Limit comparison test: Worked examples — Hinglish deep dive

4.3.9 · D3 · Maths › Calculus III — Sequences & Series › Limit comparison test

Ye deep dive Limit comparison test ki training ground hai. Parent note ne rule bataya aur prove kiya. Yahan hum har tarah ki series dhundhte hain jinpe LCT apply ho sakta hai, taki jab exam mein koi nayi fraction mile, tumhe already pata ho ki wo table ki kaun si cell mein aati hai.

Shuru karne se pehle: yaad karo wo ek number jis par poora test tika hai, $L = lim_{n \to \infty} \frac{a _{n}}{b _{n}},$ jahan $a_{n}$ tumhari series ka term hai aur $b_{n}$ wo skeleton hai jo tum top aur bottom ke strongest power rakhke banate ho. $L$ ko ek verdict ki tarah padho:

$0 < L < \infty$ — dono ka same fate, aur $\sum b_{n}$ ka fate dono decide karta hai.
$L = 0$ — $a_{n}$ chhota hai; tab hi kaam aata hai jab $\sum b_{n}$ converge kare.
$L = \infty$ — $a_{n}$ bada hai; tab hi kaam aata hai jab $\sum b_{n}$ diverge kare.

The scenario matrix

Har LCT problem jo tum kabhi bhi dekho ge, in cells mein se ek mein aata hai. Last column us example ka naam deta hai jo us cell ko perfectly cover karta hai.

Cell	Kya alag banata hai	Comparison $b_{n}$	Verdict style	Example
A polynomial / polynomial	leading powers, clean $L$	p-Series	main $0 < L < \infty$	Ex 1
B denominator mein root	fractional dominant power	$p$ -series with $p = \frac{1}{2}$	main, diverges	Ex 2
C exponential dominates	$2^{n}$ polynomial ko dabaata hai	Geometric Series	main, converges	Ex 3
D slow log factor	$ln n$ vs power of $n$	$p$ -series, $L = 0$	edge $L = 0$	Ex 4
E wrong-way direct comparison	$L = \infty$ phir bhi decide karta hai	Harmonic Series	edge $L = \infty$	Ex 5
F degenerate / trap	$a_{n} \neq \to 0$ , ya bad $b_{n}$	—	test misused	Ex 6
G real-world word problem	ek rate se total model karo	$p$ -series	main	Ex 7
H exam twist	do unknown parameters	$p$ -series in $p$	boundary tuning	Ex 8

Cell A — polynomial over polynomial

Worked example Example 1 (Cell A) —

$n = 1 \sum \infty \frac{5 n ^{3} - 2 n + 7}{2 n ^{5} + n ^{2} + 1}$ Forecast: weak terms ko thumb se dhako. Top $5 n^{3}$ jaisa lagta hai, bottom $2 n^{5}$ jaisa. Ratio $\sim \frac{5 n ^{3}}{2 n ^{5}} = \frac{5}{2 n ^{2}}$ — ye $1/ n^{2}$ jaisa smell karta hai, ek convergent $p$ -series. Guess: converges.

$b_{n}$ choose karo. Top aur bottom se sirf strongest power rakho: $b_{n} = \frac{n ^{3}}{n ^{5}} = \frac{1}{n ^{2}} .$ Ye step kyun? Bade $n$ ke liye $- 2 n + 7$ aur $+ n^{2} + 1$ leading powers ke saamne dust hain; unhi powers ka ratio decay ko govern karta hai.
$L$ compute karo. $L = lim_{n \to \infty} \frac{( 5 n ^{3} - 2 n + 7 ) / ( 2 n ^{5} + n ^{2} + 1 )}{1/ n ^{2}} = lim_{n \to \infty} \frac{n ^{2} ( 5 n ^{3} - 2 n + 7 )}{2 n ^{5} + n ^{2} + 1} .$ Top aur bottom ko $n^{5}$ se divide karo: $L = lim_{n \to \infty} \frac{5 - 2/ n ^{2} + 7/ n ^{5}}{2 + 1/ n ^{3} + 1/ n ^{5}} = \frac{5}{2} .$ Ye step kyun? Highest power $n^{5}$ se divide karne par har "weak" term $0$ ho jaata hai, sirf leading coefficients bachte hain.
Conclude karo. $0 < L = \frac{5}{2} < \infty$ , aur $\sum 1/ n^{2}$ converge karta hai ( $p = 2 > 1$ ). Series converge karti hai. ✔

Verify: $L$ finite aur positive hai, isliye $\sum a_{n}$ aur $\sum b_{n}$ ek hi fate share karte hain; $b_{n}$ ka fate convergence hai. Sanity check: $\sum \frac{5 n ^{3} - 2 n + 7}{2 n ^{5} + n ^{2} + 1}$ ke partial sums ek fixed number ke paas settle hote hain, jo convergence ke consistent hai.

