Direct comparison test: Worked examples — Hinglish deep dive

4.3.8 · D3 · Maths › Calculus III — Sequences & Series › Direct comparison test

Direct Comparison Test (DCT) ka poora game yeh hai: apni unknown series ko, term by term, ek aisi series se compare karo jiska fate tumhe pehle se pata ho. Do benchmarks se lagbhag saara kaam ho jaata hai:

Recall Do benchmark families (parent note se)

$p$ -series ::: $\sum \frac{1}{n ^{p}}$ converge karta hai jab $p > 1$ , diverge karta hai jab $p \leq 1$ . Dekho p-series test. Geometric series ::: $\sum r^{n}$ converge karta hai jab $∣ r ∣ < 1$ , diverge karta hai jab $∣ r ∣ \geq 1$ . Dekho Geometric series. Harmonic series ::: $\sum \frac{1}{n}$ — borderline case ( $p = 1$ ), diverge karta hai. Dekho Harmonic series.

Scenario matrix

Har DCT problem inhi cells mein se ek mein aati hai. Hum har ek par kam se kam ek worked example lagaenge.

Cell	Situation	Kya mushkil banata hai	Example
A	Denominator $p$ -series se thoda bada	direction: upar se trap karo	Ex 1
B	Denominator harmonic se thoda chhota	direction: neeche se push karo	Ex 2
C	Bounded wiggling numerator ( $sin$ , $cos$ )	numerator ko uske max/min se replace karo	Ex 3
D	Geometric-type terms ( $r^{n}$ )	benchmark geometric hai, $p$ -series nahi	Ex 4
E	Denominator mein extra positive term use bana deta hai chota	naive direction kaam ki nahi hoti	Ex 5
F	Sign-changing terms	DCT ko $a_{n} \geq 0$ chahiye — pehle repair karo	Ex 6
G	Degenerate / limiting input (term $\to$ constant)	test apply nahi ho sakta — pehchano	Ex 7
H	Word problem (real-world total)	pehle translate karo, phir compare karo	Ex 8
I	Exam twist: comparison galat direction mein chuni	mistake ko steel-man karo, phir fix karo	Ex 9

Ek rule jo sab kuch decide karta hai:

Upar wali figure poore page ke liye tumhara compass hai. Left side mein, ek convergent series $\sum b_{n}$ ka finite total hai; us finite total ko $B$ kaho — green dashed roof. Jo series iske neeche trapped hai (mint bars) woh kabhi $B$ se aage sum nahi ho sakti → converges. Right side mein, ek divergent series ek infinite floor (butter bars) banati hai jo hamesha chadhta rehta hai; jo series iske upar push hoti hai (coral bars) woh bhi upar khichti jaati hai → diverges. Neeche har example inhi do pictures mein se ek hai.

Cell A — denominator $p$ -series se thoda bada

Worked example Example 1 —

$n = 1 \sum \infty \frac{1}{n ^{3} + 2 n}$ (Cell A) Yahan general term hai $a_{n} = \frac{1}{n ^{3} + 2 n}$ . Forecast: neeche dominant power $n^{3}$ hai, toh yeh $\sum 1/ n^{3}$ ( $p$ -series with $p = 3 > 1$ ) ki tarah behave karna chahiye. Guess: converges. Humein kis direction ki zaroorat hai? Prove karne ke liye convergence, humein ek bada convergent partner chahiye.

Step 1. Note karo ki $n^{3} + 2 n > n^{3}$ for all $n \geq 1$ . Yeh step kyun? Positive quantity $2 n$ add karne se denominator bada ho jaata hai, jisse poora fraction chhota ho jaata hai.

Step 2. Positive numbers ke reciprocals lete waqt inequality flip ho jaati hai: $0 \leq a_{n} = \frac{1}{n ^{3} + 2 n} < \frac{1}{n ^{3}} .$ Yeh step kyun? Positive $x < y$ ke liye $\frac{1}{x} > \frac{1}{y}$ hota hai; reciprocation order reverse kar deta hai. Ab humari unknown series ek badi series ke neeche trapped hai.

