Exercises — Alternating series test — Leibniz test, proof

(a) Signs $+,-,+,-,\dots$ chalti hain toh haan, alternating hai. Yahan $a_n=\frac{(-1{)}^{n-1}}{n^2}$ aur $b_n=|a_n|=\frac{1}{n^2}>0$. ✓

(b) Trick yeh hai: $\cos (n\pi )$ equals $-1$ jab $n$ odd ho aur $+1$ jab $n$ even ho, yaani $\cos (n\pi )=(-1{)}^n$. Toh term hai $a_n=\frac{(-1{)}^n}{\sqrt{n}}$, jo hai alternating, jahan $b_n=|a_n|=\frac{1}{\sqrt{n}}$ (leading sign negative hai, jo theek hai — test bas yahi care karta hai ki signs alternate karein). ✓

(c) Saare terms positive hain — sign kabhi flip nahi hoti. Alternating nahi hai. Yeh converge karta hai (ek $p$-series jahan $p=2$ hai), lekin Leibniz se nahi.

$b_n=\frac{1}{n^2}$.

• Decreasing? Bada $n$ ⟹ bada $n^2$ ⟹ chota $\frac{1}{n^2}$. Decreasing lagta hai. ✓
• Limit 0? Jab $n\to \infty$, $\frac{1}{n^2}\to 0$. ✓ Dono hold karte hain, toh Leibniz apply hoga (L2 par rigorously verify karenge).

$b_n=\frac{1}{n^2}$.

• Condition 1 (decreasing): $b_{n+1}$ aur $b_n$ compare karo. $\frac{1}{(n+1{)}^2}<\frac{1}{n^2}$ kyunki $(n+1{)}^2>n^2$ for $n\ge 1$. Toh $b_{n+1}\le b_n$. ✓
• Condition 2 (limit 0): $\underset{n\to \infty }{\lim }\frac{1}{n^2}=0$. ✓

Dono hold karte hain ⟹ Leibniz test se series converges. $■$

$b_n=\frac{1}{\sqrt{n}}>0$ aur signs alternate karti hain.

• Decreasing: $\sqrt{n+1}>\sqrt{n}$ ⟹ $\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}$. ✓
• Limit 0: $\frac{1}{\sqrt{n}}\to 0$. ✓

Leibniz se Converges. (Ek baat: $\sum 1/\sqrt{n}$ ek $p$-series hai jahan $p=\frac{1}{2}\le 1$, toh yeh diverge karta hai — isliye yeh sirf conditionally convergent hai, dekho Absolute vs Conditional Convergence.)

$b_n=\frac{n}{2n+1}$. Condition 2 pehle check karo (yeh sasta kill-switch hai): ${\lim }_{n\to \infty }\frac{n}{2n+1}={\lim }_{n\to \infty }\frac{1}{2+\frac{1}{n}}=\frac{1}{2}\ne 0.$ Terms vanish nahi karti, toh nth-Term Test for Divergence se series diverges. Leibniz apply bhi nahi hota — condition 2 ek prerequisite hai, option nahi.

$b_n=\frac{\ln n}{n}$. $b_{n+1}$ aur $b_n$ ka direct comparison messy hai, toh hum woh tool use karte hain jo "kya yeh shape neeche ja rahi hai?" ke liye bana hai: $b_n$ ko ek continuous function $f(x)=\frac{\ln x}{x}$ maano aur uska derivative check karo ($f^{\prime }$ ka sign batata hai ki graph upar jaata hai ya neeche — yahi exactly "decreasing" ka matlab hai). $f^{\prime }(x)=\frac{\frac{1}{x}\cdot x-\ln x\cdot 1}{x^2}=\frac{1-\ln x}{x^2}.$ Denominator $x^2>0$ hai, toh sign $1-\ln x$ ka sign hai. Yeh negative hai jab $\ln x>1$, yaani $x>e\approx 2.718$. Toh $f$ $x\ge 3$ ke liye decreasing hai — sequence $b_n$ eventually decreasing hai ($n=3$ se). Leibniz ko sirf "sab large $n$ ke liye" chahiye, toh ✓.

• Limit 0: $\frac{\ln n}{n}\to 0$ (logarithm dhire chhalta hai; $n$ tez bhaagta hai). ✓

Dono conditions eventually hold karti hain ⟹ converges.

Nahi. Sequence $0$ ki taraf ja rahi hai lekin monotone decreasing nahi hai: jaise $b_2=\frac{1}{3}<b_3=\frac{1}{2}$, ek upar step — aur yeh infinitely baar hota hai, sirf shuru mein nahi, toh yeh "eventually decreasing" bhi nahi hai. Proof ka bracket argument require karta hai $b_{n+1}\le b_n$ taaki har $(b_{2k-1}-b_{2k})\ge 0$; iske bina brackets negative ho sakte hain aur "increasing, bounded" squeeze collapse ho jaata hai.

