Exercises — Power series — centre, radius of convergence, interval of convergence

KYA dekh rahe hain: $(x-a{)}^n$ ki shape. $x$ mein se jo number ghata hai woh $a=7$ hai, yani centre. WHY wahan hamesha converge karta hai: $x=7$ plug karo. Tab $(x-7{)}^n=0^n=0$ har $n\ge 1$ ke liye, isliye sirf $n=0$ waali term $c_0$ bachti hai. Ek akela number finite sum hai, jo trivially converge karta hai. Answer: centre $a=7$; $x=7$ par guaranteed convergence (opening figure mein black dot).

WHY yeh formula apply hota hai: Ratio Test version $R={\lim }_{n\to \infty }|c_n/c_{n+1}|$ tabhi valid hai jab woh limit exist kare — aur yahan uski value directly di hui hai. KYA milta hai: $R=5$. Centre $a=0$ hai ($x^n$ mein koi shift nahin), isliye open interval $(0-5,0+5)=(-5,5)$ hai. Answer: $R=5$, open interval $(-5,5)$. (Endpoints tab tak unknown hain jab tak $c_n$ nahin pata.)

Step 1 — KYA (ratio): $a_n=\frac{(x-4{)}^n}{n{5}^n}$ ke saath, $\left|\frac{a_{n+1}}{a_n}\right|=\frac{|x-4{|}^{n+1}/((n+1)5^{n+1})}{|x-4{|}^n/(n{5}^n)}=\frac{|x-4|}{5}\cdot \frac{n}{n+1}.$ WHY yeh step: $|x-4{|}^n$ cancel hokar ek factor bachta hai, $5^n$ se ek factor $5$ bachta hai, aur $\frac{n}{n+1}\to 1$. Step 2 — test ke conclusion se $R$ tak: limit $L=\frac{|x-4|}{5}$ hai. Ratio Test kehta hai series converge karti hai jab $L<1$ aur diverge karti hai jab $L>1$. Toh $x$ ka woh exact set jo converge karta hai milega $L<1$ ki inequality $|x-4|$ ke liye solve karke: $\frac{|x-4|}{5}<1⟺|x-4|<5.$ Yeh aakhri line hi radius ki defining condition $|x-a|<R$ hai, isliye seedha $R=5$ padh lete hain. Yahi wajah hai ki $L<1$ solve karne se $R$ milta hai: boundary $L=1$ exactly $|x-a|=R$ ki boundary hai. Centre $a=4$, open interval $(-1,9)$. Step 3 — endpoints (substitute karo!):

• $x=9$: $\sum \frac{5^n}{n{5}^n}=\sum \frac{1}{n}$ — harmonic series, ek p-series jismein $p=1$: diverges.
• $x=-1$: $\sum \frac{(-5{)}^n}{n{5}^n}=\sum \frac{(-1{)}^n}{n}$ — Alternating Series Test se terms $1/n$ decrease hokar $0$ par jaate hain: converges. Answer: $R=5$, interval $\boxed{[-1,9)}$.

Ratio: $\left|\frac{(2n+2)!x^{n+1}}{(2n)!x^n}\right|=(2n+2)(2n+1)|x|.$ WHY: $\frac{(2n+2)!}{(2n)!}=(2n+2)(2n+1)$, jo $\to \infty$ hai. Kisi bhi $x\ne 0$ ke liye yeh limit $\infty >1$ hai, isliye series diverge karti hai sivaay $x=0$ ke. Answer: $R=0$; sirf $x=0$ par converge karta hai.

Ratio: $\left|\frac{a_{n+1}}{a_n}\right|=|x-1|\cdot \sqrt{\frac{n}{n+1}}.$ WHY $\sqrt{\frac{n}{n+1}}\to 1$: root ke andar $\frac{n}{n+1}=\frac{1}{1+1/n}$ hai, aur $n\to \infty$ ke saath $1/n$ shrink hokar $0$ ho jaata hai, isliye fraction $\to 1$. Kisi cheez ka square root jo $1$ ki taraf ja raha ho woh bhi $1$ ki taraf jaata hai (root function $1$ par continuous hai). Isliye ratio $\to |x-1|$ hota hai. $L<1$ se $R$ tak: $|x-1|<1$ solve karne par $R=1$ milta hai, centre $a=1$, open interval $(0,2)$. Endpoint $x=2$: $(x-1)=1$, milta hai $\sum \frac{(-1{)}^n}{\sqrt{n}}$. Terms $1/\sqrt{n}$ decrease hokar $0$ par jaate hain ⇒ Alternating Series Test se convergence. (Note: yeh absolutely convergent nahin hai, kyunki $\sum 1/\sqrt{n}$ ek $p$-series hai jismein $p=\frac{1}{2}<1$: yeh conditional hai.) Endpoint $x=0$: $(x-1)=-1$, toh $(-1{)}^n(x-1{)}^n=(-1{)}^n(-1{)}^n=1$. Series hai $\sum \frac{1}{\sqrt{n}}$ — ek p-series jismein $p=\frac{1}{2}<1$: diverges. Answer: $R=1$, interval $\boxed{(0,2]}$.

