Worked examples — Integral test — proof, p-series

4.3.7 · D3 · Maths › Calculus III — Sequences & Series › Integral test — proof, p-series

Yeh deep dive ek case-by-case gym hai integral test ke liye. Parent note ne machinery banayi; yahan hum use har kone mein le jaate hain: $p$ ki har sign, exact boundary, degenerate starting points, log-tweaked series, ek word problem, aur ek exam trap. Har answer ko steps padhne se pehle guess karo.

Shuru karne se pehle, ek reminder us tool ka jo hum sabse zyada use karte hain. Jab hum likhte hain ${\int }_1^{\infty }f(x)dx={\lim }_{b\to \infty }{\int }_1^{b}f(x)dx,$ woh limit-of-an-area ek improper integral hai — dekho Improper integrals. Yeh sawaal ka jawaab deta hai "kya curve ke neeche shaded area ka badhna ruk jaata hai?" Humne integral choose kiya — sirf terms ko ghoorte rehne ki bajaye — kyunki area kuch aisa hai jise hum antiderivatives se exactly compute kar sakte hain, jabki terms ka infinite sum nahi.

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The scenario matrix

Integral test ke har problem ko in cells mein se ek mein daala ja sakta hai. Last column mein us example ka naam hai jo ise pakka samjhaata hai.

Cell	Kya badalta hai	Kaun sa sawaal test hota hai	Example
A. $p>1$	steep decay	converges?	Ex 1
B. $p<1$	gentle decay	diverges?	Ex 2
C. $p=1$ (harmonic)	exact boundary	razor edge	Ex 3
D. $0<p<1$ fractional / negative $p$	bahut slow ya growing terms	degenerate decay	Ex 4
E. Log-modified	$\frac{1}{n(\ln n{)}^q}$	second-order boundary	Ex 5
F. Shifted start / pehle decreasing nahi	$n=N>1$ se shuru	kya start matter karta hai?	Ex 6
G. Sum ki value bound karna	trapping inequality use karo	numbers, sirf fate nahi	Ex 7
H. Word problem	real quantity	pehle model banao phir test karo	Ex 8
I. Exam twist	disguised p-series	$p$ ko pehchano	Ex 9

Ab hum har cell chalte hain. Neeche ki figure do decay speeds ko side by side dikhati hai taaki tumhein feel ho kyun cutoff wahan hai jahan hai.

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Example 1 — Cell A: steep decay, $p>1$

1. $f$ identify karo. $f(x)=x^{-3}$ set karo, toh $a_n=f(n)=1/n^3$. Yeh step kyun? Integral test ko ek continuous stand-in $f$ chahiye discrete terms ke liye.
2. 3 hypotheses check karo. $[1,\infty )$ par $f>0$; yeh continuous hai ($x\ge 1$ ke liye division by zero nahi); aur $f^{\prime }(x)=-3x^{-4}<0$, toh decreasing hai. Yeh step kyun? "Decreasing" skip karna classic mistake hai — rectangle comparison sirf ek saaf direction mein kaam karta hai jab $f$ girta hai.
3. Improper integral compute karo. ${\int }_1^{\infty }{x}^{-3}dx={\lim }_{b\to \infty }\frac{x^{-2}}{-2}{|}_1^b={\lim }_{b\to \infty }(\frac{-1}{2b^2}+\frac{1}{2})=\frac{1}{2}.$ Yeh step kyun? Yahan $p=3>1$ exponent $1-p=-2$ ko negative banata hai, isliye $b^{1-p}\to 0$ — area badhna band ho jaata hai.
4. Conclude karo. Finite integral ⇒ converges.

Verify: general rule deta hai ${\int }_1^{\infty }{x}^{-p}=\frac{1}{p-1}=\frac{1}{3-1}=\frac{1}{2}$. ✓ Match karta hai.

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Example 2 — Cell B: gentle decay, $p<1$

1. Power ke roop mein rewrite karo. $\frac{1}{\sqrt{n}}=n^{-1/2}$, toh $p=\frac{1}{2}$. Yeh step kyun? $p$ ko pehchaanna series ko turant classify kar deta hai.
2. $f(x)=x^{-1/2}$ integrate karo. ${\int }_1^{\infty }{x}^{-1/2}dx={\lim }_{b\to \infty }2\sqrt{x}{|}_1^b={\lim }_{b\to \infty }(2\sqrt{b}-2)=\infty .$ Yeh step kyun? $1-p=\frac{1}{2}>0$, toh $b^{1-p}\to \infty$ — area kabhi nahi rukta.
3. Conclude karo. Divergent integral ⇒ diverges, chahe $a_n\to 0$ hi kyun na ho.

Verify: $p=0.5\le 1$, toh p-series rule divergence predict karta hai. ✓ Aur yeh tempting "terms $\to 0$" wali galti ko contradict karta hai — dekho n-th term test for divergence (yeh sirf divergence prove kar sakta hai, convergence nahi).