Toh Leibniz test apply nahi hota — yeh yahan convergence na confirm kar sakta hai na deny. (Tumhe koi alag tool chahiye hoga.) Lesson: monotonicity optional nahi hai.

Conditions hold karti hain ($b_n=1/n^3$ decreasing, $\to 0$), toh remainder bound apply hota hai: $|s-s_N|\le b_{N+1}=\frac{1}{(N+1{)}^3}.$ Hum chahte hain yeh $<0.001=\frac{1}{1000}$ ho: $\frac{1}{(N+1{)}^3}<\frac{1}{1000}⟺(N+1{)}^3>1000⟺N+1>10⟺N>9.$ Toh $N=10$ terms kaafi hain. (Compare karo alternating harmonic series se jo $0.005$ ke liye $N=200$ maangta hai — $n$ ke extra powers $1/n^3$ ko tez shrink karte hain.)

Do-step routine (dekho Absolute vs Conditional Convergence):

1. Kya yeh converge bhi karta hai? Leibniz se (L2.2), $b_n=1/\sqrt{n}$ decreasing $\to 0$ hai ⟹ converges.
2. Kya yeh absolutely converge karta hai? $\sum |a_n|=\sum \frac{1}{\sqrt{n}}$ dekho. Yeh ek $p$-series hai jahan $p=\frac{1}{2}\le 1$ ⟹ diverges.

Converge karta hai lekin absolutely nahi ⟹ conditionally convergent. Alternation ne ise bachaya.

$b_n=\frac{1}{n^p}$ jahan $p>0$.

• Convergence (Leibniz): kisi bhi $p>0$ ke liye, $n^p$ increases ⟹ $b_n$ decreases, aur $b_n=n^{-p}\to 0$. Dono conditions hold karti hain ⟹ sab $p>0$ ke liye converges.
• Absolute convergence: $\sum \frac{1}{n^p}$ ek $p$-series hai, jo converge karta hai iff $p>1$.

Summary:

• $0<p\le 1$: conditionally converges (jaise $p=1$ alternating harmonic series hai).
• $p>1$: absolutely converges.

Har $p>0$ converge karta hai; line $p=1$ conditional aur absolute ko alag karti hai.

Har endpoint substitute karo aur resulting number series padho.

• $x=1$ par: $\sum \frac{1^n}{n}=\sum \frac{1}{n}$ — harmonic series, jo diverges.
• $x=-1$ par: $\sum \frac{(-1{)}^n}{n}$ — alternating harmonic series (overall sign ke alawa). Yahan $b_n=1/n$ decreasing $\to 0$ hai, toh Leibniz se yeh converges (conditionally).

Toh interval of convergence $[-1,1)$ hai: $-1$ par closed (Leibniz ne bachaya), $+1$ par open (harmonic diverges). Yahi classic reason hai ki Leibniz power-series endpoint analysis mein aata hai.

Galti yeh hai: $b_n$ ek positive, decreasing sequence hona chahiye — lekin yahan proposed $b_n=\frac{1}{n}+\frac{(-1{)}^{n-1}}{n^2}$ mein khud ek hidden sign flip $(-1{)}^{n-1}$ hai, toh yeh monotone nahi hai (wobble karta hai). Leibniz ki hypotheses violate hoti hain; tum yeh test cite nahi kar sakte.

Lekin kya converge karta hai? Series ko split karo (legal hai kyunki hum dikhayenge ki dono pieces converge karti hain): $\sum (-1{)}^{n-1}\frac{1}{n}+\sum (-1{)}^{n-1}\cdot \frac{(-1{)}^{n-1}}{n^2}=\sum \frac{(-1{)}^{n-1}}{n}+\sum \frac{1}{n^2}.$ Pehli alternating harmonic series hai (converges, $\ln 2$ tak); doosri $\sum \frac{1}{n^2}$ hai ($p$-series jahan $p=2$ converges). Do convergent series ka sum converges. Toh series converge karti hai — bas naive Leibniz argument se nahi. Sahi jawab, galat justification — yeh phir bhi galat hai.

• L1 — alternation pehchanno, $a_n$ aur $b_n$ name karo.
• L2 — dono conditions chalao; fast kill-switch ke roop mein $\to 0$ use karo.
• L3 — derivatives se monotonicity prove karo; non-monotone failure pehchano.
• L4 — accuracy ke liye remainder bound $|s-s_N|\le b_{N+1}$; conditional vs absolute.
• L5 — parameters, power-series endpoints, aur disguised non-monotone terms.