Sequences — convergence, divergence, boundedness, monotonicity: Worked examples

4.3.1 · D3 · Maths › Calculus III — Sequences & Series › Sequences — convergence, divergence, boundedness, monotonici

Scenario matrix

Har sequence jo tumhe milegi, in case classes mein se kisi ek mein aati hai. Har row ek alag trap ya technique hai; neeche diye worked examples mein woh cell tag ki gayi hai jise wo cover karta hai, taaki tum dekh sako ki poora space fill hua hai.

#	Case class	Kya mushkil hai	Weapon of choice	Example
A	Rational in $n$ (polynomial over polynomial)	kaun si power dominate karti hai?	top power of $n$ se divide karo	Ex 1
B	Do cheezon ka difference jo dono $\to \infty$ hain	$\infty - \infty$ undefined hai	rationalise / rewrite karo	Ex 2
C	Trapped oscillation ( $sin$ , $cos$ numerator mein)	numerator kabhi settle nahi karta	Squeeze Theorem	Ex 3
D	Pure oscillation (shrinking nahi)	koi limit hi nahi	$ε$ – $N$ contradiction	Ex 4
E	Factorial / exponential race	kaun zyada fast grow karta hai?	ratio $a_{n + 1} / a_{n}$	Ex 5
F	Recursive (khud se define hota hai)	limit ek fixed point mein chhupa hai	MCT + solve $L = f (L)$	Ex 6
G	Number $e$ shape $(1 + x / n)^{n}$	$1^{\infty}$ jaisa lagta hai	known $e$ -limit + logs	Ex 7
H	Zero / degenerate & sign cases	$n = 0$ ?, negative terms, $r = 1$ ?	parameter ki har value check karo	Ex 8
I	Real-world word problem	words ko sequence mein translate karo	model banao phir limit lo	Ex 9
J	Exam twist (piecewise / hidden divergence)	ek "seedha" formula jo phir bhi diverge karta hai	subsequences mein split karo	Ex 10

Ab hum har cell ko walk karenge.

Case A — rational in $n$

Worked example

(Cell A) $n \to \infty lim \frac{3 n ^{2} + 1}{n ^{2} - 5}$ find karo. Forecast: Top aur bottom dono blow up karte hain. Kya woh same rate par blow up karte hain? Aage padhne se pehle ek finite number guess karo.

Top aur bottom dono ko $n^{2}$ se divide karo (jo sabse badi power present hai). $\frac{3 n ^{2} + 1}{n ^{2} - 5} = \frac{3 + \frac{1}{n ^{2}}}{1 - \frac{5}{n ^{2}}} .$ Yeh step kyun? Dominant power se divide karne par har "runaway" term ek shrinking $1/ n^{2}$ term mein badal jaati hai jiska limit hum pehle se jaante hain, yaani $0$ .
Algebra of limits apply karo. Jab $n \to \infty$ , tab $\frac{1}{n ^{2}} \to 0$ aur $\frac{5}{n ^{2}} \to 0$ , toh fraction $\to \frac{3 + 0}{1 - 0} = 3$ . Yeh step kyun? Convergent sequences ke sums, quotients aur constants, unke limits ke sum/quotient pe converge karte hain (algebra of limits parent note se).

Verify karo: $n = 1000$ daalo: $\frac{3 000 001}{999 995} \approx 3.00002$ — pehle se $3$ ke kaafi kareeb. ✓

Case B — do infinities ka difference ( $\infty - \infty$ )

Worked example

(Cell B) $n \to \infty lim (n + 1 - n)$ find karo. Forecast: $n + 1$ aur $n$ dono $\infty$ ki taraf badhte hain. Unka difference $\infty$ , ek constant, ya $0$ ho sakta hai. Guess karo.

