Exercises — Divergence test (necessary but not sufficient)

Kya karenge: $L=\underset{n\to \infty }{\lim }\frac{n}{n+1}$ compute karenge. Kyun: test ka poora input yahi ek limit hai. Top aur bottom dono ko $n$ se divide karo (highest power) — kyun: yeh leading behaviour ko isolate karta hai aur chote $\frac{1}{n}$ pieces ko $0$ par bhej deta hai: $\frac{n}{n+1}=\frac{1}{1+\frac{1}{n}}⟶\frac{1}{1+0}=1.$ Kyunki $L=1\ne 0$, test fire karta hai. Answer: series diverges by the Divergence Test.

Kya karenge: $L=\underset{n\to \infty }{\lim }\frac{5}{n^2}$. Jaise $n\to \infty$, denominator explode karta hai, toh $\frac{5}{n^2}\to 0$, jo $L=0$ deta hai. Yeh kyun matter karta hai: $L=0$ exactly woh case hai jahan test chup ho jaata hai. Answer: inconclusive — Divergence Test kuch nahi kehta. (Ek alag tool — jaise p-Series Test with $p=2>1$ — dikhayega ki yeh converge karta hai, lekin yeh is test ka kaam nahi hai.)

Kya karenge: $n=1,2,3,\dots$ ke liye terms $\cos (\pi n)$ dekho. $\cos \pi =-1,\cos 2\pi =1,\cos 3\pi =-1,\dots$ Terms $-1,+1,-1,+1,\dots$ bounce karte rehte hain hamesha aur kisi ek number par kabhi settle nahi karte. Yeh kyun decide karta hai: agar terms kisi single value ki taraf nahi jaate, toh $\lim a_n$ Does Not Exist (DNE) hota hai. Test ke liye, DNE ko exactly "$\ne 0$" ki tarah treat kiya jaata hai. Answer: diverges (limit DNE $\ne 0$). Dekho Limits of Sequences kyun ek oscillating sequence ki koi limit nahi hoti.

Kya karenge: same-degree rational function → top aur bottom dono ko $n^3$ (highest power) se divide karo. $n^3$ kyun: yeh har lower-power term ko $\frac{1}{n},\frac{1}{n^2},\dots$ bana deta hai jo infinity par zero ho jaate hain, sirf leading coefficients bach jaate hain. $\frac{4n^3-7n}{9n^3+2n^2+1}=\frac{4-\frac{7}{n^2}}{9+\frac{2}{n}+\frac{1}{n^3}}⟶\frac{4-0}{9+0+0}=\frac{4}{9}.$ $L=\frac{4}{9}\ne 0$. Answer: diverges.

Kya karenge: ek standard exponential limit pehchaano. Template hai ${\left(1+\frac{x}{n}\right)}^n\to e^x$. Yeh template kyun: base $\to 1$ jaata hai jabki power $\to \infty$ — ek $1^{\infty }$ tug-of-war jo exponential se resolve hota hai, $1$ se nahi. Yahan $x=3$ hai: ${\lim }_{n\to \infty }{\left(1+\frac{3}{n}\right)}^n=e^3\approx \mathrm{20.09.}$ $L=e^3\ne 0$. Answer: diverges.

Kya karenge: $L=\underset{n\to \infty }{\lim }\frac{\ln n}{n}$. Dono parts grow karte hain, lekin $n$, $\ln n$ se bahut tezi se grow karta hai, toh ratio $\to 0$ jaata hai. $\frac{\ln n}{n}⟶0.$ $L=0$. Yahi point kyun hai: terms genuinely vanish karte hain, toh test chup ho jaata hai. Answer: inconclusive (Integral Test baad mein dikhayega ki yeh actually diverge karta hai — ek sundar reminder ki "$L=0$" dono fates chhupa sakta hai).

Kya karenge: dono limits lo. $\frac{1}{\sqrt{n}}\to 0,\frac{1}{n^2}\to 0.$ Dono $L=0$ dete hain, toh dono ke liye Divergence Test inconclusive hai. Yeh pairing kyun instructive hai: in dono ke true fates opposite hain — p-Series Test ke according, (a) $\sum n^{-1/2}$ diverges ($p=\frac{1}{2}\le 1$) jabki (b) $\sum n^{-2}$ converges ($p=2>1$). Same "$L=0$", opposite outcomes. Yahi ek fact hai jo batata hai ki test kabhi convergence detector nahi ban sakta. Answer: dono ke liye inconclusive.

Kya karenge: $|r|$ ke cases ke hisaab se ${\lim }_{n\to \infty }{r}^n$ examine karo.

• $|r|>1$: $|r{|}^n\to \infty$, toh $r^n\nrightarrow 0$. Test fires → diverges.
• $r=1$: terms hain $1,1,1,\cdots \to 1\ne 0$. Test fires → diverges.
• $r=-1$: terms $1,-1,1,\dots$ → limit DNE $\ne 0$. Test fires → diverges.
• $|r|<1$: $r^n\to 0$, toh $L=0$. Test silent hai → inconclusive. Answer: Divergence Test divergence prove karta hai sabhi $\boxed{|r|\ge 1}$ ke liye. $|r|<1$ ke liye yeh inconclusive hai (geometric-series formula, ek stronger tool, phir convergence $\frac{1}{1-r}$ dikhata hai). Boundary $1$ par kyun hoti hai: $|r|<1$ exactly woh regime hai jahan repeated multiplication ek number ko kuch nahi kar deta; $|r|=1$ par shrinking rukk jaati hai.