Cell B — ek root ek fractional power chhupata hai

Worked example Example 2 (Cell B) —

$n = 1 \sum \infty \frac{n + 1}{n + 4}$ Forecast: top $\sim n = n^{1/2}$ , bottom $\sim n$ . Ratio $\sim n^{1/2} / n = 1/ n^{1/2}$ — ek $p$ -series with $p = \frac{1}{2} \leq 1$ . Guess: diverges.

$b_{n} = \frac{1}{n ^{1/2}}$ pick karo. Kyun? Top par dominant power $n^{1/2}$ hai, bottom par $n^{1}$ ; unka ratio $n^{1/2 - 1} = n^{- 1/2}$ hai.
$L$ compute karo. $L = lim_{n \to \infty} \frac{( n + 1 ) / ( n + 4 )}{1/ n} = lim_{n \to \infty} \frac{n ( n + 1 )}{n + 4} = lim_{n \to \infty} \frac{n + n}{n + 4} .$ Top aur bottom ko $n$ se divide karo: $L = lim_{n \to \infty} \frac{1 + 1/ n}{1 + 4/ n} = 1.$ Ye step kyun? $n$ se multiply karne par awkward root ek clean polynomial ratio mein badal jaata hai.
Conclude karo. $0 < L = 1 < \infty$ , aur $\sum 1/ n^{1/2}$ diverge karta hai ( $p = \frac{1}{2} \leq 1$ ). Series diverge karti hai. ✔

Verify: $L = 1$ matlab terms bade $n$ ke liye $1/ n$ ke identically jaisi behave karti hain; yahan Direct Comparison Test se divergence prove karna painful hoga (inequalities galat direction mein point karti hain), aur exactly yahi reason hai ki LCT sahi tool hai.

Cell C — exponential sab ko haraata hai (geometric skeleton)

Worked example Example 3 (Cell C) —

$n = 1 \sum \infty \frac{3 ^{n} + n ^{2}}{4 ^{n} - n}$ Forecast: exponentials polynomials ko crush karte hain. Top $\sim 3^{n}$ , bottom $\sim 4^{n}$ . Ratio $\sim (3/4)^{n}$ — ek Geometric Series with $∣ r ∣ = 3/4 < 1$ . Guess: converges.

Figure dekho: blue $3^{n}$ curve aur yellow $4^{n}$ curve dono rocket ki tarah upar jaate hain, lekin red ratio $3^{n} / 4^{n}$ geometrically collapse ho jaata hai zero ki taraf — ye collapse hi convergence jaisi dikhti hai.

$b_{n} = (\frac{3}{4})^{n} = \frac{3 ^{n}}{4 ^{n}}$ pick karo. Kyun? Top aur bottom ka dominant exponential rakho; $n^{2}$ aur $- n$ , $3^{n}$ aur $4^{n}$ ke saamne negligible hain (base $> 1$ wala koi bhi exponential eventually kisi bhi polynomial se bada ho jaata hai).
$L$ compute karo. $L = lim_{n \to \infty} \frac{( 3 ^{n} + n ^{2} ) / ( 4 ^{n} - n )}{3 ^{n} / 4 ^{n}} = lim_{n \to \infty} \frac{4 ^{n} ( 3 ^{n} + n ^{2} )}{3 ^{n} ( 4 ^{n} - n )} .$ Top aur bottom ko $3^{n} 4^{n}$ se divide karke pieces mein split karo: $L = lim_{n \to \infty} \frac{1 + n ^{2} / 3 ^{n}}{1 - n / 4 ^{n}} = \frac{1 + 0}{1 - 0} = 1.$ Ye step kyun? $n^{2} / 3^{n} \to 0$ aur $n / 4^{n} \to 0$ kyunki exponential growth polynomial growth ko outpace karti hai — ye wahi fact hai jo figure dikhata hai.
Conclude karo. $0 < L = 1 < \infty$ , aur $\sum (3/4)^{n}$ converge karta hai (geometric, $∣ r ∣ < 1$ ). Series converge karti hai. ✔

Verify: $\sum_{n = 1}^{\infty} (3/4)^{n} = \frac{3/4}{1 - 3/4} = 3$ , ek finite total — reference series genuinely converge karti hai, isliye LCT ka verdict trustworthy hai.

