Exercises — Direct comparison test

Goal = convergence, isliye humein apni series ko ek finite roof ke neeche trap karna hoga. Iska matlab hai $b_n$ bigger hona chahiye (ek upper bound) aur use converge karna chahiye. Concretely: $\frac{1}{n^3+1}<\frac{1}{n^3}$, aur $\sum 1/n^3$ ek $p$-series hai jisme $p=3>1$ → converges. Answer: bigger aur convergent.

Goal = divergence, isliye hum ek infinite floor se upar push karte hain: $b_n$ smaller hona chahiye (ek lower bound) aur divergent hona chahiye. Concretely $\frac{1}{\sqrt{n}}=\frac{1}{n^{1/2}}$ khud ek $p$-series hai jisme $p=\frac{1}{2}\le 1$, toh yeh directly diverge karta hai — lekin ek comparison ke roop mein, note karo ki $\frac{1}{\sqrt{n}}\ge \frac{1}{n}$, aur smaller harmonic $\sum \frac{1}{n}$ diverge karta hai. Answer: smaller aur divergent.

Haan. $\sum \frac{1}{2^n}$ geometric hai jisme $r=\frac{1}{2}$, $|r|<1$, isliye yeh converges karta hai. Inequality ek bigger convergent partner provide karti hai — convergence claim ke liye correct arrow. Valid. ✅

Guess: dominant power $n^2$ → $\frac{1}{n^2}$ jaisa behave karta hai → converges. Kyunki $3n^2+2>n^2$ sabhi $n\ge 1$ ke liye, hume milta hai $0\le \frac{1}{3n^2+2}<\frac{1}{n^2}$. $\sum 1/n^2$ ek $p$-series hai, $p=2>1$ → converges. Bigger convergent roof ⇒ converges. ✅

Guess: $\frac{n}{n^2}=\frac{1}{n}$ → diverges. Humein ek smaller divergent partner chahiye. Kyunki $n^2+1\le n^2+n^2=2n^2$ for $n\ge 1$, hamare paas hai $\frac{n}{n^2+1}\ge \frac{n}{2n^2}=\frac{1}{2n}\ge 0$. $\sum \frac{1}{2n}=\frac{1}{2}\sum \frac{1}{n}$ diverge karta hai (harmonic). Smaller divergent floor ⇒ diverges. ✅ $2n^2$ kyun? Humne deliberately denominator ko $n^2$ ke ek clean multiple tak inflate kiya taaki term ek harmonic-type lower bound tak collapse ho jaye.

$\cos n$ $[-1,1]$ mein wiggle karta hai, isliye $3\le 4+\cos n\le 5$; khaas taur pe term $\ge 0$ hai. Numerator ko uske maximum se bound karo: $\frac{4+\cos n}{n^3}\le \frac{5}{n^3}$. $\sum \frac{5}{n^3}=5\sum \frac{1}{n^3}$, $p=3>1$ → converges. Bigger convergent roof ⇒ converges. ✅

Denominator mein $3^n$ dominant hai, aur $2^n+3^n>3^n$, isliye $0\le \frac{1}{2^n+3^n}<\frac{1}{3^n}$. $\sum {\left(\frac{1}{3}\right)}^n$ geometric hai jisme $r=\frac{1}{3}$, $|r|<1$ → converges. Bigger convergent roof ⇒ converges. ✅

Guess: $\sim \frac{1}{n^2}$ → converges. Ek bigger convergent partner chahiye. $n\ge 2$ ke liye, $n^2-n=n^2\left(1-\frac{1}{n}\right)\ge n^2\cdot \frac{1}{2}=\frac{n^2}{2}$ (kyunki $1-\frac{1}{n}\ge \frac{1}{2}$). Isliye $\frac{1}{n^2-n}\le \frac{2}{n^2}$. $\sum \frac{2}{n^2}$ converges karta hai ($p=2$). ⇒ converges. ✅ $\frac{1}{2}$ wala move kyun? Directly $\frac{1}{n^2-n}>\frac{1}{n^2}$ convergence ke liye wrong direction point karta hai, isliye hum ek valid upper bound manufacture karte hain by controlling karte hue ki $n^2-n$ kitna $n^2$ se neeche jaata hai.

