$ (Step 0) hai isliye $(a_n+1)>0$; unka product $\le 0$ hai, isliye $-(a_n-2)(a_n+1)\ge 0$. Isliye $a_{n+1}^2\ge a_n^2$. Kyunki dono $a_{n+1}>0$ aur $a_n>0$ hain, squaring ne order preserve kiya, isliye $a_{n+1}\ge a_n$. Square kyun? Recursion mein square root hai; squaring se woh hat jaata hai aur sign padhne layak ho jaata hai. Step 3 — MCT. Increasing + bounded above $\Rightarrow$ kisi $L$ pe converge karta hai. Step 4 — fixed point solve karo. Dono $a_n\to L$ aur $a_{n+1}\to L$, aur $a_{n+1}=\sqrt{2+a_n}$ se milta hai $L=\sqrt{2+L}$, yaani $L^2-L-2=0\Rightarrow (L-2)(L+1)=0$. Kyunki terms positive hain, $L=\boxed{2}$.

Staircase figure padhna: horizontal axis pe $a_1=1$ pe amber dot se shuru karo. Cyan curve $y=\sqrt{2+x}$ tak vertically upar jao — woh height $a_2$ hai. White diagonal $y=x$ tak horizontally jao taaki $a_2$ axis pe copy ho jaaye, phir $a_3$ ke liye phir curve tak vertically jao, aur aise hi aage. Amber staircase seedha aur upar climb karta hai, uske steps chhote hote jaate hain jab woh white dot ki taraf aata hai jahan curve diagonal se milti hai — fixed point $L=2$.

Boundedness: $\left|\frac{(-1{)}^{n}n}{n+1}\right|=\frac{n}{n+1}<1$, isliye $(b_n)$ bounded hai. Lekin yeh akele convergence guarantee nahi karta. Parity ke hisaab se split karo. Even $n$ ke liye: $b_n=\frac{n}{n+1}\to 1$. Odd $n$ ke liye: $b_n=\frac{-n}{n+1}\to -1$. Even-indexed subsequence $1$ ke paas jaata hai, odd-indexed $-1$ ke paas. Ek convergent sequence har subsequence ko same limit pe force karta hai; yahan do subsequences alag-alag answer de rahe hain, isliye $(b_n)$ diverge karta hai. Bolzano–Weierstrass angle: sequence bounded hai, isliye B–W guarantee karta hai ki koi na koi convergent subsequence exist karta hai — hamare paas toh do (limits $1$ aur $-1$) hain. Boundedness convergent subsequences kharidti hai, poore sequence ki convergence nahi.

Goal: given $\epsilon >0$, aisa $N$ dhundho ki $n>N\Rightarrow |a_n-L|<\epsilon$. Kyunki $b_n\to L$: $\exists N_1$ jahan $n>N_1\Rightarrow L-\epsilon <b_n$. Kyunki $c_n\to L$: $\exists N_2$ jahan $n>N_2\Rightarrow c_n<L+\epsilon$. Do cutoffs phir max kyun? Har limit apni wall deta hai; $N=\max (N_1,N_2)$ lene par dono promises ek saath hold karte hain. $n>N=\max (N_1,N_2)$ ke liye, sandwich chain karo: $L-\epsilon <b_n\le a_n\le c_n<L+\epsilon ⟹|a_n-L|<\epsilon .$ Yahi exactly $\epsilon$–$N$ condition hai, isliye $a_n\to L$. $■$

Strategy (proof by contradiction): maano $A\ne A^{\prime }$. $\epsilon =\frac{|A-A^{\prime }|}{2}>0$ lo — do candidate limits ke beech ke gap ka aadha. Aadha kyun? Taaki do bands $(A-\epsilon ,A+\epsilon )$ aur $(A^{\prime }-\epsilon ,A^{\prime }+\epsilon )$ overlap na kar sakein. Convergence to $A$: $\exists N_1$, $n>N_1\Rightarrow |a_n-A|<\epsilon$. Convergence to $A^{\prime }$: $\exists N_2$, $n>N_2\Rightarrow |a_n-A^{\prime }|<\epsilon$. Ek index par combine karo. Koi bhi $n>N=\max (N_1,N_2)$ pick karo; tab dono inequalities us single $n$ par hold karti hain. Ab triangle inequality $|x+y|\le |x|+|y|$ use karo $x=A-a_n$ aur $y=a_n-A^{\prime }$ ke saath (note karo $x+y=A-A^{\prime }$): $|A-A^{\prime }|=|(A-a_n)+(a_n-A^{\prime })|\le |A-a_n|+|a_n-A^{\prime }|<\epsilon +\epsilon =2\epsilon =|A-A^{\prime }|.$ Yeh kehta hai $|A-A^{\prime }|<|A-A^{\prime }|$, jo impossible hai (koi number apne aap se strictly chhota nahi hota). Yeh impossibility sirf $A\ne A^{\prime }$ assume karne se aayi (wahi toh $\epsilon >0$ banata tha). Isliye $A=A^{\prime }$: ek sequence ki zyada se zyada ek hi limit hoti hai. $■$

