Exercises — Absolute vs conditional convergence

Step — signs hatao. $|a_n|=\frac{1}{n^3}$, toh $\sum |a_n|=\sum \frac{1}{n^3}$. Kyun: absolute convergence poori tarah positive series $\sum |a_n|$ se decide hoti hai.

Step — type pehchano. Yeh ek p-series hai jisme $p=3$. Ek $p$-series tab converge karti hai jab $p>1$. Yahan $p=3>1$, toh $\sum \frac{1}{n^3}$ converge karta hai.

Verdict: kyunki $\sum |a_n|$ converge karta hai, series absolutely convergent hai. Alternating signs irrelevant hain.

Step — absolute test. $|a_n|=\frac{1}{\sqrt{n}}=\frac{1}{n^{1/2}}$, ek $p$-series jisme $p=\frac{1}{2}\le 1$ → diverges. Toh yeh absolutely convergent nahi hai.

Step — $\sum a_n$ ko khud test karo Alternating Series Test (AST) se. Magnitudes $b_n=\frac{1}{\sqrt{n}}$ hain (i) positive, (ii) decreasing, (iii) $\to 0$. Teeno conditions satisfy hoti hain, toh $\sum a_n$ converge karta hai.

Verdict: converge karta hai par absolutely nahi → conditionally convergent.

Step — sabse sasta test pehle: kya terms zero ki taraf jaate hain? ${\lim }_{n\to \infty }\left|(-1{)}^n\frac{n}{n+1}\right|={\lim }_{n\to \infty }\frac{n}{n+1}=1\ne 0.$ Ye test pehle kyun: agar terms $0$ ki taraf nahi shrink hote, toh koi bhi series converge nahi kar sakti — yeh Term Test for Divergence hai, sabse tezi se kaam aane wala tool.

Verdict: $a_n\nrightarrow 0$, toh $\sum a_n$ diverges. (Na absolute na conditional apply hota hai — yeh simply convergent nahi hai.)

Step — tool chuno. Denominator mein $2^n$ jaise factorials ya powers Ratio Test ki demand karte hain, kyunki consecutive terms ka ratio powers ko cleanly collapse kar deta hai. Hum ise $|a_n|=\frac{n}{2^n}$ pe apply karte hain.

Step — ratio compute karo. $L={\lim }_{n\to \infty }\frac{|a_{n+1}|}{|a_n|}={\lim }_{n\to \infty }\frac{(n+1)/2^{n+1}}{n/2^n}={\lim }_{n\to \infty }\frac{n+1}{2n}=\frac{1}{2}.$ Kyun: $2^n$ cancel ho ke sirf ek factor $\frac{1}{2}$ bachta hai; polynomial ratio $\frac{n+1}{n}\to 1$.

Step — verdict padho. $L=\frac{1}{2}<1$, aur $|a_n|$ pe apply kiya gaya ratio test seedha absolute convergence decide karta hai.

Verdict: absolutely convergent.

Step — absolute test. $|a_n|=\frac{1}{n\ln n}$. Yeh textbook integral-test series hai: ${\int }_2^{\infty }\frac{dx}{x\ln x}=[\ln (\ln x){]}_2^{\infty }=\infty$, toh $\sum \frac{1}{n\ln n}$ diverges. Absolutely convergent nahi hai. Integral test kyun: $\frac{1}{x\ln x}$ positive aur decreasing hai, aur iska antiderivative $\ln (\ln x)$ known hai — integral sum ko ek aisi growth mein convert karta hai jise hum padh sakte hain.

Step — $\sum a_n$ ko test karo AST se. Magnitudes $\frac{1}{n\ln n}$ positive hain, decreasing hain (dono $n$ aur $\ln n$ badhte hain), aur $\to 0$ hain. AST → converge karta hai.

Verdict: conditionally convergent.

Step — comparison se absolute test. $|a_n|=\frac{1}{n^2+1}\le \frac{1}{n^2}$ for all $n\ge 1$. Comparison Test kyun: $\frac{1}{n^2+1}$ ka koi useful closed antiderivative nahi hai, par yeh ek known convergent $p$-series $\sum \frac{1}{n^2}$ ($p=2>1$) se dominate hota hai.