Step 3. Benchmark: $\sum \frac{1}{n ^{3}}$ ek $p$ -series hai $p = 3 > 1$ ke saath → converges (p-series test). Yeh step kyun? Humein ek partner chahiye jiska fate humein pehle se pata ho; $p$ -series test woh verdict fauran de deta hai.

Step 4. Conclusion: $0 \leq a_{n} \leq \frac{1}{n ^{3}}$ aur roof converges → humari series converges. ✅

Verify: partial sum $\sum_{n = 1}^{10} \frac{1}{n ^{3} + 2 n} \approx 0.4789$ aur yeh $\sum_{n = 1}^{10} \frac{1}{n ^{3}} \approx 1.1975$ se neeche rehna chahiye. Rehta hai — bars roof ke neeche hain, trap confirm ho gaya.

Cell B — denominator harmonic se thoda chhota

Worked example Example 2 —

$n = 1 \sum \infty \frac{1}{n + 3}$ (Cell B) Yahan $a_{n} = \frac{1}{n + 3}$ . Forecast: neeche $n = n^{1/2}$ ki tarah badhta hai, toh terms $\sum 1/ n^{1/2}$ ki tarah behave karte hain, ek $p$ -series with $p = \frac{1}{2} \leq 1$ → diverges. Prove karne ke liye divergence humein ek chhota divergent partner chahiye.

Step 1. Hum chahte hain ki humara fraction se ek chhota fraction ho, toh denominator ko saaf, controlled tarike se badhaate hain. Har $n \geq 1$ ke liye $n \geq 1$ hai, isliye $3 \leq 3 n$ , aur therefore $n + 3 \leq n + 3 n = 4 n .$ Yeh step kyun? Awkward $n + 3$ ko bade, saaf $4 n$ se replace karne se fraction chhota ho jaata hai lekin ek pehchaana $p$ -series partner milta hai. Note karo yeh bound all $n \geq 1$ ke liye hold karta hai, toh koi "eventually" caveat nahi chahiye.

Step 2. Reciprocate karo (bada denominator → chhota fraction): $a_{n} = \frac{1}{n + 3} \geq \frac{1}{4 n} = \frac{1}{4} \cdot \frac{1}{n ^{1/2}} \geq 0.$ Yeh step kyun? Yeh inequality ko divergence ke liye sahi direction mein point karta hai — humari series partner ke upar baitti hai.

Step 3. Benchmark: $\sum \frac{1}{4} \cdot \frac{1}{n ^{1/2}} = \frac{1}{4} \sum n^{- 1/2}$ ek constant times $p$ -series with $p = \frac{1}{2} \leq 1$ hai → diverges. Yeh step kyun? Nonzero constant kabhi convergence status nahi badalta, toh yeh floor genuinely divergent hai.

Step 4. Humari series ek divergent floor ke upar baitti hai → diverges. ✅

Verify: $n = 100$ par, $\frac{1}{100 + 3} = \frac{1}{13} \approx 0.0769$ aur $\frac{1}{4} \cdot \frac{1}{100} = \frac{1}{40} = 0.025$ . Indeed $0.0769 \geq 0.025$ — floor confirm ho gaya.

Cell C — bounded wiggling numerator

Worked example Example 3 —

$n = 1 \sum \infty \frac{3 + cos ( 2 n )}{n ^{2}}$ (Cell C) Yahan $a_{n} = \frac{3 + cos ( 2 n )}{n ^{2}}$ . Forecast: numerator wiggle karta hai lekin bounded rehta hai; $n^{2}$ dominate karta hai → $\sum 1/ n^{2}$ ( $p = 2 > 1$ ) ki tarah behave karna chahiye → converges. Convergence prove karo → bada convergent partner chahiye.

Step 1. Wiggle bound karo: kyunki $- 1 \leq cos (2 n) \leq 1$ hai, $2 \leq 3 + cos (2 n) \leq 4$ hoga. Yeh step kyun? Hum $cos$ ko directly sum nahi kar sakte, toh numerator ko uske maximum $4$ se replace karte hain taaki mess khatam ho lekin ek clean upper bound mile.