Ratio: $\left|\frac{a_{n+1}}{a_n}\right|=\frac{|x+2|}{3}\cdot \frac{n^2}{(n+1{)}^2}.$ WHY $\frac{n^2}{(n+1{)}^2}\to 1$: isse ${\left(\frac{n}{n+1}\right)}^2={\left(\frac{1}{1+1/n}\right)}^2$ likho. Kyunki $1/n\to 0$, andar wala $\to 1$, aur $1$ ko square karne par bhi $1$ milta hai. Isliye ratio $\to \frac{|x+2|}{3}$ hota hai. $L<1$ se $R$ tak: converge karta hai jab $\frac{|x+2|}{3}<1⟺|x+2|<3$: $R=3$, centre $a=-2$, open interval $(-5,1)$. Endpoint $x=1$: $(x+2)=3$, milta hai $\sum \frac{3^n}{n^2{3}^n}=\sum \frac{1}{n^2}$ — p-series $p=2>1$: converges. Endpoint $x=-5$: $(x+2)=-3$, milta hai $\sum \frac{(-3{)}^n}{n^2{3}^n}=\sum \frac{(-1{)}^n}{n^2}$ — absolutely converge karta hai: converges. Answer: $R=3$, interval $\boxed{[-5,1]}$.

KYA subtle hai: sirf even powers aate hain, isliye $x^{2n+1}$ ka coefficient $0$ hai. $c_n/c_{n+1}$ naively likhne par zero aur nonzero terms mix ho jaate hain — dangerous. Fix: Ratio Test directly actual terms par apply karo, poore $x^{2n}$ ko object maan ke. $\left|\frac{x^{2(n+1)}/4^{n+1}}{x^{2n}/4^n}\right|=\frac{|x{|}^2}{4}.$ WHY: $x^{2n+2}/x^{2n}=x^2$ aur $4^{n+1}/4^n=4$. Yahan limit $L=\frac{|x{|}^2}{4}$ mein koi $n$ nahin bacha. Test ki convergence condition $L<1$ solve karke: $\frac{|x{|}^2}{4}<1⟺|x{|}^2<4⟺|x|<2$. Toh $R=2$, centre $0$, open interval $(-2,2)$. Sanity check via Geometric Series: series $\sum (x^2/4{)}^n$ hai, geometric with ratio $r=x^2/4$; converge karta hai exactly jab $|r|<1$, yaani $|x^2/4|<1$, yaani $|x|<2$ — match karta hai. Endpoints $x=\pm 2$: ratio $r=4/4=1$, toh har ek $\sum 1=1+1+1+\cdots$ hai — terms $0$ ko nahin jaate: diverges (divergence test). Answer: $R=2$, interval $\boxed{(-2,2)}$.

Step 1 — term by term differentiate karo: $\frac{d}{dx}\frac{x^n}{n}=\frac{n{x}^{n-1}}{n}=x^{n-1}.$ Toh $f^{\prime }(x)={\sum }_{n=1}^{\infty }{x}^{n-1}=1+x+x^2+\cdots ={\sum }_{k=0}^{\infty }{x}^k.$ Step 2 — WHY radius unchanged rehta hai: yeh term-by-term differentiation theorem hai Term-by-term Differentiation and Integration se. Iska proof Root Test master formula $\frac{1}{R}={\mathrm{limsup}}_n\sqrt[n]{|c_n|}$ par tika hai. Differentiate karne par coefficient $c_n$ roughly $n{c}_n$ ban jaata hai, isliye naya radius ${\mathrm{limsup}}_n\sqrt[n]{n|c_n|}$ use karta hai. Lekin $\sqrt[n]{n}\to 1$ ($n$ ka $n$-th root $1$ ki taraf jaata hai, kyunki $n$ kisi bhi exponential $r^n$ se bahut dheere badhta hai), isliye $n$ ka extra factor $\mathrm{limsup}$ mein gayab ho jaata hai — $\frac{1}{R}$ bilkul pehle jaisa rehta hai. Yahi reason hai ki differentiation $R$ preserve karta hai. Yahan $R=1$ abhi bhi hai, aur waqai $f^{\prime }$ ek Geometric Series hai ratio $x$ ke saath: $|x|<1$ ke liye converge karta hai. Step 3 — endpoints BADAL sakte hain:

• $x=1$: $\sum 1=1+1+\cdots$ — diverges ($f$ ke liye bhi excluded tha).
• $x=-1$: $\sum (-1{)}^k=1-1+1-\cdots$ — partial sums oscillate karte hain, terms $\to 0$ nahin jaate: diverges (original $f$ mein convergent tha!). Lesson: $f$, $x=-1$ par converge karta tha lekin uska derivative $f^{\prime }$ nahin karta. Differentiation $R$ rakhta hai lekin endpoints kho sakta hai. Answer: $f^{\prime }(x)=\frac{1}{1-x}$ on $(-1,1)$; interval $\boxed{(-1,1)}$.

