Exercises — Limit comparison test

Maano $a_n>0$ aur $b_n>0$ sabhi bade $n$ ke liye, aur set karo $L=\underset{n\to \infty }{\lim }\frac{a_n}{b_n}$.

• Main case $0<L<\infty$: $\sum a_n$ aur $\sum b_n$ both converge or both diverge (same fate).
• Edge $L=0$: agar $\sum b_n$ converges, toh $\sum a_n$ converge karta hai. (Kuch nahi kehta agar $\sum b_n$ diverge kare.)
• Edge $L=\infty$: agar $\sum b_n$ diverges, toh $\sum a_n$ diverge karta hai. (Kuch nahi kehta agar $\sum b_n$ converge kare.)

Skeleton. Top mein strongest power $5n$ hai (behaves like $n$). Bottom mein $2n^2$ hai (behaves like $n^2$). Toh $b_n=\frac{n}{n^2}=\frac{1}{n}.$ Yeh $b_n$ kyun? Bade $n$ ke liye $+3$ aur $+n$ leading powers ke saamne dab jaate hain — decay rate sirf har slot ka sabse bada power decide karta hai. Fate. $\sum 1/n$ Harmonic Series hai jisme $p=1$ hai, jo diverge karta hai. Kyunki $a_n$, $1/n$ ko track karta hai, toh series diverge karta hai. ✔

Top $\sim n^2$, bottom $\sim n^5$, toh $b_n=\frac{n^2}{n^5}=\frac{1}{n^3}.$ $\sum 1/n^3$ ek p-Series hai jisme $p=3>1$ hai, jo converge karta hai. Toh $\sum a_n$ converge karta hai. ✔

Pick. Top $\sim 4n^2$, bottom $\sim n^3$: $b_n=\frac{n^2}{n^3}=\frac{1}{n}$. $L$ compute karo. $L={\lim }_{n\to \infty }\frac{(4n^2-n)/(n^3+5)}{1/n}=\lim \frac{n(4n^2-n)}{n^3+5}=\lim \frac{4n^3-n^2}{n^3+5}.$ Top aur bottom ko $n^3$ se divide karo: $\frac{4-1/n}{1+5/n^3}\to 4.$ Conclude. $0<L=4<\infty$, aur $\sum 1/n$ diverge karta hai (harmonic). Toh $\sum a_n$ diverge karta hai. ✔

Pick. Root ke andar strongest power $4n^3$ hai, toh $\sqrt{4n^3+n}\approx \sqrt{4n^3}=2n^{3/2}$. Skeleton ke liye $2$ constant drop kar sakte hain: $b_n=\frac{1}{n^{3/2}}.$ $L$ compute karo. $L=\lim \frac{1/\sqrt{4n^3+n}}{1/n^{3/2}}=\lim \frac{n^{3/2}}{\sqrt{4n^3+n}}.$ Root se $n^3$ factor karo: $\sqrt{4n^3+n}=n^{3/2}\sqrt{4+1/n^2}$, toh $L=\lim \frac{n^{3/2}}{n^{3/2}\sqrt{4+1/n^2}}=\frac{1}{\sqrt{4}}=\frac{1}{2}.$ Conclude. $0<L=\frac{1}{2}<\infty$, aur $\sum 1/n^{3/2}$ converge karta hai ($p=\frac{3}{2}>1$). Toh $\sum a_n$ converge karta hai. ✔

Pick. Bade $n$ ke liye, $4^n$ $n^2$ ko crush kar deta hai, toh denominator $4^n$ jaisa behave karta hai, aur milta hai $b_n=\frac{3^n}{4^n}={\left(\frac{3}{4}\right)}^n,$ ek Geometric Series jisme ratio $r=\frac{3}{4}$ hai. $L$ compute karo. $L=\lim \frac{3^n/(4^n+n^2)}{3^n/4^n}=\lim \frac{4^n}{4^n+n^2}=\lim \frac{1}{1+n^2/4^n}=1,$ kyunki $n^2/4^n\to 0$ (exponential polynomial ko beat karta hai). Conclude. $0<L=1<\infty$, aur $\sum (3/4{)}^n$ converge karta hai ($|r|=\frac{3}{4}<1$). Toh $\sum a_n$ converge karta hai. ✔