(i) Definition se convergence (Partial Sums and Series Convergence se): ek series apne partial sums ki limit hai. ${\lim }_{N\to \infty }{S}_N={\lim }_{N\to \infty }\frac{3N}{N+2}=\frac{3}{1+0}=3.$ Partial sums $3$ par settle karte hain, toh series $3$ par converge karti hai. (ii) Term recover karo Telescoping Series trick $a_N=S_N-S_{N-1}$ use karke: $a_N=\frac{3N}{N+2}-\frac{3(N-1)}{N+1}=\frac{3N(N+1)-3(N-1)(N+2)}{(N+2)(N+1)}.$ Numerator expand karo: $3(N^2+N)-3(N^2+N-2)=6$. Toh $a_N=\frac{6}{(N+1)(N+2)}⟶0.$ $L=0$, toh Divergence Test inconclusive hai. Seekh: bhale hi hum jaante hain ki yeh converge karta hai (partial sums se), Divergence Test akela sirf yahi kehta "I don't know." Theory ke saath consistent — $L=0$ yahan necessary hai aur indeed hold karta hai, lekin yeh kabhi convergence certify nahi karta.

Kya karenge: dikhao ki terms $0$ ki taraf nahi jaate, actually blow up karte hain. Term ko $n$ fractions ke product ki tarah likho: $\frac{n^n}{n!}=\underset{=1}{\underbrace{\frac{n}{n}}}\cdot \underset{\ge 1}{\underbrace{\frac{n}{n-1}}}\cdot \underset{\ge 1}{\underbrace{\frac{n}{n-2}}}\cdots \underset{=n}{\underbrace{\frac{n}{1}}}.$ Har factor $\ge 1$ hai aur aakhri $n$ hai, toh $\frac{n^n}{n!}\ge n\to \infty$. Yeh bound kyun: hume sirf yeh jaanna hai ki terms vanish karna fail karte hain; ek clean lower bound jo khud $\to \infty$ ho, kaafi hai. Isliye $L=\infty \ne 0$. Answer: Divergence Test se diverges.

Pehli series, $\sin (1/n)$: jaise $n\to \infty$, $\frac{1}{n}\to 0$, aur $\sin (\text{small})\to \sin 0=0$. Toh $L=0$ → inconclusive. (Kyunki bade $n$ ke liye $\sin (1/n)\approx \frac{1}{n}$, baad mein ek comparison dikhata hai ki yeh actually harmonic series ki tarah diverge karta hai — lekin Divergence Test yeh nahi dekh sakta.) Doosri series, $\sin (n)$: yahan $n$ radians mein hai aur circle ke around bina kabhi settle kiye ghoomta rehta hai. Values $\sin (n)$ $[-1,1]$ mein dense hain aur kisi ek number ki taraf nahi jaate → $\lim$ DNE $\ne 0$. Answer: pehla inconclusive hai; doosra diverges (limit DNE).

Proof (parent derivation ko mirror karta hai): Maano $S_N={\sum }_{n=1}^N{a}_n$. "$S$ par converge karta hai" ka matlab hai ${\lim }_{N\to \infty }{S}_N=S$ (finite). Identity $a_N=S_N-S_{N-1}$ use karo. Index ko ek se shift karne par limit nahi badlti, toh $S_{N-1}\to S$ bhi. Kyunki dono limits exist karte hain aur finite hain, difference ki limit, limits ke difference ke barabar hai: ${\lim }_{N\to \infty }{a}_N={\lim }_{N\to \infty }(S_N-S_{N-1})=S-S=0.■$ Reverse kyun fail karta hai (ek sentence): argument sirf ek direction mein chalta hai — Harmonic Series $\sum \frac{1}{n}$ mein $a_n\to 0$ hota hai phir bhi uske partial sums $S_N\to \infty$ jaate hain, toh "$a_n\to 0$" convergence force nahi kar sakta.

Jo hum jaante hain: kyunki $\sum a_n$ converge karta hai, L5·Q2 se zaroor $a_n\to 0$ hoga. Dhyan raho — test sirf tab fire karta hai jab $(a_n+b_n)$ ki limit nonzero/DNE ho. Hum hamesha sirf Divergence Test se conclude nahi kar sakte, kyunki $b_n$ khud $0$ ki taraf ja sakta hai (ek divergent series ke vanishing terms ho sakte hain, jaise harmonic series). Tab $a_n+b_n\to 0$ aur test chup ho jaata hai.

• Agar $b_n\nrightarrow 0$: tab $a_n+b_n\to 0+(\text{nonzero ya DNE})\ne 0$, toh Divergence Test prove karta hai ki $\sum (a_n+b_n)$ diverge karta hai.
• Agar $b_n\to 0$ (still divergent, jaise harmonic): $a_n+b_n\to 0$, Divergence Test inconclusive — phir bhi sum diverge karta hai, linearity se prove hota hai (convergent + divergent = divergent), yeh ek alag argument hai. Answer: sum hamesha diverge karta hai, lekin Divergence Test ise sirf tabhi prove kar sakta hai jab $b_n\nrightarrow 0$ ho. Jab $b_n\to 0$ ho toh linearity theorem ki zaroorat padti hai.