Conjugate se multiply karo — yeh $1$ ki ek aisi form hai jo kuch bhi nahi bigaadti: $n + 1 - n = (n + 1 - n) \cdot \frac{n + 1 + n}{n + 1 + n} = \frac{( n + 1 ) - n}{n + 1 + n} = \frac{1}{n + 1 + n} .$ Yeh step kyun? $(a - b) (a + b) = a^{2} - b^{2}$ numerator se square roots khatam kar deta hai, ek undefined $\infty - \infty$ ko ek clean "chhota over bada" mein badal deta hai.
Limit padh lo. Denominator $\to \infty$ hai, toh poora fraction $\to 0$ . Yeh step kyun? Constant over "runaway" hamesha $0$ ki taraf jaata hai.

Verify karo: $n = 100$ : $101 - 100 = 10.0499 - 10 = 0.0499$ , aur $\frac{1}{101 + 100} = \frac{1}{20.0499} = 0.0499$ . ✓ Dono same shrinking value dete hain.

Case C — oscillation jo ek shrinking envelope mein trapped hai

Worked example

(Cell C) $n \to \infty lim \frac{sin n}{n}$ find karo. Forecast: $sin n$ $- 1$ aur $1$ ke beech hamesha wobble karta rehta hai. $n$ se divide karne par — kya wobble survive karta hai?

Wobble ko bound karo. Har $n$ ke liye, $- 1 \leq sin n \leq 1$ , toh positive number $n$ se divide karne par: $- \frac{1}{n} \leq \frac{s i n n}{n} \leq \frac{1}{n} .$ Yeh step kyun? Hum $sin n$ ko term by term compute nahi kar sakte, lekin hum usse do aisi sequences ke beech cage kar SAKTE hain jo humein samajh aati hain. Figure dekho: sequence isolated dots par exist karti hai (term "between" $n = 3$ aur $n = 4$ ke beech exist nahi karta), har magenta dot violet–orange funnel $\pm \frac{1}{n}$ ke andar baitha hai.
Squeeze karo. $- \frac{1}{n} \to 0$ aur $\frac{1}{n} \to 0$ dono. Squeeze Theorem se, $\frac{s i n n}{n} \to 0$ . Yeh step kyun? Agar funnel ki dono walls $0$ par close ho rahi hain, toh jo bhi andar trapped hai uske paas $0$ ke siwa kahi jaane ki jagah nahi. Squeeze Theorem dekho.

Verify karo: $n = 50$ : $\frac{s i n 50}{50} = \frac{- 0.2624}{50} = - 0.00525$ , jo $\pm \frac{1}{50} = \pm 0.02$ ke andar hai. ✓

Case D — pure oscillation, koi shrinking nahi

Worked example

(Cell D) Dikhao ki $a_{n} = cos (nπ)$ diverge karta hai. Forecast: $cos (nπ)$ ka value $n = 1, 2, 3, 4$ ke liye kya hoga? Calculate karo, phir guess karo.

Simplify karo. $cos (0) = 1, cos (π) = - 1, cos (2 π) = 1, \dots$ toh $a_{n} = (- 1)^{n}$ : sequence $- 1, 1, - 1, 1, \dots$ Yeh step kyun? Disguise ko pehchaanna ( $cos (nπ) = (- 1)^{n}$ ) ek "trig mystery" ko classic oscillator mein reduce kar deta hai.
$ε$ – $N$ contradiction. Maan lo yeh kisi $L$ par converge karta hai. Tolerance $ε = 1$ lo. Convergence demand karta hai ki kuch $N$ ke baad saare terms band $(L - 1, L + 1)$ mein hon, jiska width $2$ hai. Lekin consecutive terms $1$ aur $- 1$ hain — exactly $2$ ki distance par. Do points jo $2$ ki distance par hain woh dono strictly width $2$ ke band ke andar nahi aa sakte. Contradiction. Yeh step kyun? Convergence force karta hai ki tail cluster kare; $2$ ka fixed gap clustering ko forbid karta hai. Koi bhi $L$ kaam nahi kar sakta, toh yeh diverge karta hai.