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Example 3 — Cell C: exact boundary, harmonic series

1. Note karo $p=1$. Antiderivative ke liye power rule ($\frac{x^{1-p}}{1-p}$) yahan break ho jaata hai kyunki $1-p=0$ se division by zero hota hai. Yeh step kyun? $p=1$ genuinely ek alag case hai; $1/x$ ka antiderivative $\ln x$ hai, koi power nahi.
2. Log ke saath integrate karo. ${\int }_1^{\infty }\frac{dx}{x}={\lim }_{b\to \infty }\ln b=\infty .$ Logarithm kyun? Kyunki $\int x^{-1}dx=\ln x$ — woh ek power jiska antiderivative log hai. Yahi unique reason hai ki $p=1$ converge karne ki bajaye diverge karta hai.
3. Conclude karo. Diverges — yeh Harmonic series hai.

Verify: $\ln b\to \infty$, aur $p=1\ngtr 1$. ✓ Dono agree karte hain.

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Example 4 — Cell D: degenerate decay ($0<p<1$ fractional, aur $p\le 0$)

1. (a) Classify karo. $p=0.9<1$ ⇒ rule ke hisaab se, diverges. Yeh step kyun? Chahe $1$ se kitna bhi neeche se ho, $1-p>0$ phir bhi $b^{1-p}\to \infty$ bhejta hai.
2. (a) Integral check. ${\int }_1^{\infty }{x}^{-0.9}dx=\frac{x^{0.1}}{0.1}{|}_1^{\infty }=\infty$. Divergence confirm hoti hai.
3. (b) Degenerate case $p=-2$. Toh $a_n=1/n^{-2}=n^2\to \infty$, toh terms $0$ tak bhi nahi pahunchte. Yeh step kyun? Integral test ko decreasing $f$ chahiye. Yahan $f(x)=x^2$ increase karta hai — test ki hypotheses fail ho jaati hain, toh hum iske paas jaate hi nahi.
4. (b) Sasta tool use karo. n-th term test for divergence se: $a_n=n^2\nrightarrow 0$ ⇒ turant diverges.

Verify: (a) $p=0.9\le 1$ ⇒ diverges. ✓ (b) ${\lim }_{n\to \infty }{n}^2=\infty \ne 0$ ⇒ term test se diverges. ✓

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Example 5 — Cell E: second-order boundary $\frac{1}{n(\ln n{)}^q}$

1. $f$ set karo aur $n=2$ se shuru karo. $f(x)=\frac{1}{x(\ln x{)}^q}$; hum $2$ se shuru karte hain kyunki $\ln 1=0$ division by zero dega. Yeh step kyun? Convergence ek finite front chunk ki kabhi parwaah nahi karta, toh start ko $2$ shift karna free hai.
2. Substitute karo $u=\ln x$, $du=dx/x$. ${\int }_2^{\infty }\frac{dx}{x(\ln x{)}^q}={\int }_{\ln 2}^{\infty }\frac{du}{u^q}.$ Substitution kyun? Yeh messy integrand ko $u$ mein plain p-integral mein convert kar deta hai — poora problem collapse ho jaata hai "kya $\int u^{-q}du$ finite hai?" par.
3. $u$ mein p-integral rule apply karo. ${\int }^{\infty }{u}^{-q}du$ converge karta hai $⟺q>1$. Yeh step kyun? Humne exactly yahi p-series section mein prove kiya tha — ab variable $u$ par reuse kar rahe hain.
4. Conclude karo. Converges $⟺q>1$. $q=1$ ke liye yeh $\sum \frac{1}{n\ln n}$ divergence hai, parent note se.

Verify: $q=2$ par: ${\int }_{\ln 2}^{\infty }{u}^{-2}du=\frac{1}{\ln 2}$ (finite) ⇒ converges. $q=1$ par: $\int u^{-1}du=\infty$ ⇒ diverges. ✓ Numerically $\frac{1}{\ln 2}\approx 1.4427$.

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Example 6 — Cell F: shifted start / pehle decreasing nahi

1. $n=1$ ko alag handle karo. $n=1$ par, $\ln 1=0$, toh woh term $0$ hai; harmless. Yeh step kyun? Ek finite term kabhi convergence nahi badalta.
2. Eventual decrease check karo. $f(x)=\frac{\ln x}{x^2}$, aur $f^{\prime }(x)=\frac{1-2\ln x}{x^3}<0⟺\ln x>\frac{1}{2}⟺x>e^{1/2}\approx \mathrm{1.65.}$ Toh $f$, $x\ge 2$ ke liye decrease karta hai. Yeh step kyun? Integral test ko sirf $f$ ka eventually decreasing hona chahiye — test $n=2$ se shuru karo.
3. Integration by parts se integrate karo, $u=\ln x,dv=x^{-2}dx$: $\int \frac{\ln x}{x^2}dx=-\frac{\ln x}{x}-\frac{1}{x}+C.$ Yeh step kyun? Log aur power ke products integration by parts maangte hain.
4. $2$ se $\infty$ tak limit lo. Jab $x\to \infty$, $\frac{\ln x}{x}\to 0$ aur $\frac{1}{x}\to 0$, toh ${\int }_2^{\infty }\frac{\ln x}{x^2}dx=[0-0]-(-\frac{\ln 2}{2}-\frac{1}{2})=\frac{\ln 2}{2}+\frac{1}{2},$ ek finite number ⇒ converges.