Step 2. $n^{2} > 0$ se divide karo (inequality preserve hoti hai): $0 \leq a_{n} = \frac{3 + c o s ( 2 n )}{n ^{2}} \leq \frac{4}{n ^{2}} .$ Yeh step kyun? Lower bound $2 > 0$ non-negativity guarantee karta hai (DCT ke liye zaroori hai), aur upper bound humari roof deta hai.

Step 3. Benchmark: $\sum \frac{4}{n ^{2}} = 4 \sum \frac{1}{n ^{2}}$ converges ( $p = 2 > 1$ ). Yeh step kyun? Constant $4$ convergence affect nahi karta, toh yeh ek certified finite roof hai.

Step 4. Finite roof ke neeche trapped → converges. ✅

Verify: $\sum_{n = 1}^{20} \frac{3 + c o s ( 2 n )}{n ^{2}} \approx 3.9036$ , aur yeh $\sum_{n = 1}^{20} \frac{4}{n ^{2}} \approx 6.1854$ ke neeche hona chahiye. Hai.

Cell D — geometric-type terms

Worked example Example 4 —

$n = 1 \sum \infty \frac{2 ^{n}}{3 ^{n} + 1}$ (Cell D) Yahan $a_{n} = \frac{2 ^{n}}{3 ^{n} + 1}$ . Forecast: yeh $p$ -series nahi hai — variable exponent mein hai. Ratio $2^{n} / 3^{n} = (2/3)^{n}$ with $∣ r ∣ = 2/3 < 1$ ek geometric benchmark suggest karta hai → converges. Convergence prove karo → bada convergent partner.

Step 1. $+ 1$ drop karo denominator chhota karne ke liye, jo fraction ko bada banata hai: $a_{n} = \frac{2 ^{n}}{3 ^{n} + 1} < \frac{2 ^{n}}{3 ^{n}} = (\frac{2}{3})^{n} .$ Yeh step kyun? Chhota denominator bada fraction deta hai — exactly woh direction jo upper trap ke liye chahiye. Aur $3^{n}$ akela ek clean geometric term hai.

Step 2. Non-negativity confirm karo: $2^{n} > 0$ aur $3^{n} + 1 > 0$ , toh har term $\geq 0$ hai. Accha.

Step 3. Benchmark: $\sum (2/3)^{n}$ geometric series hai $r = \frac{2}{3}$ , $∣ r ∣ < 1$ ke saath → converges. Yeh step kyun? Geometric-series test us terms ke liye fauran, pakka verdict deta hai jisme index exponent mein ho — yahan sahi benchmark family yahi hai, $p$ -series family nahi.

Step 4. Convergent geometric roof ke neeche trapped → converges. ✅

Verify: tail $(2/3)^{n}$ ki tarah decay karta hai. Numerically $\sum_{n = 1}^{\infty} (2/3)^{n} = \frac{2/3}{1 - 2/3} = 2$ , aur $\sum_{n = 1}^{15} \frac{2 ^{n}}{3 ^{n} + 1} \approx 1.2494 < 2$ — roof ke neeche.

Cell E — extra positive term use bana deta hai chhota (arrow ke liye trap)

Worked example Example 5 —

$n = 1 \sum \infty \frac{n}{n ^{2} + 1}$ (Cell E) Yahan $a_{n} = \frac{n}{n ^{2} + 1}$ . Forecast: neeche $n^{2}$ , upar $n$ , toh terms $\frac{n}{n ^{2}} = \frac{1}{n}$ ki tarah behave karte hain → harmonic-like → diverges. Divergence prove karo → chhota divergent partner. Dhyan do: " $+ 1$ " fraction ko $\frac{1}{n}$ se chhota bana deta hai, toh obvious comparison kaam ki nahi wali direction mein point karta hai.

Step 1. Pehle trap notice karo. $n^{2} + 1 > n^{2} \Rightarrow \frac{n}{n ^{2} + 1} < \frac{n}{n ^{2}} = \frac{1}{n}$ . Yahi hai $a_{n} < \frac{1}{n}$ — ek chhoti divergent partner humein kuch nahi batati. Yeh step kyun? Kaam ki nahi wali direction ko pehle pehchanna ek galat conclusion se bachata hai (parent note mein yahi mistake batai gayi hai).