Verify karo: Values ka set exactly ${- 1, 1}$ hai — do accumulation points, toh koi single limit nahi. ✓

Case E — factorial vs power race

Worked example

(Cell E) $n \to \infty lim \frac{n !}{n ^{n}}$ find karo. Forecast: $n!$ bahut bada hai. Lekin $n^{n}$ mein $n$ ke $n$ copies multiply hain, jabki $n! = 1 \cdot 2 \dots n$ mein bahut saare chhote factors hain. Kaun jeeta hai?

EK typical term dekho. $\frac{n !}{n ^{n}} = \frac{1 \cdot 2 \cdot 3 \dots n}{n \cdot n \cdot n \dots n} = \frac{1}{n} \cdot \frac{2}{n} \dots \frac{n}{n} \leq \frac{1}{n} \cdot 1 \cdot 1 \dots 1 = \frac{1}{n} .$ Yeh step kyun? Har factor $\frac{k}{n} \leq 1$ hai, aur pehla factor $\frac{1}{n}$ poore product ko neeche kheench leta hai. Toh $0 < \frac{n !}{n ^{n}} \leq \frac{1}{n}$ .
Squeeze karo. $0 \to 0$ aur $\frac{1}{n} \to 0$ , toh sequence $\to 0$ . Yeh step kyun? Wahi funnel logic jaisi Cell C mein — $0$ aur $\frac{1}{n}$ ke beech caged.
Term-ratio shrink factor se cross-check karo. $\frac{a _{n + 1}}{a _{n}} = (\frac{n}{n + 1})^{n} \to e^{- 1} < 1$ dekho. (Yeh series ratio test nahi hai — hum sirf yeh pooch rahe hain ki har term pichle se kaise compare karta hai.) Jab $\frac{a _{n + 1}}{a _{n}}$ eventually $1$ se chhote kisi fixed number se neeche rehta hai, toh terms kam se kam geometrically shrink karti hain, toh unhe $0$ ki taraf jaana hi hoga. The number e dekho.

Verify karo: $n = 10$ : $\frac{10 !}{1 0 ^{10}} = \frac{3 628 800}{1 0 ^{10}} = 0.00036288$ , aur $\frac{1}{10} = 0.1$ — safely trapped aur tiny. ✓

Case F — recursive sequence

Worked example

(Cell F) Maan lo $a_{1} = 2, a_{n + 1} = \frac{1}{2} (a_{n} + \frac{2}{a _{n}})$ . $lim a_{n}$ find karo. Forecast: Yeh Newton's iteration disguise mein hai. $a_{2}$ haath se calculate karo aur guess karo yeh kahan jaata hai.

Neeche $2$ se bounded hai. AM–GM se, $\frac{1}{2} (a_{n} + \frac{2}{a _{n}}) \geq a_{n} \cdot \frac{2}{a _{n}} = 2$ , toh $a_{n + 1} \geq 2$ sabhi $n \geq 1$ ke liye. Yeh step kyun? MCT ko ek floor chahiye. AM–GM ( $\frac{1}{2} (x + y) \geq x y$ ) use deta hai kyunki product $a_{n} \cdot \frac{2}{a _{n}} = 2$ constant hai.
Decreasing hai. $a_{n} - a_{n + 1} = a_{n} - \frac{1}{2} (a_{n} + \frac{2}{a _{n}}) = \frac{a _{n}^{2} - 2}{2 a _{n}} \geq 0$ kyunki $a_{n} \geq 2 \Rightarrow a_{n}^{2} \geq 2$ . Yeh step kyun? MCT ko monotonicity chahiye; difference ka sign use deta hai.
MCT apply karo aur fixed point solve karo. Decreasing + bounded below ⟹ kisi $L$ par converge karta hai. $a_{n}$ aur $a_{n + 1}$ dono same $L$ ki taraf jaate hain, toh $L = \frac{1}{2} (L + \frac{2}{L}) \Rightarrow 2 L^{2} = L^{2} + 2 \Rightarrow L^{2} = 2 \Rightarrow L = 2$ (positive branch, kyunki terms $> 0$ hain). Yeh step kyun? Limit ko wahi relation satisfy karni chahiye jo sequence follow karti hai — yahi fixed-point equation hai. Completeness of the Real Numbers par rests karti hai (jo guarantee karta hai ki $in f$ exist karta hai).