Verify: $\frac{\ln 2}{2}+\frac{1}{2}\approx 0.8466$ (finite). ✓ Cutoff $x=e^{1/2}$ satisfy karta hai $\ln (e^{1/2})=\frac{1}{2}$. ✓

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Example 7 — Cell G: trapping inequality se value bound karna

1. Tail-trap yaad karo. Ek convergent decreasing $f$ ke liye, remainder $R_N={\sum }_{n=N+1}^{\infty }{a}_n$ obey karta hai ${\int }_{N+1}^{\infty }f\le R_N\le {\int }_N^{\infty }f.$ Yeh step kyun? Wahi rectangle logic hai jaisa parent proof mein tha, $N$ terms ke baad baake par apply kiya.
2. $f=x^{-2}$, $N=10$ plug karo. ${\int }_N^{\infty }{x}^{-2}dx=\frac{1}{N}$, toh $\frac{1}{11}\le R_{10}\le \frac{1}{10}.$ Yeh step kyun? ${\int }_a^{\infty }{x}^{-2}=1/a$ — ek clean closed form.
3. Known partial sum $S_{10}={\sum }_{n=1}^{10}\frac{1}{n^2}$ add karo. Toh $S=S_{10}+R_{10}$ deta hai $S_{10}+\frac{1}{11}\le S\le S_{10}+\frac{1}{10}.$ Yeh step kyun? Yahi payoff hai — integral test trap karta hai lekin sum ke barabar kabhi nahi hota (yeh doosri classic mistake hai).

Verify: $S_{10}\approx 1.549768$, toh bounds hain $[1.640677,1.649768]$; aur sach mein ${\pi }^2/6\approx 1.644934$ andar hai. ✓

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Example 8 — Cell H: word problem

1. Model banao. Total ink $={\sum }_{n=1}^{\infty }\frac{1}{n^{1.5}}$ mL — ek p-series jahan $p=1.5$ hai. Yeh step kyun? Test karne se pehle physical quantity ko series mein translate karo.
2. Classify karo. $p=1.5>1$ ⇒ converges: total ink ek finite number of mL hai.
3. Bound karo (kitna?). Tail argument se, $S\le a_1+{\int }_1^{\infty }{x}^{-1.5}dx=1+\frac{1}{1.5-1}=1+2=3$ mL. Yeh step kyun? Abstract "finite" ko ek usable units-wale upper bound of $3$ mL mein convert karta hai.

Verify: ${\int }_1^{\infty }{x}^{-1.5}dx=\frac{1}{1.5-1}=2$, toh crude bound $1+2=3$ mL holds. (True value $\zeta (1.5)\approx 2.612$ mL — dekho Riemann zeta function — jo $3$ se kam hai.) ✓

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Example 9 — Cell I: exam twist (disguised p-series)

1. Dominant behaviour dhundho. Large $n$ ke liye, $\frac{n+3}{n^3+n}\approx \frac{n}{n^3}=\frac{1}{n^2}$. Yeh step kyun? Leading powers dominate karte hain; yeh hint deta hai $p=2$.
2. Limit comparison test se rigorously confirm karo. $b_n=1/n^2$ se compare karo: ${\lim }_{n\to \infty }\frac{(n+3)/(n^3+n)}{1/n^2}={\lim }_{n\to \infty }\frac{n^2(n+3)}{n^3+n}={\lim }_{n\to \infty }\frac{n^3+3n^2}{n^3+n}=1.$ Yeh step kyun? Finite, nonzero limit ka matlab hai dono series ka same fate hoga — directly rational function integrate karne se zyada clean.
3. Conclude karo. Kyunki $\sum 1/n^2$ converge karta hai ($p=2>1$), toh given series bhi karta hai. Converges.

Verify: limit-comparison ratio limit $1$ (finite, nonzero) ke barabar hai, aur yardstick $p=2>1$ converge karta hai. ✓

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Active recall

Recall Kaun sa cell, kaun sa tool?

• Terms grow karte hain ($p\le 0$) ::: n-th term test — ek line, diverges.
• Denominator mein log ::: $u=\ln x$ substitute karo, $u$ mein p-integral mein reduce ho jaata hai.
• $n$ ka rational function ::: limit comparison with $1/n^{(\mathrm{deg}\text{denom}-\mathrm{deg}\text{numer})}$.
• Actual number bound chahiye ::: trapping / tail inequality, kabhi equality nahi.

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Connections

• Integral test — proof, p-series — parent note: proof aur p-series rule.
• Comparison test — kisi messy series ko p-series se upar/neeche bound karo.
• Limit comparison test — Ex 9 mein rational function ko tame karne ke liye use kiya.
• Improper integrals — har example ke andar ka limit-of-area engine.
• Harmonic series — Ex 3, exact boundary.
• Riemann zeta function — woh true values jo humne Ex 7 aur Ex 8 mein bound kiye.
• n-th term test for divergence — Ex 4 mein fast filter.

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