Naive skeleton kehta $b_n=1/n^3$. Lekin $\ln n\to \infty$, toh $\frac{(\ln n)/n^3}{1/n^3}=\ln n\to \infty$ — yeh $L=\infty$ case hai ek convergent $b_n$ ke against, jo inconclusive hai. Humein ek aisa $b_n$ chahiye jo thoda bada ho phir bhi convergent ho, taaki uska extra power of $n$ slow $\ln n$ ko swallow kar sake. Pick. $b_n=\frac{1}{n^{2.5}}$ ($p=2.5>1$, still convergent). $L$ compute karo. $L=\lim \frac{(\ln n)/n^3}{1/n^{2.5}}=\lim \frac{\ln n}{n^{0.5}}=0,$ kyunki $n$ ki koi bhi positive power eventually $\ln n$ ko beat karti hai. Conclude. $L=0$ aur $\sum 1/n^{2.5}$ converge karta hai, toh $L=0$ rule se $\sum a_n$ converge karta hai. ✔

Skeleton simplify karo. $\ln$ ignore karte hue, fraction $\frac{n}{n^2+1}$ $\frac{1}{n}$ jaisa behave karta hai, lekin $\ln (n+1)$ divide kar raha hai jo convergence mein thodi madad karta hai. Clean verdict ke liye $b_n=1/n$ se compare karo aur $L=0$ expect karo. $L=\lim \frac{n/[\ln (n+1)(n^2+1)]}{1/n}=\lim \frac{n^2}{\ln (n+1)(n^2+1)}=\lim \frac{1}{\ln (n+1)\cdot (1+1/n^2)}=0,$ kyunki $\ln (n+1)\to \infty$. Lekin $\sum 1/n$ diverge karta hai — toh $L=0$ paired with ek divergent $b_n$ inconclusive hai. Yeh galat yardstick hai. Repair. Denominator mein $\ln$ weak hai; term barely $1/n$ se chhota hai. Iske bajaye $b_n=\frac{1}{n\ln (n+1)}$ se compare karo. $\ln (n+1)$ vs $\ln n$ ke baare mein ek quick note: ye asymptotically equal hain, kyunki $\frac{\ln (n+1)}{\ln n}=\frac{\ln n+\ln (1+1/n)}{\ln n}=1+\frac{\ln (1+1/n)}{\ln n}\to 1,$ toh $\sum \frac{1}{n\ln (n+1)}$ aur $\sum \frac{1}{n\ln n}$ ka same fate hai. Baad wale ka fate Integral Test se aata hai: $\int \frac{dx}{x\ln x}=\ln \ln x\to \infty$, toh $\sum \frac{1}{n\ln n}$ diverge karta hai. $L=\lim \frac{n/[\ln (n+1)(n^2+1)]}{1/[n\ln (n+1)]}=\lim \frac{n^2}{n^2+1}=1.$ Conclude. $0<L=1<\infty$ ek divergent reference ke against, toh $\sum a_n$ diverge karta hai. ✔

Diagnosis. Factorials mein koi "leading power" nahi hoti strip karne ke liye, toh koi natural $p$-series ya geometric skeleton nahi hai — LCT yahan theek nahi baithta. Yeh Ratio Test ke liye chillata hai (consecutive terms ke ratios factorials ko khatam karte hain). Ratio Test. $a_n=\frac{n!}{(2n)!}$ ke saath, $\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{(2n+2)!}\cdot \frac{(2n)!}{n!}=\frac{(n+1)}{(2n+2)(2n+1)}=\frac{n+1}{(2n+2)(2n+1)}.$ Jab $n\to \infty$ toh yeh $\to 0<1$. Converge karta hai. ✔ Lesson: LCT algebraic (power/root/exponential-base) terms ke liye hai; factorials Ratio Test mein jaate hain.