Verify karo: $a_{2} = \frac{1}{2} (2 + 1) = 1.5$ , $a_{3} = \frac{1}{2} (1.5 + \frac{2}{1.5}) = 1.41 \overline{6}$ , $a_{4} \approx 1.414216$ . Indeed $\to 2 \approx 1.414214$ . ✓

Case G — $e$ shape $(1 + x / n)^{n}$

Worked example

(Cell G) $n \to \infty lim (1 + \frac{3}{n})^{n}$ find karo. Forecast: Base $1 + \frac{3}{n} \to 1$ hai lekin exponent $n \to \infty$ . Yeh indeterminate $1^{\infty}$ hai — $1$ ho sakta hai, $\infty$ bhi ho sakta hai. Guess karo.

Defining $e$ -limit yaad karo. The number e se, $(1 + \frac{x}{n})^{n} \to e^{x}$ . Yeh step kyun? Poori mushkil $1^{\infty}$ mein hai; $e$ -limit ek akela tool hai jo ise cleanly resolve karta hai, kyunki $e$ ko define hi bilkul isi tarah ke limit se kiya jaata hai.
$x = 3$ match karo. Directly, $(1 + \frac{3}{n})^{n} \to e^{3}$ . Yeh step kyun? Hamara expression $x = 3$ ke saath template hai — shape pehchaanne ke baad koi algebra nahi chahiye.
(Optional log proof, rigorously kiya gaya) Log lo: $ln a_{n} = n ln (1 + \frac{3}{n})$ . $t = \frac{3}{n}$ likho, toh $t \to 0$ jab $n \to \infty$ , aur $ln a_{n} = 3 \cdot \frac{l n ( 1 + t )}{t}$ . Key fact yeh standard limit hai $t \to 0 lim \frac{ln ( 1 + t )}{t} = 1$ (yeh $1$ par $ln$ ki derivative hai, kyunki $\frac{d}{d x} ln x_{x = 1} = 1$ — Limits of functions dekho). Isliye $ln a_{n} \to 3 \cdot 1 = 3$ , aur kyunki $exp$ continuous hai, $a_{n} = e^{l n a_{n}} \to e^{3}$ . Yeh step kyun? Log lena scary exponent ko ek product $3 \cdot \frac{l n ( 1 + t )}{t}$ mein badal deta hai; hum phir ek rigorously-established limit par lean karte hain, na ki kisi unjustified series truncation par.

Verify karo: $n = 1000$ : $(1 + \frac{3}{1000})^{1000} \approx 20.06$ , aur $e^{3} \approx 20.09$ . ✓ Neeche se close aa raha hai (convergence to $e^{x}$ slow hai, toh $n = 1000$ par thoda gap expected hai).

Case H — zero / degenerate & sign cases

Worked example

(Cell H) Geometric sequence $a_{n} = r^{n}$ ke liye, har real value of $r$ ke liye convergence classify karo. Forecast: Tum jaante ho ki $r = \frac{1}{2}$ shrink karke $0$ ho jaata hai aur $r = 2$ blow up karta hai. Lekin $r = - 1$ ke baare mein kya? $r = 1$ ? $r = 0$ ? $r = - 2$ ? Padhne se pehle enumerate karo.

Number line ke har region ko work through karo (figure mein har region ko color kiya gaya hai):

