Exercises — Integral test — proof, p-series

Rule yeh hai: $\sum 1/n^p$ converges $⟺p>1$.

• (a) $\frac{1}{n^3}=\frac{1}{n^3}$, to $p=3>1$ ⇒ converges.
• (b) $\frac{1}{\sqrt{n}}=\frac{1}{n^{1/2}}$, to $p=\frac{1}{2}<1$ ⇒ diverges.
• (c) $\frac{1}{n}=\frac{1}{n^1}$, to $p=1$ ⇒ diverges (yeh Harmonic series hai, exact boundary case).

Page ke top par restate ki gayi teen hypotheses check karo.

• (a) Positive ✓, continuous ✓, aur $f^{\prime }(x)=-2x^{-3}<0$ to decreasing ✓ ⇒ legal.
• (b) $\sin x$ $-1$ aur $1$ ke beech swing karta hai: na positive, na decreasing ⇒ illegal (areas cancel ho jaate).
• (c) $\frac{1}{x}-2$, $x>\frac{1}{2}$ ke liye negative hai ⇒ $[1,\infty )$ par positive nahi ⇒ illegal.

WHAT: $f(x)=\frac{1}{x^2+1}$ lo. WHY legal (teenon hypotheses): positive ✓, continuous ✓; $f^{\prime }(x)=\frac{-2x}{(x^2+1{)}^2}<0$ for $x\ge 1$ to decreasing ✓. ${\int }_1^{\infty }\frac{dx}{x^2+1}={\lim }_{b\to \infty }[\arctan x{]}_1^b={\lim }_{b\to \infty }(\arctan b-\arctan 1)=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}.$ WHY $\arctan$? $\frac{1}{x^2+1}$ ka antiderivative $\arctan x$ hai — yeh woh ek integral hai jiska value ek angle hai, aur $\arctan b\to \frac{\pi }{2}$ as $b\to \infty$. Integral finite hai ($\pi /4\approx 0.785$) ⇒ series converges.

WHAT: $f(x)=x{e}^{-x^2}$, $[1,\infty )$ par positive hai; $f^{\prime }(x)=e^{-x^2}(1-2x^2)<0$ for $x\ge 1$ ⇒ decreasing ✓. WHY substitution $u=x^2$: aage wala factor $x$ exactly $\frac{1}{2}du$ hai jahan $du=2xdx$ — yeh pure integrand ko $e^{-u}$ mein collapse kar deta hai. ${\int }_1^{\infty }x{e}^{-x^2}dx\stackrel{u=x^2}{=}\frac{1}{2}{\int }_1^{\infty }{e}^{-u}du=\frac{1}{2}[-e^{-u}{]}_1^{\infty }=\frac{1}{2}{e}^{-1}=\frac{1}{2e}.$ Finite ($\approx 0.184$) ⇒ series converges.

KAHAN decrease karta hai: $f^{\prime }(x)=\frac{1-\ln x}{x^2}$. Yeh $<0$ tab hai jab $\ln x>1$, yaani $x>e\approx 2.718$. To $f$ $x=e$ tak increase karta hai, phir decrease karta hai.

Figure s02 mein kya dekhna hai: blue curve $y=\ln x/x$ $x=1$ se climb karta hai, yellow dashed line $x=e$ par ek single peak (red dot) par pahunchta hai, phir hamesha ke liye girta hai. Peak ke left mein yeh rise karta hai (green label), isliye integral-test rectangle comparison wahan ulti direction mein jaata; peak ke right mein yeh fall karta hai (red label), jahaan test legal hai.

WHY yeh theek hai: integral test ko sirf eventually decreasing $f$ chahiye — $N=3$ se start karo (tail hi convergence decide karta hai; shuru ke finite terms fate nahi badal sakte). ${\int }_3^{\infty }\frac{\ln x}{x}dx\stackrel{u=\ln x}{=}{\int }_{\ln 3}^{\infty }udu={\lim }_{b\to \infty }\frac{u^2}{2}{|}_{\ln 3}^b=\infty .$ WHY $u=\ln x$: $du=dx/x$ $\frac{1}{x}$ ko absorb kar leta hai, sirf pure $u$ bachta hai, jo bina bound ke grow karta hai. Integral diverge karta hai ⇒ series diverges.