Step 2. Repair: denominator ko kuch aise bound karo jo $n^{2} + 1$ se bada ho lekin phir bhi $\sim n^{2}$ ho. $n \geq 1$ ke liye, $n^{2} + 1 \leq n^{2} + n^{2} = 2 n^{2}$ . Yeh step kyun? Denominator ko $2 n^{2}$ tak badhaane se fraction controlled tarike se chhota hota hai jo phir bhi divergent partner deta hai.

Step 3. Tab $a_{n} = \frac{n}{n ^{2} + 1} \geq \frac{n}{2 n ^{2}} = \frac{1}{2} \cdot \frac{1}{n} \geq 0.$ Yeh step kyun? Ab humari series $\frac{1}{2} \sum \frac{1}{n}$ ke upar baitti hai, jo ek divergent floor hai.

Step 4. Benchmark: $\frac{1}{2} \sum \frac{1}{n}$ harmonic series ka aadha hai → diverges. Yeh step kyun? Divergent series ki har term aadhi kar dene se woh divergent rehti hai, toh floor genuinely infinite hai.

Step 5. Infinite floor ke upar push kiya → diverges. ✅

Verify: $n = 50$ par, $\frac{50}{2501} \approx 0.01999$ aur $\frac{1}{2} \cdot \frac{1}{50} = 0.01$ . Indeed $0.01999 \geq 0.01$ — floor holds.

Cell F — sign-changing terms (DCT directly nahi chal sakta)

Worked example Example 6 —

$n = 1 \sum \infty \frac{( - 1 ) ^{n}}{n ^{2}}$ (Cell F) Yahan $a_{n} = \frac{( - 1 ) ^{n}}{n ^{2}}$ . Forecast: terms ka sign alternate hota hai, toh DCT — jo demand karta hai $a_{n} \geq 0$ — directly apply nahi ho sakta. Lekin hum absolute values compare kar sakte hain.

Step 1. Obstacle pehchano: $\frac{( - 1 ) ^{n}}{n ^{2}}$ kabhi kabhi negative hota hai, toh partial sums monotone increasing nahi hote aur Monotone Convergence fail karta hai. Raw DCT illegal hai. Yeh step kyun? Kyun test tootta hai yeh batana — yahi aadhaa skill hai — exactly yahi failure parent note warn karta hai.

Step 2. Iske bajaye absolute value compare karo: $∣ a_{n} ∣ = \frac{( - 1 ) ^{n}}{n ^{2}} = \frac{1}{n ^{2}} \leq \frac{1}{n ^{2}} .$ Yeh step kyun? $∣ a_{n} ∣ \geq 0$ hai, toh DCT absolute values par legal hai. Yahan $∣ a_{n} ∣ = \frac{1}{n ^{2}}$ ek convergent $p$ -series hi hai ( $p = 2 > 1$ ).

Step 3. Kyunki $\sum ∣ a_{n} ∣$ converges hai, original series absolutely converges hai → isliye converges hai (Absolute convergence).

Step 4. Conclusion: converges (absolutely). ✅

Verify: $\sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n}}{n ^{2}} = - \frac{π ^{2}}{12} \approx - 0.8225$ , ek finite value — convergence confirm. Aur $\sum ∣ a_{n} ∣ = \frac{π ^{2}}{6} \approx 1.6449$ finite hai, absolute convergence confirm karta hai.

Cell G — degenerate / limiting input (test apply nahi ho sakta)

Worked example Example 7 —

$n = 1 \sum \infty \frac{n}{n + 1}$ (Cell G) Yahan $a_{n} = \frac{n}{n + 1}$ . Forecast: $n \to \infty$ par terms dekhte hain: $\frac{n}{n + 1} \to 1$ , nahi $0$ . Jo series ke terms $0$ tak nahi shrink hote woh converge nahi kar sakti — koi comparison nahi chahiye.

Step 1. Limiting value calculate karo: $n \to \infty lim \frac{n}{n + 1} = n \to \infty lim \frac{1}{1 + 1/ n} = 1$ . Yeh step kyun? $n$ -th term test (har comparison ka prerequisite): agar $a_{n} \neq \to 0$ hai, toh series fauran diverge karti hai. DCT yahan zaroori nahi hai aur wasteful hoga.