$∣ r ∣ < 1$ (jaise $r = \frac{1}{2}$ ya $r = - \frac{1}{2}$ ): $∣ r^{n} ∣ = ∣ r ∣^{n} \to 0$ , toh $a_{n} \to 0$ . Negative case phir bhi $\to 0$ karta hai kyunki magnitude shrink karti hai chahe signs alternate karein. Kyun: $1$ se chhoti magnitude, baar baar khud se multiply hone par, vanish ho jaati hai.
$r = 1$ : har term $1$ hai, toh $a_{n} \to 1$ (converges — degenerate constant case). Yeh kyun matter karta hai: $r = 1$ knife-edge hai; sums ke liye formula $\frac{a}{1 - r}$ yahan zero se divide karta, toh yeh ek special value hai jise zaroori hai call out karo.
$r = 0$ : sabhi $n \geq 1$ ke liye $0^{n} = 0$ , toh $a_{n} \to 0$ (zero input — trivially convergent).
$r = - 1$ : $(- 1)^{n}$ oscillate karta hai $- 1, 1, - 1, \dots$ — diverges (Cell D).
$r > 1$ (jaise $r = 2$ ): $r^{n} \to + \infty$ — infinity tak diverge karta hai.
$r < - 1$ (jaise $r = - 2$ ): magnitude $\to \infty$ AUR sign alternate karta hai, giving $- 2, 4, - 8, 16, \dots$ — diverges (unbounded oscillation).

Summary: $r^{n}$ converges exactly tab jab $- 1 < r \leq 1$ (limit $0$ agar $∣ r ∣ < 1$ , limit $1$ agar $r = 1$ ); diverges otherwise.

Verify karo: $(0.5)^{20} = 0.00000095$ (→0), $(- 0.5)^{20} = + 0.00000095$ (→0), $1^{20} = 1$ , $(- 1)^{20} = 1$ lekin $(- 1)^{21} = - 1$ (koi limit nahi), $2^{20} = 1048576$ (→∞). ✓

Case I — real-world word problem

Worked example

(Cell I) Ek drug body se aise nikalti hai ki har ghante present amount ka $30%$ clear ho jaata hai, aur ek fresh $10$ mg dose add hoti hai. Agar $a_{n}$ $n$ -vi dose ke turant baad present mg hai (jahan $a_{1} = 10$ ), toh kya drug level stabilise hoga, aur kis value par? Forecast: Baar baar add-then-decay — kya yeh hamesha climb karta rahega, ya ek steady "therapeutic" level par settle hoga? Ek ceiling guess karo.

Model banao. Dose ke baad, $70%$ next ghante tak survive karta hai, phir $10$ mg add hota hai: $a_{n + 1} = 0.7 a_{n} + 10$ . Yeh step kyun? " $30%$ cleared" ka matlab $70% = 0.7$ bachta hai; $+ 10$ naya dose hai. Words → recurrence.
Steady state (fixed point) guess karo. Agar yeh $L$ par stabilise hota hai, toh $L = 0.7 L + 10 \Rightarrow 0.3 L = 10 \Rightarrow L = \frac{10}{0.3} = 33. \overline{3}$ mg. Yeh step kyun? Equilibrium par level change karna band kar deta hai, toh $a_{n + 1} = a_{n} = L$ — wahi fixed-point trick jo Cell F mein thi.
Convergence confirm karo. Fixed point subtract karo: $a_{n + 1} - L = 0.7 (a_{n} - L)$ , toh gap har step mein $0.7$ factor se shrink karta hai: $a_{n} - L = (0.7)^{n - 1} (a_{1} - L) \to 0$ kyunki $∣0.7∣ < 1$ (Cell H). Isliye $a_{n} \to L$ . Yeh step kyun? Ceiling se distance khud ek geometric sequence hai jiska ratio $0.7$ hai — aur humne prove kiya ki woh $0$ tak converge karte hain.

Verify karo: $a_{1} = 10, a_{2} = 0.7 (10) + 10 = 17, a_{3} = 0.7 (17) + 10 = 21.9, a_{4} = 25.33$ , $33. \overline{3}$ mg ki taraf climb kar raha hai. Units: poori tarah mg mein — consistent. ✓

Case J — exam twist (hidden divergence)

Worked example

(Cell J) Kya $a_{n} = \frac{( - 1 ) ^{n} n}{n + 1}$ converge karta hai? Forecast: $\frac{n}{n + 1} \to 1$ hai, jo achha aur settled lagta hai. Poori cheez converge karti hai bolne ka mann karta hai. Commit karne se pehle thoda aur dhyaan se dekho.