WHAT: $f(x)=\frac{1}{x(\ln x{)}^p}$ on $[2,\infty )$. WHY positive: $x\ge 2$ ke liye, $\ln x>0$ to denominator ke dono factors positive hain ✓. WHY continuous: yeh continuous functions ka quotient hai jiska denominator $[2,\infty )$ par kabhi zero nahi hota ✓. WHY decreasing: $g(x)=x(\ln x{)}^p$ (denominator) likho; phir $g^{\prime }(x)=(\ln x{)}^p+x\cdot p(\ln x{)}^{p-1}\cdot \frac{1}{x}=(\ln x{)}^{p-1}(\ln x+p)>0\text{for }x\ge 2,$ kyunki $\ln x>0$ aur $\ln x+p>0$. Ek growing positive denominator ka matlab hai $f=1/g$ decreasing hai ✓. WHY $u=\ln x$: $du=dx/x$ integrand ko $u$ mein pure p-integral bana deta hai. ${\int }_2^{\infty }\frac{dx}{x(\ln x{)}^p}\stackrel{u=\ln x}{=}{\int }_{\ln 2}^{\infty }\frac{du}{u^p}.$ Ab yeh $u$ mein ek p-integral hai, jiska convergence hum jaante hain:

• $p>1$: $\int u^{-p}du$ converges ⇒ series converges.
• $p\le 1$: diverges ⇒ series diverges. To answer wohi same razor cutoff hai jaise p-series: convergence $⟺p>1$. ($p=1$ par hum parent ke Example 3 par wapas aate hain, jo diverge karta hai.)

WHY integrate nahi karte: $\frac{1}{\sqrt{x^3+1}}$ ka koi clean antiderivative nahi hai. Iski jagah ek p-series se compare karo jiska fate hum integral test se jaante hain. Sabhi $n\ge 1$ ke liye: $\sqrt{n^3+1}>\sqrt{n^3}=n^{3/2}$, isliye $\frac{1}{\sqrt{n^3+1}}<\frac{1}{n^{3/2}}.$ WHY yeh bound help karta hai: $\sum \frac{1}{n^{3/2}}$ ek p-series hai jisme $p=\frac{3}{2}>1$, jise integral test converges prove karta hai. Ek convergent series ke neeche chhota positive series bhi converge karta hai (Comparison test). ⇒ converges. (Cross-check Limit comparison test se: $\lim \frac{1/\sqrt{n^3+1}}{1/n^{3/2}}=1$, ek finite nonzero limit, isliye same fate.)

Trapping inequality yaad karo jisme $S_N={\sum }_{n=1}^{N}f(n)$ $N$-term partial sum hai: ${\int }_1^{N+1}f\le S_N\le f(1)+{\int }_1^{N}f.$ Yahan $f(x)=x^{-2}$, $f(1)=1$, aur ${\int }_1^{\infty }{x}^{-2}dx=1$ (parent Example 1). $N\to \infty$ lo to $S_N\to S$: $\underset{=1}{\underbrace{{\int }_1^{\infty }\frac{dx}{x^2}}}\le S\le \underset{=1}{\underbrace{f(1)}}+\underset{=1}{\underbrace{{\int }_1^{\infty }\frac{dx}{x^2}}}.$ To $1\le S\le 2$. WHY yahi test ka point hai: yeh sum ko evaluate kiye bina bound karta hai. True value ${\pi }^2/6\approx 1.645$ hai, jo wakai $[1,2]$ mein aata hai — kabhi bhi kisi integral ke equal nahi (yeh classic mistake hai). Tighten karne ke liye: tail ${\sum }_{n=2}^{\infty }$ ko ${\int }_1^{\infty }{x}^{-2}=1$ se bound karo, jo $S\le 1+1=2$ aur $S\ge 1+{\int }_2^{\infty }{x}^{-2}=1+\frac{1}{2}=1.5$ deta hai.