Step 2. Kyunki terms $1 \neq = 0$ ke paas jaate hain, running total unbounded badhta hai. Yeh step kyun? Tum hamesha ke liye aise numbers add kar rahe ho jo $1$ ke paas hain — sum $\approx n$ ki tarah chadhta hai.

Step 3. Conclusion: diverges by term test. ✅ (DCT yahan galat tool hai — yeh pehchaanna hi seekhna hai.)

Verify: $\sum_{n = 1}^{100} \frac{n}{n + 1} \approx 94.808$ , already $100$ ke paas aur linearly badh raha hai — clearly unbounded.

Cell H — ek real-world word problem

Worked example Example 8 — Ink on an infinite ruler

(Cell H) Ek printer ek ruler par mark lagaata hai: position $n$ (cm) par woh ek line print karta hai jiska ink area $\frac{1}{n ^{2} + n}$ mm² hai. Kya infinitely many marks pe istemaal ki gayi total ink finite rehti hai?

Yahan $a_{n} = \frac{1}{n ^{2} + n}$ . Forecast: areas $1/ n^{2}$ ki tarah shrink karte hain → total finite hona chahiye (converges). Convergence prove karo → bada convergent partner.

Step 1. Translate karo: total ink $= n = 1 \sum \infty \frac{1}{n ^{2} + n}$ , sab terms $> 0$ (areas). Accha — DCT legal. Yeh step kyun? Physical quantities (area, mm²) non-negative hote hain, toh sign condition automatic hai.

Step 2. Roof ke neeche bound karo: $n^{2} + n > n^{2} \Rightarrow \frac{1}{n ^{2} + n} < \frac{1}{n ^{2}}$ . Yeh step kyun? Bada denominator → chhota fraction → hum $\frac{1}{n ^{2}}$ ke neeche hain.

Step 3. Benchmark $\sum \frac{1}{n ^{2}}$ converges ( $p = 2 > 1$ ) → total ink finite hai. Yeh step kyun? Finite roof ka matlab hai ink ka dabba kabhi overflow nahi hoga.

Step 4. Conclusion: total ink is finite (converges). ✅ Units: har term mm² hai, areas ka sum → total mm² mein.

Verify: fact mein $\frac{1}{n ^{2} + n} = \frac{1}{n} - \frac{1}{n + 1}$ telescope karke exactly $\sum_{n = 1}^{\infty} = 1$ mm² deta hai. Yeh converges karta hai, aur $1 < \frac{π ^{2}}{6} \approx 1.645$ — $p$ -series roof ke neeche, jaisa promise kiya tha.

Cell I — exam twist (galat-arrow steel-man)

Worked example Example 9 — Galat argument dhundho aur theek karo:

$n = 1 \sum \infty \frac{1}{2 n}$ (Cell I) Yahan $a_{n} = \frac{1}{2 n}$ . Ek student likhta hai: "Kyunki $\frac{1}{2 n} \leq \frac{1}{n}$ hai aur $\sum \frac{1}{n}$ ... converges? — toh $\sum \frac{1}{2 n}$ converges karta hai." Do errors dhundho.

Forecast: $\frac{1}{2 n} = \frac{1}{2} \cdot \frac{1}{n}$ harmonic series ka aadha hai → diverges. Student ka argument do jagah toot gayi hai.

Step 1. Error 1 (galat benchmark): $\sum \frac{1}{n}$ harmonic series hai aur woh diverge karta hai, converge nahi. Yeh step kyun? Aisi series se compare karna jiska behavior tumne galat yaad kar liya hai sab kuch poison kar deta hai (Harmonic series).

Step 2. Error 2 (galat arrow): bhalaa maan lo $\sum \frac{1}{n}$ diverge karta hai, $\frac{1}{2 n} \leq \frac{1}{n}$ dikhana humari series ko divergent ek ke neeche rakhta hai — kaam ki nahi wali direction. Divergent series se chhota hona kuch prove nahi karta. Yeh step kyun? Yeh classic "arrow mismatch" hai: divergence neeche se dikhana padta hai (chhota divergent partner).