Cell D — ek slow logarithm, $L = 0$ rule use karke

Worked example Example 4 (Cell D) —

$n = 2 \sum \infty \frac{ln n}{n ^{3}}$ Forecast: $ln n$ itna slowly grow karta hai ki $\frac{l n n}{n ^{3}}$ almost $1/ n^{3}$ ki tarah decay karta hai. Guess: converges — lekin hum ek clean verdict chahte hain, isliye deliberately aisa comparison choose karte hain jo $L$ ko $0$ par bhej de.

$b_{n} = \frac{1}{n ^{2}}$ pick karo. Kyun? Hum ek thoda bada lekin phir bhi convergent $p$ -series se compare karte hain ( $p = 2 > 1$ ). $\frac{1}{n ^{2}}$ mein $\frac{1}{n ^{3}}$ ke mukable extra $n$ factor denominator mein ek $n$ chhodta hai jo slow $ln n$ ko swallow kar leta hai, aur $L = 0$ force ho jaata hai.
$L$ compute karo. $L = lim_{n \to \infty} \frac{( l n n ) / n ^{3}}{1/ n ^{2}} = lim_{n \to \infty} \frac{l n n}{n} .$ Ye classic hai: $n$ ki positive power se divided logarithm $0$ ki tarah tend karta hai ( $n$ ki power hamesha jeetti hai). To $L = 0$ . Ye step kyun? Kyunki $ln n$ , $n^{ε}$ se slower grow karta hai kisi bhi $ε > 0$ ke liye; yahan $ε = 1$ hai isliye ratio khatam ho jaata hai.
Conclude karo. $L = 0$ aur $\sum 1/ n^{2}$ converge karta hai. LCT ke $L = 0$ rule se, series converge karti hai. ✔

Verify: $L = 0$ case sirf tab convergence deta hai jab $\sum b_{n}$ converge kare — aur $p = 2 > 1$ ye guarantee karta hai. Agar humne carelessly $b_{n} = 1/ n$ (jo diverge karta hai) $L = 0$ ke saath use kiya hota, to test kuch nahi kehta (Cell F dekho).

Cell E — $L = \infty$ jo phir bhi decide karta hai

Worked example Example 5 (Cell E) —

$n = 1 \sum \infty \frac{n + 1}{n ^{3/2}}$ Forecast: $\frac{n + 1}{n ^{3/2}} \approx \frac{n}{n ^{3/2}} = \frac{1}{n ^{1/2}}$ , aur $p = \frac{1}{2} \leq 1$ diverge karta hai. Guess: diverges. Hum dikhayenge ki $L = \infty$ edge case bhi kaam karta hai agar hum harmonic series se compare karein.

$b_{n} = \frac{1}{n}$ pick karo (Harmonic Series, jo diverge karti hai). Kyun? Hum deliberately aisa $b_{n}$ choose karte hain jo $a_{n}$ se faster decay karta hai, isliye ratio blow up ho jaata hai — aur kyunki $\sum b_{n}$ diverge karta hai, $L = \infty$ rule phir bhi apply hota hai.
$L$ compute karo. $L = lim_{n \to \infty} \frac{( n + 1 ) / n ^{3/2}}{1/ n} = lim_{n \to \infty} \frac{n ( n + 1 )}{n ^{3/2}} = lim_{n \to \infty} \frac{n ^{2} + n}{n ^{3/2}} .$ Top aur bottom ko $n^{3/2}$ se divide karo: $L = lim_{n \to \infty} (n^{1/2} + n^{- 1/2}) = \infty.$ Ye step kyun? Leading term $n^{2} / n^{3/2} = n^{1/2}$ bina bound ke grow karta hai — to $a_{n}$ eventually harmonic term se bahut bada hai.
Conclude karo. $L = \infty$ aur $\sum 1/ n$ diverge karta hai. $L = \infty$ rule se, series diverge karti hai. ✔

Verify: kyunki $a_{n}$ eventually divergent $1/ n$ se bada hai, to ye bhi zaroor diverge karega. Ek friendly $b_{n} = 1/ n^{1/2}$ se cross-check karo: to $L = lim \frac{n + 1}{n} = 1$ , main case, aur $\sum 1/ n^{1/2}$ diverge karta hai — same verdict, do raaste.