Hum sirf elementary algebra plus MCT use karte hain — koi external inequality assume nahi ki; jo ek bound chahiye woh scratch se prove ki gayi hai neeche.

Positivity. $x_1=2>0$; agar $x_n>0$ toh $x_{n+1}=\frac{1}{2}(x_n+\frac{2}{x_n})$ positive numbers ka average hai, isliye $x_{n+1}>0$. Induction se sab terms positive hain.

Step 1 — $\sqrt{2}$ se neeche bounded (seedha prove kiya, AM–GM quote kiye bina). Kisi bhi $x_n>0$ ke liye, $x_{n+1}^2-2={\left(\frac{1}{2}\left(x_n+\frac{2}{x_n}\right)\right)}^2-2=\frac{x_n^2}{4}+1+\frac{1}{x_n^2}-2=\frac{x_n^2}{4}-1+\frac{1}{x_n^2}={\left(\frac{x_n}{2}-\frac{1}{x_n}\right)}^2\ge 0.$ Key move yeh hai ki expression ek perfect square ke roop mein recognize karo, jo free mein $\ge 0$ hota hai. Isliye $x_{n+1}^2\ge 2$, aur kyunki $x_{n+1}>0$ hai, $x_{n+1}\ge \sqrt{2}$ har $n\ge 1$ ke liye. Step 2 — decreasing ($n\ge 2$ se). $x_n^2\ge 2$ (Step 1) use karke: $x_n-x_{n+1}=x_n-\frac{1}{2}\left(x_n+\frac{2}{x_n}\right)=\frac{x_n^2-2}{2x_n}\ge 0,$ kyunki numerator $x_n^2-2\ge 0$ aur denominator $2x_n>0$. Isliye sequence decrease karti hai. Step 3 — MCT. Decreasing + bounded below $\Rightarrow$ $L=\inf x_n$ pe converge karta hai; is infimum ka existence exactly Completeness of $\mathbb{R}$ (greatest-lower-bound property) hai. Solve karo $L=\frac{1}{2}(L+\frac{2}{L})\Rightarrow L^2=2\Rightarrow L=\boxed{\sqrt{2}}$ (positive root, kyunki sab terms $>0$ hain). Cauchy angle: $\mathbb{R}$ mein "convergent" aur "Cauchy" ka matlab ek hi hota hai (phir completeness se). Yeh sequence Cauchy hai — uske terms arbitrarily close ho jaate hain — lekin agar hum sirf $\mathbb{Q}$ mein rehte, toh target $\sqrt{2}$ missing hota, aur sequence ek Cauchy sequence hoti jiska koi limit nahi. Completeness exactly woh hole band karti hai.

Ek recursion converge karta hai yeh prove karne ke liye, kaunse order mein? ::: (1) positivity agar zaroori ho, (2) bounded, (3) monotone $\Rightarrow$ MCT, (4) $L=f(L)$ solve karo. Bounded alone tumhe kya deta hai (convergence nahi)? ::: Ek convergent subsequence (Bolzano–Weierstrass). $\epsilon$–$N$ proofs mein $N$ kis par depend kar sakta hai? ::: Sirf $\epsilon$ (aur $L$) par — running index $n$ par kabhi nahi. Do subsequences ke alag limits ka matlab kya hai? ::: Poora sequence diverge karta hai. Heron's sequence ko completeness kyun chahiye? ::: Uski limit $\sqrt{2}\notin \mathbb{Q}$ hai; sirf complete $\mathbb{R}$ hi ise supply kar sakta hai.