Step 3. Sahi comparison — SAHI inequality banao. Humein ek chhota divergent partner chahiye jo humari series ke neeche ho. $b_{n} = \frac{1}{2 n + 2}$ lo. Kyunki $2 n + 2 > 2 n$ hai isliye $a_{n} = \frac{1}{2 n} > \frac{1}{2 n + 2} = \frac{1}{2} \cdot \frac{1}{n + 1} \geq 0.$ Yeh step kyun? Ab humari series genuinely ek partner ke upar hai, aur woh partner $\frac{1}{2} \sum \frac{1}{n + 1}$ hai — ek shifted half-harmonic series.

Step 4. Partner ko benchmark karo. $\frac{1}{2} \sum_{n = 1}^{\infty} \frac{1}{n + 1} = \frac{1}{2} \sum_{m = 2}^{\infty} \frac{1}{m}$ harmonic series (sirf pehla term miss) times $\frac{1}{2}$ hai → diverges. Yeh step kyun? Ek term drop karna aur baki sab aadha karna ek divergent series ko nahi bacha sakta, toh floor sach mein infinite hai.

Step 5. Conclusion. $\frac{1}{2 n} \geq \frac{1}{2 n + 2} \geq 0$ aur chhota partner diverge karta hai → $\sum \frac{1}{2 n}$ diverges. ✅ (Equivalently, $\sum \frac{1}{2 n} = \frac{1}{2} \sum \frac{1}{n}$ harmonic series ka nonzero constant times hai, jo diverge karta hai.)

Verify: $\sum_{n = 1}^{N} \frac{1}{2 n} = \frac{1}{2} H_{N}$ jahan $H_{N} \to \infty$ ; e.g. $\frac{1}{2} H_{1000} \approx 3.742$ aur $\frac{1}{2} ln N$ ki tarah badh raha hai — unbounded.

Recall checkpoint

Recall Quick self-test

Is page par, symbol $a_{n}$ hamesha kya denote karta hai? ::: Jis series ko test kiya ja raha hai uska general term — exactly woh expression jo summation sign ke andar hai. $\sum \frac{1}{n ^{3} + 2 n}$ trap karne ke liye humne kis convergent series se compare kiya, aur kis inequality ke zariye? ::: $\frac{1}{n ^{3} + 2 n} < \frac{1}{n ^{3}}$ (bada denominator → chhota fraction); $\sum \frac{1}{n ^{3}}$ converges karta hai. $\sum \frac{( - 1 ) ^{n}}{n ^{2}}$ par DCT directly kyun nahi chal sakta? ::: Terms ka sign change hota hai, toh partial sums monotone nahi hote; iske bajaye absolute values compare karo (absolute convergence). $\sum \frac{n}{n + 1}$ — kaunsa ek-line test isse khatam karta hai? ::: Terms $\to 1 \neq = 0$ , toh $n$ -th term test divergence force karta hai; koi comparison nahi chahiye. Example 5 mein, $\frac{n}{n ^{2} + 1} < \frac{1}{n}$ kaam ki nahi wali direction kyun hai? ::: Yeh dikhata hai ki humari series ek divergent series se chhoti hai, jo kuch prove nahi karta; humne $\leq 2 n^{2}$ se repair kiya taaki chhota divergent floor $\frac{1}{2 n}$ mile. Bounded-numerator series $\frac{3 + c o s 2 n}{n ^{2}}$ ke liye, convergent roof paane ke liye numerator ki jagah kya value aata hai? ::: Uska maximum, $4$ , jo $\leq \frac{4}{n ^{2}}$ deta hai.

Connections

Direct comparison test — parent rule jis par yeh examples exercise karte hain.
p-series test — Cells A, B, C, E, H ke liye benchmark.
Geometric series — Cell D ke liye benchmark.
Harmonic series — Cells B, E, I ke liye benchmark.
Absolute convergence — sign-changing Cell F ke liye repair.
Limit comparison test — smoother alternative jab Cells B/E ka algebra mushkil ho jaaye.
Integral test — $p$ -series benchmarks classify karne ka doosra tarika.
Monotone convergence theorem — engine, aur woh reason jis se Cell F fail hoti hai.