Step — conclude karo. Kyunki $0\le |a_n|\le \frac{1}{n^2}$ aur $\sum \frac{1}{n^2}$ converge karta hai, $\sum |a_n|$ converge karta hai.

Verdict: absolutely convergent.

Step — convergence type identify karo. $\sum \frac{1}{n}$ (Harmonic series) diverge karta hai, jabki alternating version converge karta hai. Toh yeh conditionally convergent hai.

Step — theorem invoke karo. Riemann Rearrangement Theorem kehta hai: ek conditionally convergent series ke liye, rearranging se koi bhi real value produce ki ja sakti hai. Toh koi order-independent "true value" nahi hai — dono students ne legitimately apni ordering ka sum compute kiya. Aisa kyun hota hai: positive-only part $\sum \frac{1}{2k-1}$ aur negative-only part $\sum \frac{-1}{2k}$ dono diverge karte hain ($+\infty$ aur $-\infty$ ki taraf). Rearranging redistribute karta hai ki doosre reservoir ki taraf jaane se pehle har infinite reservoir se kitna draw karo, running total ko kahin bhi steer karta hai.

Step — Student B ki value. "Ek positive, do negatives" pattern woh classic rearrangement hai jiska sum $\frac{1}{2}\ln 2$ hai. Sketch: group as $\left(1-\frac{1}{2}\right)-\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{6}\right)-\frac{1}{8}+\cdots =\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\cdots =\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{3}-\cdots \right)=\frac{1}{2}\ln 2.$

Verdict: Student A ko $\ln 2\approx 0.693$ milta hai; Student B ko $\frac{1}{2}\ln 2\approx 0.347$ milta hai. Dono apni ordering ke liye sahi hain — yahi conditional convergence ka poora lesson hai.

Step — general claim decide karo: yeh SAHI hai. Maano $\sum a_n$ converge karta hai aur $\sum a_n^2$ diverge karta hai. Hum dikhate hain ki $\sum a_n$ absolutely convergent nahi ho sakta, jo use conditional force karta hai.

Step — key comparison. Kyunki $\sum a_n$ converge karta hai, iske terms $\to 0$ hain, toh eventually $|a_n|<1$. Un $n$ ke liye, $a_n^2=|a_n{|}^2\le |a_n|.$ Yeh inequality kyun: $1$ se chhoti kisi number ko square karne se woh aur chhoti ho jaati hai, toh $a_n^2$, $|a_n|$ ke neeche hota hai.

Step — Comparison Test se contrapositive. Agar $\sum |a_n|$ converge karta, toh comparison se $0\le a_n^2\le |a_n|$ force karta ki $\sum a_n^2$ bhi converge kare. Par hume bataya gaya hai ki $\sum a_n^2$ diverge karta hai — contradiction. Isliye $\sum |a_n|$ diverge karta hai.

Step — conclude karo. $\sum a_n$ converge karta hai par $\sum |a_n|$ diverge karta hai → conditionally convergent. Yeh implication SAHI hai.

Step — concrete case check karo. $a_n=\frac{(-1{)}^n}{\sqrt{n}}$ ke liye: $\sum a_n$ AST se converge karta hai (Problem 1.2), aur $a_n^2=\frac{1}{n}$ se $\sum a_n^2=\sum \frac{1}{n}$ milta hai jo diverge karta hai. Dono premises hold karte hain, aur yeh series conditionally convergent hai (Problem 1.2 verdict) — exactly jaisa theorem predict karta hai. ✓

Takeaway: yahan $\sum a_n^2$ ka diverge karna non-absolute convergence certify karne ka ek valid shortcut hai — par sirf tab jab tum pehle se jaante ho ki $\sum a_n$ khud converge karta hai.

Step — absolute behaviour. $\sum |a_n|=\sum \frac{1}{n^p}$ ek $p$-series hai: converge karta hai iff $p>1$.

• Toh $p>1$ ke liye: absolutely convergent.

Step — jab absolute fail ho ($p\le 1$), $\sum a_n$ ko AST se test karo. Magnitudes $\frac{1}{n^p}$ tab $0$ ki taraf decrease hote hain jab $p>0$.

• $0<p\le 1$ ke liye: AST apply hota hai (decreasing $\to 0$), toh $\sum a_n$ converge karta hai, par $\sum |a_n|$ diverge karta hai → conditionally convergent.