Guess: $\sqrt{n}\cdot (n+1)\sim n^{3/2}$ → $\frac{1}{n^{3/2}}$ jaisa → converges ($p=\frac{3}{2}>1$). Kyunki $n+1>n$, $\sqrt{n}(n+1)>\sqrt{n}\cdot n=n^{3/2}$, isliye $0\le \frac{1}{\sqrt{n}(n+1)}<\frac{1}{n^{3/2}}$. $\sum n^{-3/2}$ converges karta hai. Bigger convergent roof ⇒ converges. ✅

Guess: ratio $\to (2/3{)}^n$ → geometric-jaisa, converges. Ek bigger convergent partner chahiye. $n\ge 1$ ke liye, $3^n-1\ge 3^n-3^{n-1}=3^{n-1}\cdot 2=\frac{2}{3}{3}^n$. Actually simpler: $3^n-1\ge \frac{1}{2}{3}^n$ for $n\ge 1$ (kyunki $3^n\ge 2$ deta hai $3^n-1\ge \frac{3^n}{2}$). Toh $\frac{2^n}{3^n-1}\le \frac{2^n}{\frac{1}{2}{3}^n}=2{\left(\frac{2}{3}\right)}^n$. $\sum 2(2/3{)}^n$ geometric hai jisme $r=\frac{2}{3}<1$ → converges. Bigger convergent roof ⇒ converges. ✅

Key fact: $\ln n<n$ sabhi $n\ge 1$ ke liye (log, line se slower grow karta hai). Toh $\frac{1}{\ln n}>\frac{1}{n}\ge 0$ for $n\ge 3$. $\sum \frac{1}{n}$ diverge karta hai (harmonic). Smaller divergent floor ⇒ diverges. ✅ $1/n$ se compare kyun? Hum divergence suspect karte hain kyunki $\ln n$, $n$ se chhota hai, isliye $\frac{1}{\ln n}$, $\frac{1}{n}$ se bada hai — hum ek known-infinite floor ke upar baithe hain.

Dono pieces non-negative hain, isliye ek valid single upper bound upper bounds ka sum hai. Har piece converge karti hai: $\sum \frac{1}{n^2}$ ($p=2$) aur $\sum \frac{1}{2^n}$ (geometric $r=\frac{1}{2}$). Do convergent series ka sum converge karta hai. DCT ke saath directly: $\frac{1}{n^2}+\frac{1}{2^n}\le \frac{1}{n^2}+\frac{1}{n^2}=\frac{2}{n^2}$ for $n\ge 1$ (kyunki $2^n\ge n^2$ for $n\ge 4$; chhote $n$ alag check karo — finitely many terms matter nahi karte). $\sum \frac{2}{n^2}$ converge karta hai. ⇒ converges. ✅

Kyunki $-1\le \sin n\le 1$, numerator $n+\sin n\ge n-1\ge 0$ for $n\ge 1$ (equality sirf ise shrink karti hai). $n\ge 2$ ke liye, $n+\sin n\ge n-1\ge \frac{n}{2}$. Toh $\frac{n+\sin n}{n^2}\ge \frac{n/2}{n^2}=\frac{1}{2n}\ge 0$. $\sum \frac{1}{2n}$ diverge karta hai (harmonic). Smaller divergent floor ⇒ diverges. ✅ Insight: dominant term $\frac{n}{n^2}=\frac{1}{n}$ hai; $\sin n$ ek bounded nuisance hai jo hum lower bound ke saath absorb karte hain.