Magnitude strip karo. $∣ a_{n} ∣ = \frac{n}{n + 1} = \frac{1}{1 + \frac{1}{n}} \to 1$ , toh terms ki size $1$ ke paas settle karti hai — woh shrink nahi karti. Yeh step kyun? Limit dhundhne se pehle, check karo ki terms chhoti bhi ho rahi hain ya nahi. Yahan woh size $1$ ke aas paas hover karti hain, jo ek red flag hai: ek term-magnitude jo $1$ ke paas rehe, woh sirf tabhi converge kar sakti hai jab sign bhi settle kare.
Parity ke hisaab se subsequences mein split karo. Even aur odd indices ko alag alag dekho, kyunki factor $(- 1)^{n}$ evens par $+ 1$ aur odds par $- 1$ hota hai: $even n : a_{n} = + \frac{n}{n + 1} \to + 1, odd n : a_{n} = - \frac{n}{n + 1} \to - 1.$ Yeh step kyun? Ek convergent sequence har subsequence ko apna ek hi limit share karne par force karti hai. Agar hum do subsequences exhibit karein jinke different limits hain, toh koi single $L$ exist nahi kar sakta.
$ε$ – $N$ argument se seal karo. Maan lo $a_{n} \to L$ . $ε = \frac{1}{2}$ lo. Kuch $N$ ke baad saare terms $(L - \frac{1}{2}, L + \frac{1}{2})$ mein hain, jo width $1$ ka band hai. Lekin bade $n$ ke liye ek even term $+ 1$ ke paas hai aur next (odd) term $- 1$ ke paas — kareeban $2$ ki distance, jo width $1$ ke band mein fit nahi ho sakti. Contradiction. Yeh step kyun? Yeh wahi clustering argument hai jo Cell D mein thi, ab ek rational disguise mein — yeh "two subsequence limits" ko ek hard contradiction mein convert karta hai.
Conclude karo: do limit points $+ 1$ aur $- 1$ ka matlab hai ki $a_{n}$ oscillation se diverge karta hai.

Verify karo: $a_{100} = + \frac{100}{101} = 0.9901$ , $a_{101} = - \frac{101}{102} = - 0.9902$ — consecutive terms kareeban $2$ ki distance par, hamesha ke liye, toh koi single limit nahi. ✓

Matrix, revisited

Recall Kaun si weapon kaun si cell ke liye? (khud test karo)

Rational $\frac{p ( n )}{q ( n )}$ ::: $n$ ki highest power se divide karo; degrees compare karo. Difference $n + 1 - n$ (ek $\infty - \infty$ ) ::: roots khatam karne ke liye conjugate se multiply karo. $\frac{s i n n}{n}$ ya trapped oscillation ::: $\pm \frac{1}{n}$ ke beech Squeeze karo. Pure oscillation $(- 1)^{n}$ , $cos (nπ)$ ::: diverges — $ε$ – $N$ contradiction (fixed gap). Factorial vs power $\frac{n !}{n ^{n}}$ ::: ek term bound karo, phir squeeze karo; ya term-ratio shrink factor. Recursive $a_{n + 1} = f (a_{n})$ ::: monotone + bounded prove karo (MCT), phir $L = f (L)$ solve karo. $(1 + \frac{x}{n})^{n}$ shape ::: yeh defining limit se $e^{x}$ hai. Geometric $r^{n}$ ::: converges iff $- 1 < r \leq 1$ ; $r = \pm 1, 0$ alag alag check karo. Word problem "decay + dose" ::: $a_{n + 1} = r a_{n} + c$ model banao, fixed point $L = \frac{c}{1 - r}$ . Rational $\times (- 1)^{n}$ ::: even/odd subsequences mein split karo; different limits ⟹ diverges.

Bolzano–Weierstrass Theorem bhi dekho (har bounded sequence ki ek convergent subsequence hoti hai — woh escape hatch jab monotonicity fail ho), Cauchy Sequences (limit naam liye bina convergence), Limits of functions (in ideas ka continuous cousin), aur Series — convergence tests (jahan yeh term-limits pehli cheez hain jo tum check karte ho).