Step — $p\le 0$ wala tail. Agar $p\le 0$ toh $\frac{1}{n^p}=n^{-p}\nrightarrow 0$ ($\ge 1$ rehta hai ya badhta hai). Term Test → divergent.

Full answer:

• $p>1$: absolutely convergent.
• $0<p\le 1$: conditionally convergent.
• $p\le 0$: divergent.

Figure 2 ise ek number line ke roop mein dikhata hai taaki teeno regions aur unki boundaries ek nazar mein dikhen: divergent stretch $p\le 0$ (white, terms vanish nahi hote), conditional band $0<p\le 1$ (amber), aur absolute ray $p>1$ (cyan). Endpoint markers note karo — $p=1$ conditional mein aata hai, jabki $p=0$ isse excluded hai (divergent).

Step — $|a_n|$ pe Ratio Test se radius. $a_n=\frac{(x-3{)}^n}{n}$ ke saath, $L={\lim }_{n\to \infty }\left|\frac{(x-3{)}^{n+1}/(n+1)}{(x-3{)}^n/n}\right|=|x-3|{\lim }_{n\to \infty }\frac{n}{n+1}=|x-3|.$ Absolute convergence ke liye $L<1$ chahiye: $|x-3|<1$, yaani $2<x<4$. Radius $R=1$, centre $3$.

Step — right endpoint $x=4$. Series ban jaati hai $\sum \frac{1^n}{n}=\sum \frac{1}{n}$ = harmonic → diverges.

Step — left endpoint $x=2$. Series ban jaati hai $\sum \frac{(-1{)}^n}{n}$ → AST se converge karta hai, par $\sum \frac{1}{n}$ diverge karta hai → conditionally convergent.

Full answer:

• $2<x<4$: absolutely convergent.
• $x=2$: conditionally convergent.
• $x=4$: divergent.
• Warna ($x<2$ ya $x>4$): divergent. Interval of convergence: $[2,4)$.

(a) — mechanism. $S$ ko uske positive terms $P=1+\frac{1}{3}+\frac{1}{5}+\cdots$ aur negatives ke (magnitudes) $N=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots$ mein split karo. Har ek ek scaled Harmonic series hai: $P\ge \frac{1}{2}\sum \frac{1}{n}=\infty ,N=\frac{1}{2}\sum \frac{1}{n}=\infty .$ Yeh kyun matter karta hai: tumhare paas do infinite reservoirs hain, ek $+$'s ka aur ek $-$'s ka. Target $T$ reach karne ke liye: positive terms tab tak add karo jab tak running total pehli baar $T$ se zyada na ho jaaye, phir negatives add karo jab tak $T$ se neeche na aa jaaye, phir positives, aur aisa karte raho. Kyunki har reservoir infinite hai tum kabhit khatam nahi hote, aur kyunki individual terms $\to 0$ hain har baar overshoot $0$ ki taraf shrink karta hai — toh running total exactly $T$ pe converge karta hai. Yahi Riemann Rearrangement Theorem ke peeche constructive proof hai.

(b) — AST error bound. Ek alternating series ke liye jisme decreasing magnitudes $b_n\to 0$ hain, $N$ terms ke baad error at most first omitted term hoti hai: $|S-S_N|\le b_{N+1}=\frac{1}{N+1}$. Kyun: consecutive partial sums limit ko straddle karte hain, toh gap agla step se chhota hota hai.

Chahiye $\frac{1}{N+1}\le 0.01\Rightarrow N+1\ge 100\Rightarrow N\ge 99$.

Answer: $N=99$ terms guarantee karte hain ki partial sum $\ln 2$ ke $0.01$ ke andar ho. (Contrast karo absolutely convergent $\sum \frac{1}{n^2}$-type series se, jiske tails kahin zyada tezi se shrink hote hain — yeh ek vivid measure hai ki conditional convergence kitni dheere settle hoti hai.)

Pehla sawaal jo hamesha poochhte ho ::: Kya $\sum |a_n|$ converge karta hai? Agar haan ::: Absolutely convergent — ruko, koi doosra test nahi. Agar nahi ::: $\sum a_n$ khud test karo (usually AST): converge kare → conditional; diverge kare → divergent. Power series ke endpoints ::: Hamesha haath se substitute aur test karo (Ratio Test $L=1$ pe khamosh hota hai).