Guess: top $\sim 2^n$, bottom $\sim 3^n$ → $(2/3{)}^n$ jaisa → converges. Upper bound: $1+2^n\le 2\cdot 2^n$ (kyunki $1\le 2^n$) aur $1+3^n\ge 3^n$. Toh $\frac{1+2^n}{1+3^n}\le \frac{2\cdot 2^n}{3^n}=2{\left(\frac{2}{3}\right)}^n.$ Geometric $r=\frac{2}{3}<1$ → converges. Bigger convergent roof ⇒ converges. ✅

DCT try karo: $\frac{1}{n\ln n}<\frac{1}{n}$ — lekin $\frac{1}{n}$ diverge karta hai, toh smaller-than-divergent comparison useless hai. Aur $\frac{1}{n\ln n}>\frac{1}{n^2}$ large $n$ ke liye — lekin $\frac{1}{n^2}$ converge karta hai, toh bigger-than-convergent comparison bhi useless hai. DCT dono taraf se squeeze ho jaata hai. Correct tool — Integral test: ${\int }_2^{\infty }\frac{dx}{x\ln x}=[\ln (\ln x){]}_2^{\infty }=\infty$. Integral diverge karta hai, isliye series diverges. ✅ Lesson: DCT ko correct side par ek benchmark chahiye; yeh series do benchmark families ke beech rehti hai, isliye DCT ise reach nahi kar sakta — tum integral test tak escalate karo.

Terms sign change karte hain, isliye DCT directly apply nahi hota (ise $a_n\ge 0$ chahiye). Iske bajaye absolute values compare karo: $\left|\frac{(-1{)}^n}{n^2}\right|=\frac{1}{n^2}$, aur $\sum \frac{1}{n^2}$ converge karta hai. DCT se absolute values ki series converge karti hai, isliye original absolutely convergent hai (Absolute convergence), aur isliye convergent. ✅ Detour kyun? DCT ka proof monotone partial sums par rests karta hai, jiske liye non-negative terms chahiye — isliye hum ise $|a_n|$ par apply karte hain.

Claim: converges iff $p>1$ (same threshold jaise pure $p$-series mein).

• Agar $p>1$: $n\ge 2$ ke liye, $n^p-1\ge \frac{1}{2}{n}^p$ (kyunki $n^p\ge 2$ deta hai $n^p-1\ge \frac{n^p}{2}$), isliye $\frac{1}{n^p-1}\le \frac{2}{n^p}$. $\sum \frac{2}{n^p}$ converge karta hai ($p>1$). Bigger convergent roof ⇒ converges.
• Agar $p\le 1$: $n^p-1<n^p$, isliye $\frac{1}{n^p-1}>\frac{1}{n^p}\ge 0$. $\sum \frac{1}{n^p}$ diverge karta hai ($p\le 1$). Smaller divergent floor ⇒ diverges. Dono directions DCT se pin hue. Answer: converges $⟺p>1$. ✅

$x\ge 0$ ke liye hamare paas $0\le \sin x\le x$ hai (sine $[0,\infty )$ par apna input kabhi overshoot nahi karta jahan yeh non-negative hai; yahan $\frac{1}{n^2}\in (0,1]$, isliye $\sin (\frac{1}{n^2})\ge 0$). Isliye $0\le \sin \left(\frac{1}{n^2}\right)\le \frac{1}{n^2}$. $\sum \frac{1}{n^2}$ converge karta hai. Bigger convergent roof ⇒ converges. ✅ $\sin x\le x$ kyun? Yeh clean upper bound hai jo ek transcendental term ko ek $p$-series comparison mein convert karta hai — yahi wajah hai ki DCT itna powerful hai.

Har problem ek sawaal tak reduce hui: kaunsa trusted dost, kaunsi side par? Convergence ⇒ ek bigger convergent roof ke neeche trap karo (numerator upar, denominator neeche). Divergence ⇒ ek smaller divergent floor ke upar push karo (numerator neeche, denominator upar). Jab target benchmarks ke beech squeeze ho jaaye (logs!) ya sign change kare, DCT akela nahi kar sakta — Integral test, Limit comparison test, ya Absolute convergence ke liye reach karo.