Cell F — degenerate traps

Worked example Example 6 (Cell F) — LCT todne ke do tarike

Trap 6a — terms zero tak jaate hi nahi. Consider karo $n = 1 \sum \infty \frac{n + 1}{2 n + 3}$ .

Forecast: LCT touch karne se pehle, term check karo. Kya $a_{n} \to 0$ ?

Term limit compute karo. $lim_{n \to \infty} \frac{n + 1}{2 n + 3} = lim_{n \to \infty} \frac{1 + 1/ n}{2 + 3/ n} = \frac{1}{2} \neq = 0.$
Ye step kyun? Ek convergent series mein zaroor $a_{n} \to 0$ hona chahiye (divergence test / term test). Yahan $a_{n} \to \frac{1}{2}$ , isliye series turant diverge karti hai — koi comparison needed nahi. LCT ink waste hoga.

Verify: $\frac{1}{2} \neq = 0$ , to convergence ke liye necessary condition fail hoti hai. Diverges. ✔

Trap 6b — mismatched $L$ . Lo convergent $\sum \frac{1}{n ^{2}}$ , lekin bewaqoofi se $b_{n} = \frac{1}{n}$ (harmonic, divergent) se compare karo.

$L$ compute karo. $L = lim_{n \to \infty} \frac{1/ n ^{2}}{1/ n} = lim_{n \to \infty} \frac{1}{n} = 0.$
Ye step kyun? $L = 0$ tab hi convergence conclude karta hai jab $\sum b_{n}$ converge kare. Yahan $\sum 1/ n$ diverge karta hai, isliye $L = 0$ rule koi information nahi deta — test is $b_{n}$ ke saath inconclusive hai.
Fix: $b_{n} = 1/ n^{2}$ khud lo (ya koi bhi convergent $p$ -series with $p > 1$ ); tab $L = 1$ , main case, converges cleanly.

Verify: $\sum 1/ n^{2} = π^{2} /6$ converge karta hai — sahi comparison ye confirm karta hai; galat wala bas stall karta hai. Lesson: $0 < L < \infty$ aim karo, aur $L = 0$ ko hamesha convergent $b_{n}$ ke saath match karo, $L = \infty$ ko divergent $b_{n}$ ke saath.

Cell G — ek real-world word problem

Worked example Example 7 (Cell G) — total signal energy

Ek sensor, ping $n$ par, $a_{n} = \frac{4 n + 1}{n ^{3} + 2 n}$ joules energy receive karta hai. Question: kya infinitely many pings par total energy $\sum_{n = 1}^{\infty} a_{n}$ finite rehti hai (converge) ya blow up hoti hai (diverge)?

Forecast: top $\sim 4 n$ , bottom $\sim n^{3}$ , ratio $\sim 4/ n^{2}$ — $1/ n^{2}$ ki tarah, convergent. Guess: total energy finite hai.

$b_{n} = \frac{1}{n ^{2}}$ pick karo. Kyun? Top ka strongest power $n$ hai, bottom ka $n^{3}$ ; ratio $n / n^{3} = 1/ n^{2}$ .
$L$ compute karo. $L = lim_{n \to \infty} \frac{( 4 n + 1 ) / ( n ^{3} + 2 n )}{1/ n ^{2}} = lim_{n \to \infty} \frac{n ^{2} ( 4 n + 1 )}{n ^{3} + 2 n} = lim_{n \to \infty} \frac{4 n ^{3} + n ^{2}}{n ^{3} + 2 n} .$ Top aur bottom ko $n^{3}$ se divide karo: $L = lim_{n \to \infty} \frac{4 + 1/ n}{1 + 2/ n ^{2}} = 4.$
Conclude karo. $0 < L = 4 < \infty$ , aur $\sum 1/ n^{2}$ converge karta hai. Total energy converge karti hai — ye finite hai. ✔

Verify: units check — har $a_{n}$ joules mein hai, ek convergent joule series ka sum ek finite joule total hai; physically received energy per ping $1/ n^{2}$ ki tarah decay karta hai, jo bounded amount mein sum karne ke liye kaafi fast hai. $L = 4$ finite aur positive hai, isliye verdict valid hai.

