Exercises — Ratio test — proof, limitations

$\frac{a_{n+1}}{a_n}=\frac{3^{n+1}/(n+1)!}{3^n/n!}=\frac{3^{n+1}}{3^n}\cdot \frac{n!}{(n+1)!}=\frac{3}{n+1}.$ KYA: humne term $n+1$ ko term $n$ se divide kiya. YE CANCEL KYUN HOTE HAIN: $3^{n+1}/3^n=3$ aur $n!/(n+1)!=1/(n+1)$ — factorials telescope kar jaate hain. $L={\lim }_{n\to \infty }\frac{3}{n+1}=0<1\Rightarrow \text{converges absolutely}.$

$\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{5^{n+1}}\cdot \frac{5^n}{n!}=\frac{(n+1)!}{n!}\cdot \frac{5^n}{5^{n+1}}=\frac{n+1}{5}.$ $L={\lim }_{n\to \infty }\frac{n+1}{5}=\infty >1\Rightarrow \text{diverges}.$ Factorials hamesha eventually kisi bhi fixed exponential ko crush kar dete hain.

$\frac{a_{n+1}}{a_n}=\frac{1/(n+1{)}^2}{1/n^2}=\frac{n^2}{(n+1{)}^2}\to 1.$ $L=1\Rightarrow \text{inconclusive}.$ Yeh series actually converge karti hai (yeh ek $p$-series hai jisme $p=2$ hai), lekin ratio test ise dekh nahi sakta — power-law terms hamesha $L=1$ dete hain.

$\frac{a_{n+1}}{a_n}=\frac{(n+1{)}^2}{2^{n+1}}\cdot \frac{2^n}{n^2}={\left(\frac{n+1}{n}\right)}^2\cdot \frac{1}{2}.$ KYUN: $2^n$ se $1/2$ milta hai; polynomial part squares ka ratio hai. $L={\lim }_{n\to \infty }\frac{1}{2}{\left(1+\frac{1}{n}\right)}^2=\frac{1}{2}<1\Rightarrow \text{converges absolutely}.$

$\frac{a_{n+1}}{a_n}=\frac{(2n+2)!}{((n+1)!{)}^2}\cdot \frac{(n!{)}^2}{(2n)!}.$ Dono factorial families ko group karo. Upar ke liye: $\frac{(2n+2)!}{(2n)!}=(2n+2)(2n+1)$. Neeche ke liye: $\frac{((n+1)!{)}^2}{(n!{)}^2}=(n+1{)}^2$. Toh $\frac{a_{n+1}}{a_n}=\frac{(2n+2)(2n+1)}{(n+1{)}^2}=\frac{2(n+1)(2n+1)}{(n+1{)}^2}=\frac{2(2n+1)}{n+1}.$ $L={\lim }_{n\to \infty }\frac{4n+2}{n+1}=4>1\Rightarrow \text{diverges}.$

$\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{(n+1{)}^{n+1}}\cdot \frac{n^n}{n!}=(n+1)\cdot \frac{n^n}{(n+1{)}^{n+1}}=\frac{n^n}{(n+1{)}^n}={\left(\frac{n}{n+1}\right)}^n.$ YEH LIMIT KYUN: ise ${\left(1+\frac{1}{n}\right)}^{-n}$ likhte hain. Classic limit ${\left(1+\frac{1}{n}\right)}^n\to e$ se milta hai: $L=\frac{1}{e}\approx 0.3679<1\Rightarrow \text{converges absolutely}.$

$\left|\frac{a_{n+1}}{a_n}\right|=\frac{|x{|}^{n+1}}{(n+1{)}^2}\cdot \frac{n^2}{|x{|}^n}=|x|{\left(\frac{n}{n+1}\right)}^2\to |x|.$ Toh $L=|x|$ hai. Ratio test: $|x|<1$ ke liye converge, $|x|>1$ ke liye diverge — radius of convergence $R=1$ hai. Endpoints ($L=1$, ratio test blind):

• $x=1$: $\sum \frac{1}{n^2}$ — ek convergent $p$-series ($p=2$). Converge karta hai.
• $x=-1$: $\sum \frac{(-1{)}^n}{n^2}$ — absolutely converge karta hai (kyunki $\sum 1/n^2$ karta hai). Converge karta hai. Answer: sabhi $x\in [-1,1]$ ke liye converge karta hai.

Exercise 2.3 ke $\frac{n!}{n^n}$ wale ratio ko use karte hue aur $|x|$ attach karte hue: $\left|\frac{a_{n+1}}{a_n}\right|=|x|{\left(\frac{n}{n+1}\right)}^n\to \frac{|x|}{e}.$ $L=\frac{|x|}{e}<1⟺|x|<e$. Toh radius $R=e$ hai: $|x|<e$ ke liye converge, $|x|>e$ ke liye diverge karta hai. Endpoint $|x|=e$ par $L=1$ milta hai (inconclusive); ek finer tool (Raabe's test ya Stirling) chahiye — ratio test ne radius pin karne ka apna kaam kar diya hai.