Diagnosis. Poori cheez $n$ ki power par raised hai; base $\frac{2}{5}$ ki taraf jaata hai. Yeh geometric lagta hai, toh $b_n=(2/5{)}^n$ banao (ek Geometric Series, $|r|=\frac{2}{5}<1$, converge karta hai). $L$ compute karo. $L=\lim \frac{{\left(\frac{2n+1}{5n+3}\right)}^n}{(2/5{)}^n}=\lim {\left(\frac{5(2n+1)}{2(5n+3)}\right)}^n=\lim {\left(\frac{10n+5}{10n+6}\right)}^n.$ Base ko $1-\frac{1}{10n+6}$ likho. Ab classical exponential limit ${\left(1+\frac{u}{m}\right)}^m\to e^u$ use karo: yahan numerator mein "$-1$" $u$ ka role play karta hai aur denominator $10n$ jaisa badhta hai, toh ${\left(1-\frac{1}{10n+6}\right)}^n={\left[{\left(1-\frac{1}{10n+6}\right)}^{10n+6}\right]}^{\frac{n}{10n+6}}\to (e^{-1}{)}^{1/10}=e^{-1/10},$ ek finite positive number (inner bracket $\to e^{-1}$, aur exponent $\frac{n}{10n+6}\to \frac{1}{10}$). Conclude. $0<L=e^{-1/10}<\infty$, aur $\sum (2/5{)}^n$ converge karta hai, toh $\sum a_n$ converge karta hai. ✔ (Root Test bhi ise ek second mein solve karta hai — LCT kaam kiya kyunki hum geometric skeleton naam de sake.)

Skeleton. Top $\sim n^p$, bottom $\sim n^3$, toh $b_n=\frac{n^p}{n^3}=\frac{1}{n^{3-p}}.$ $L$ compute karo. $L=\lim \frac{n^p/(n^3+1)}{1/n^{3-p}}=\lim \frac{n^p\cdot n^{3-p}}{n^3+1}=\lim \frac{n^3}{n^3+1}=1.$ $0<L=1<\infty$ har $p$ ke liye, toh $\sum a_n$ ka fate $\sum 1/n^{3-p}$ jaisa hai. Boundary. Yeh p-Series converge karta hai $⟺3-p>1⟺p<2$. Answer. $\sum a_n$ converge karta hai $\boxed{p<2}$ ke liye aur $p\ge 2$ ke liye diverge karta hai. ✔

LCT kyun ruk jaata hai. Har candidate $b_n=1/n^q$ deta hai $L=0$ (agar $q<1$, divergent ref, inconclusive) ya $L=\infty$ (agar $q>1$, convergent ref, inconclusive) — $(\ln n{)}^{\alpha }$ factor sub-polynomial hai, toh koi power skeleton same rate par nahi baithta. LCT logarithmic borderlines resolve nahi kar sakta. Sahi tool: Integral Test. $f(x)=\frac{1}{x(\ln x{)}^{\alpha }}$ ke saath, $u=\ln x$, $du=dx/x$ substitute karo: ${\int }_2^{\infty }\frac{dx}{x(\ln x{)}^{\alpha }}={\int }_{\ln 2}^{\infty }\frac{du}{u^{\alpha }}.$ Yeh $u$ mein ek $p$-integral hai: yeh converge karta hai $⟺\alpha >1$. Answer. $\sum a_n$ converge karta hai $\boxed{\alpha >1}$ ke liye, $0<\alpha \le 1$ ke liye diverge karta hai. ✔ Takeaway: LCT un series ko rank karta hai jo power jaisi decay karte hain; logarithmic fine-structure ke liye Integral Test chahiye.

LCT bilkul galat tool kab hota hai? ::: Jab terms factorials ya $n$-th powers hain (use karo Ratio Test/Root Test), ya jab decay power se sirf ek logarithm se alag hoti hai (use karo Integral Test).