Idea (same rectangle geometry jaise page ke top par, tail par shift ki gayi). $[n,n+1]$ par, decreasing $f$ deta hai $f(n+1)\le {\int }_n^{n+1}f\le f(n).$

Figure s03 mein kya dekhna hai: blue curve $y=1/x^2$ tail $x\ge 3$ par hai. Red dashed rectangles left-endpoint height $f(n)$ use karte hain aur curve ke upar nikalte hain (unka sum overshoot karta hai ⇒ upper bound); green rectangles right-endpoint height $f(n+1)$ use karte hain aur curve ke neeche baithe hain (unka sum undershoot karta hai ⇒ lower bound). Remainder $R_N$ dono shaded areas ke beech trap hai.

$R_N$ ke liye upper bound: left inequality $f(n+1)\le {\int }_n^{n+1}f$ use karke, $n=N,N+1,\dots$ sum karo: $R_N={\sum }_{n=N+1}^{\infty }f(n)={\sum }_{k=N}^{\infty }f(k+1)\le {\sum }_{k=N}^{\infty }{\int }_k^{k+1}f={\int }_N^{\infty }f.$ $R_N$ ke liye lower bound: right inequality ${\int }_n^{n+1}f\le f(n)$ use karke, $n=N+1,N+2,\dots$ sum karo: ${\int }_{N+1}^{\infty }f={\sum }_{n=N+1}^{\infty }{\int }_n^{n+1}f\le {\sum }_{n=N+1}^{\infty }f(n)=R_N.$ Combine karo: ${\int }_{N+1}^{\infty }f\le R_N\le {\int }_N^{\infty }f$. $■$

$\sum 1/n^2$ par apply karo, $N=3$: ${\int }_N^{\infty }{x}^{-2}dx=\frac{1}{N}$, isliye $\frac{1}{4}\le R_3\le \frac{1}{3},\text{i.e. }0.25\le R_3\le 0.333\dots$ (True remainder $R_3={\pi }^2/6-1-\frac{1}{4}-\frac{1}{9}\approx 0.2838$ — band ke andar hai.)

Partial sum $S_3=1+\frac{1}{4}+\frac{1}{9}=\frac{49}{36}\approx 1.3611$. Problem 5.1 se, $\frac{1}{4}\le R_3\le \frac{1}{3}$. Isliye $S_3+\frac{1}{4}\le S\le S_3+\frac{1}{3},1.6111\le S\le \mathrm{1.6944.}$ Midpoint estimate: $S\approx \frac{1.6111+1.6944}{2}=1.6528$, maximum error ke saath $\frac{1.6944-1.6111}{2}\approx 0.0417$. True value ${\pi }^2/6\approx 1.6449$ $[1.6111,1.6944]$ ke andar hai aur hamare midpoint se $0.008$ ke andar — Problem 4.2 ke crude $[1,2]$ se kaafi better. WHY better hai: humne kuch exact terms compute kiye aur sirf tail ko integrals se estimate kiya, jahaan rectangle error chota hota hai.

• $N$-term partial sum $S_N$ define karo. ::: $S_N={\sum }_{n=1}^{N}f(n)$, pehle $N$ terms ka running total.
• Trapping inequality state karo. ::: ${\int }_1^{N+1}f\le S_N\le f(1)+{\int }_1^{N}f$.
• $\frac{1}{2n}$ ke liye $p$ aur uska fate? ::: $p=1$, diverges (constant $\frac{1}{2}$ irrelevant hai).
• $\sum \frac{1}{n^2+1}$ ke liye kaun sa antiderivative use hua? ::: $\arctan x$; integral $=\pi /4$, converges.
• $\frac{\ln x}{x}$ kahan se decreasing start hota hai? ::: $x>e$ ke liye; test $N=3$ se karo.
• $\sum \frac{1}{n(\ln n{)}^p}$ ka cutoff? ::: Converges $⟺p>1$.
• Convergent integral-test series ke liye remainder bound? ::: ${\int }_{N+1}^{\infty }f\le R_N\le {\int }_N^{\infty }f$.