Ratio kyun fail karta hai: ratio $|c_{n+1}/c_n|$ ya toh $\frac{1}{3^n}$ hai (even→odd step) ya $3^{n+1}$ (odd→even step). Yeh near-$0$ aur huge ke beech oscillate karta hai — koi limit exist nahin karta, isliye $R=\lim |c_n/c_{n+1}|$ ke paas converge karne ko kuch nahin. Root kyun kaam karta hai: Root Test $\sqrt[n]{|c_n|}$ use karta hai, jo hai $\sqrt[n]{3^n}=3(n\text{ even}),\sqrt[n]{1}=1(n\text{ odd}).$ Yeh sequence sirf $3$ aur $1$ ke beech alternate karti hai — iska limit superior (woh sabse badi value jis par yeh baar baar return karta hai) $\beta ={\mathrm{limsup}}_{n\to \infty }\sqrt[n]{|c_n|}=3$ hai. Cauchy–Hadamard: $\frac{1}{R}=\beta =3\Rightarrow \boxed{R=\frac{1}{3}}.$ $\mathrm{limsup}$ kyun na ki $\lim$: $\mathrm{limsup}$ hamesha $[0,\infty ]$ mein exist karta hai, isliye yeh master tool hai; ratio limit sirf tab convenient special case hai jab cheezein settle ho jaayein.

KYA subtle hai: sirf odd powers $x^{2n+1}$ aate hain (even-power coefficients $0$ hain). L4·(a) ki tarah, hum gap bypass karte hain aur Ratio Test directly actual terms $a_n=\frac{(-1{)}^n{x}^{2n+1}}{(2n+1)!}$ par apply karte hain, $x^{2n+1}$ ko object maan ke. Ratio: $\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{x^{2n+3}/(2n+3)!}{x^{2n+1}/(2n+1)!}\right|=\frac{|x{|}^2}{(2n+3)(2n+2)}.$ WHY: $x^{2n+3}/x^{2n+1}=x^2$, aur $\frac{(2n+1)!}{(2n+3)!}=\frac{1}{(2n+3)(2n+2)}$ (do extra factorial factors denominator mein hain). Limit lo: har fixed $x$ ke liye, denominator $(2n+3)(2n+2)\to \infty$, isliye poora ratio $\to 0$. Kyunki $0<1$ chahe $x$ kuch bhi ho, Ratio Test saare real $x$ ke liye convergence deta hai. Conclusion: $R=\infty$; interval $\boxed{(-\infty ,\infty )}$. Kyunki $R=\infty$ hai isliye check karne ke liye koi finite endpoint nahin hai — series har jagah converge karti hai (jaisa hona chahiye, kyunki yeh $\sin x$ ke barabar hai jo saare $x$ ke liye defined hai).

Root values: $\sqrt[n]{|c_n|}$ perfect squares par ($n=0,1,4,9,\dots$) $\sqrt[n]{2^n}=2$ hai aur baaki jagah $\sqrt[n]{5^n}=5$ hai. Kaun si value $\mathrm{limsup}$ dominate karti hai? Perfect squares rarer aur rarer hote jaate hain, lekin unke beech value infinitely baar $5$ hoti hai. Limit superior woh sabse badi value hai jo infinitely baar hit hoti hai, isliye $\beta =5$. Cauchy–Hadamard: $\frac{1}{R}=5\Rightarrow \boxed{R=\frac{1}{5}}$, centred at $a=2$; open interval $\left(2-\frac{1}{5},2+\frac{1}{5}\right)=\left(\frac{9}{5},\frac{11}{5}\right)$.

$\sum c_n(x+3{)}^n$ ka centre? ::: $x=-3$ (base $x+3=0$ rakho). $\lim |c_n/c_{n+1}|=5$ aur centre $0$ diya, toh open interval kya hai? ::: $(-5,5)$. $|x-a|=R$ par ratio test limit $=$ kya deta hai? ::: exactly $1$ — inconclusive; endpoints alag se test karo. Agar coefficient ratio oscillate kare, toh kaun sa formula phir bhi kaam karta hai? ::: $1/R={\mathrm{limsup}}_{n\to \infty }\sqrt[n]{|c_n|}$ (Cauchy–Hadamard / Root). Kya term-by-term differentiation $R$ change karta hai? ::: Nahin — $R$ unchanged rehta hai (kyunki $\sqrt[n]{n}\to 1$), lekin endpoints re-test karne padenge. $\sum x^{2n}/4^n$ ka radius kya hai? ::: $R=2$; geometric with ratio $x^2/4$.