Poori term $n$-th power mein hai, isliye root test ise seedha kaat deta hai: $\sqrt[n]{a_n}=\frac{n}{2n+1}\to \frac{1}{2}<1\Rightarrow \text{converges}.$ Ratio test bhi kaam karta hai lekin ratio ${\left(\frac{n+1}{2n+3}\right)}^{n+1}/{\left(\frac{n}{2n+1}\right)}^n$ bahut painful hai — "everything-to-the-$n$" form ko pehchaan ke root test ki taraf jaana hi asli analysis skill hai.

$\frac{a_{n+1}}{a_n}=\frac{2^{n+1}+(n+1)}{3^{n+1}+(n+1{)}^2}\cdot \frac{3^n+n^2}{2^n+n}.$ Dominant-term thinking: bade $n$ ke liye, $2^n+n\approx 2^n$ aur $3^n+n^2\approx 3^n$. Toh $\frac{a_{n+1}}{a_n}\approx \frac{2^{n+1}}{3^{n+1}}\cdot \frac{3^n}{2^n}=\frac{2}{3}.$ Rigorously, top/bottom ko leading power se divide karo: limit $L=\frac{2}{3}<1$ aata hai $\Rightarrow$ absolutely converge karta hai. (Alternatively compare karo $\sum (2/3{)}^n$ se — same conclusion, aur ratio limit confirm bhi hoti hai.)

$\frac{a_{n+1}}{a_n}=\frac{n\ln n}{(n+1)\ln (n+1)}.$ Dono $\frac{n}{n+1}\to 1$ aur $\frac{\ln n}{\ln (n+1)}\to 1$ hain, isliye $L=1$ — inconclusive (power/log term ke liye expected hai). Integral test ki taraf jaao: ${\int }_2^{\infty }\frac{dx}{x\ln x}$. $u=\ln x$, $du=\frac{dx}{x}$ substitute karo: $\int \frac{du}{u}=\ln u=\ln (\ln x){|}_2^{\infty }=\infty .$ Integral diverge karta hai $\Rightarrow$ series diverge karti hai. Ratio test yeh kabhi nahi dhoondh sakta tha — logs $L=1$ bana dete hain.

$r=\frac{3}{4}$ choose karo (koi bhi number $(\frac{1}{2},1)$ mein chalega). Kyunki ratio $\to \frac{1}{2}<r$ hai, ek aisa $N$ exist karta hai jiske liye sab $n\ge N$ ke liye $|a_{n+1}|<r|a_n|$ ho. Chain karte hue (jaise parent proof mein hai): $|a_{N+k}|<r^k|a_N|.$ Jaise $k\to \infty$, $r^k\to 0$ kyunki $0<r<1$ hai, isliye $|a_{N+k}|\to 0$, hence $a_n\to 0$. $■$ Yahi wajah hai ki $L<1$ automatically term test pass kar leta hai — geometric bound terms ko zero par force karta hai.

Koi bhi do $p$-series kaam karengi jisme $p\le 1$ vs $p>1$ ho. Lo:

• Divergent: $\sum \frac{1}{\sqrt{n}}$ ($p=\frac{1}{2}\le 1$). Ratio $=\sqrt{\frac{n}{n+1}}\to 1$ hai, isliye $L=1$; $p$-test se diverge karta hai.
• Convergent: $\sum \frac{1}{n^3}$ ($p=3>1$). Ratio $={\left(\frac{n}{n+1}\right)}^3\to 1$ hai, isliye $L=1$; $p$-test se converge karta hai. Dono $L=1$ par baithe hain phir bhi alag hain — boundary sirf ratio se genuinely undecidable hai. Neeche figure dekho ki kyun har power law $1$ par flatten ho jaata hai.

Exercise 2.2 se, $\frac{(2n)!}{(n!{)}^2}$ ka ratio $\to 4$ hai, isliye iska reciprocal $\frac{(n!{)}^2}{(2n)!}$ ka ratio $\to \frac{1}{4}$ hai. $x$ attach karte hue: $\left|\frac{a_{n+1}}{a_n}\right|=x\cdot \frac{(n+1{)}^2}{(2n+2)(2n+1)}=x\cdot \frac{n+1}{2(2n+1)}\to \frac{x}{4}.$ $L=\frac{x}{4}<1⟺x<4$. Radius $R=4$: $0<x<4$ ke liye converge, $x>4$ ke liye diverge karta hai. Endpoint $x=4$: $L=1$, inconclusive — Stirling dikhata hai ki term $\frac{1}{\sqrt{\pi n}}$ ki tarah behave karta hai, isliye wahan $\sum$ diverge karta hai. Ratio test sahi $R=4$ locate karta hai.

JHOOTH. "$<1$ har $n$ ke liye" condition "$L<1$" se zyada weak hai: ratio $1$ se neeche se $1$ ki taraf approach kar sakta hai bina kisi fixed $r<1$ ke jo use trap kare. Counterexample: harmonic series $\sum \frac{1}{n}$. Yahan $\frac{a_{n+1}}{a_n}=\frac{n}{n+1}<1$ har $n$ ke liye hai, phir bhi series diverge karti hai ($L=1$). Moral: ratio test ko limit $L$ strictly $1$ se neeche chahiye hota hai, jo ek fixed gap $[L,1)$ deta hai geometric domination ke liye. Ratio jo merely $1$ ki taraf creep karta rahe woh koi aisa gap nahi chhodta.