Cell H — exam twist (tune karne ka parameter)

Worked example Example 8 (Cell H) — kis

$p$ ke liye converge karta hai? $n = 1 \sum \infty \frac{n ^{2} + 1}{n ^{p} + n}$ Forecast: top $\sim n^{2}$ , bottom $\sim n^{p}$ (assuming $p > 1$ taki $n^{p}$ , $n$ ko dominate kare). Ratio $\sim n^{2} / n^{p} = 1/ n^{p - 2}$ — ek $p$ -series with exponent $p - 2$ . Ye converge karta hai jab $p - 2 > 1$ , yaani $p > 3$ . Guess: converges iff $p > 3$ .

$b_{n} = \frac{1}{n ^{p - 2}}$ pick karo (valid for $p > 1$ taki $n^{p}$ dominant denominator term ho). Kyun? Leading powers rakho: $n^{2} / n^{p} = n^{2 - p} = 1/ n^{p - 2}$ .
$L$ compute karo. $L = lim_{n \to \infty} \frac{( n ^{2} + 1 ) / ( n ^{p} + n )}{1/ n ^{p - 2}} = lim_{n \to \infty} \frac{n ^{p - 2} ( n ^{2} + 1 )}{n ^{p} + n} = lim_{n \to \infty} \frac{n ^{p} + n ^{p - 2}}{n ^{p} + n} .$ Top aur bottom ko $n^{p}$ se divide karo: $L = lim_{n \to \infty} \frac{1 + n ^{- 2}}{1 + n ^{1 - p}} = 1 (for p > 1) .$
Conclude karo. $L = 1$ hamesha (main case) $p > 1$ ke liye, isliye $\sum a_{n}$ , $\sum 1/ n^{p - 2}$ ka fate share karta hai. Wo $p$ -series converge karta hai $⟺ p - 2 > 1 ⟺ p > 3$ . Boundary check: $p = 3$ par, $b_{n} = 1/ n$ (harmonic) diverge karta hai, isliye series $p = 3$ par diverge karti hai; $p < 1$ ke liye denominator $n$ se dominated hota hai aur term $\sim n^{2} / n = n$ zero tak jaata hi nahi (diverges). To converges exactly when $p > 3$ .

Verify: $p = 4$ plug karo: $b_{n} = 1/ n^{2}$ , converges — match. $p = 3$ plug karo: $b_{n} = 1/ n$ , diverges — boundary match. $p = 5$ plug karo: $b_{n} = 1/ n^{3}$ , converges — match. Sab "converges $⟺ p > 3$ " ke consistent hain. ✔

Recall Har series kis cell mein hai?

$\sum \frac{7 n ^{4}}{n ^{7} + 3}$ (poly/poly, converges) ::: Cell A — $b_{n} = 1/ n^{3}$ , $L = 7$ . $\sum \frac{1}{n ^{3} + n}$ ::: Cell B — $b_{n} = 1/ n^{3/2}$ , $p = 3/2 > 1$ , converges. $\sum \frac{2 ^{n} + 1}{5 ^{n}}$ ::: Cell C — geometric, $r = 2/5$ , converges. $\sum \frac{l n n}{n ^{4}}$ ::: Cell D — $L = 0$ vs convergent $1/ n^{3}$ , converges. $\sum \frac{n ^{2}}{n ^{3/2}}$ ::: Cell F — terms $\to \infty$ , turant diverges.

Dekho bhi: Ratio Test aur Integral Test un series ke liye jahan koi clean $p$ -series ya geometric skeleton exist nahi karta, aur Squeeze Theorem jo LCT ke peeche $ε = L /2$ sandwich ko power deta hai.

Recall

Recall Self-test

In Cell C, hum $n^{2}$ ko $3^{n}$ ke saamne kyun ignore kar sakte hain? ::: Exponential growth (base $> 1$ ) eventually kisi bhi polynomial ko dominate karta hai, isliye $n^{2} / 3^{n} \to 0$ . In Cell E, $L = \infty$ phir bhi verdict kyun deta hai? ::: Kyunki $\sum b_{n}$ (harmonic) diverge karta hai, aur $L = \infty$ rule tab $\sum a_{n}$ ko bhi diverge karne par force karta hai. In Cell H, boundary $p = 3$ par kya series converge karti hai? ::: Nahi — $b_{n} = 1/ n$ harmonic hai, diverge karta hai. Cell F trap 6a mein kya tootta hai? ::: Terms $1/2 \neq = 0$ ki taraf tend karte hain, isliye divergence test LCT se pehle hi use khatam kar deta hai.