Integral test — proof, p-series: Visual walkthrough

4.3.7 · D2 · Maths › Calculus III — Sequences & Series › Integral test — proof, p-series

Is parent result ka ek visual walkthrough: the integral test. Hum poora proof sirf rectangles aur ek girte hue curve se banate hain — kuch aur assume nahi kiya. Agar aapne pehle kabhi integral sign nahi dekhi, toh Step 1 se shuru karo.

Step 1 — Sum ko rectangles ki ek wall mein badlo

KYA. Har term $a_{n}$ ek rectangle ban jaata hai: width $1$ , height $a_{n} = f (n)$ . Inhe side by side line up karo. Total shaded area sum ke barabar hoti hai $S_{N} = a_{1} + a_{2} + \dots + a_{N}$ .

KYUN. Area ek aisi cheez hai jo hum dekh sakte hain aur bound kar sakte hain. Infinitely many numbers ka raw sum invisible hota hai; rectangles ki wall nahi hoti. Agar hum wall ki area ko trap kar sakein, toh hum sum ko trap kar lete hain.

PICTURE. Har bar number line par $x = n$ se $x = n + 1$ tak baitha hai. Height labels dekho: pehla bar $f (1)$ tall hai, agla $f (2)$ , aur aage aise hi — har ek pichle se chhota, kyunki haare terms shrink karte hain.

Step 2 — Bars ke upar smooth ramp bichhaao

KYA. Curve $y = f (x)$ draw karo — ek smooth ramp jo dahine taraf dhire-dhire neeche jaati hai. Uske neeche ki area $x = 1$ se $x = b$ tak $\int_{1}^{b} f (x) d x$ likhi jaati hai.

KYUN. Hum ramp ki area calculate kar sakte hain calculus se (yahi toh integral karta hai — yeh infinitely thin area ke slivers ko add karta hai). Hum bar-sum directly calculate nahi kar sakte. Toh hum dono ko compare karte hain: ramp jo bhi kare, bars wahi follow karne par majboor ho jaate hain.

PICTURE. Bars aur ramp ek hi downhill shape share karte hain. Kuch bars ramp ke upar niklte hain; kuch neeche chhup jaate hain. Yeh upar nikalna / neeche chhupna hi poora game hai — Steps 3 aur 4 ise exactly pin karte hain.

Step 3 — Lambe bars: ek OVER-estimate (upper sum)

KYA. Strip $[n, n + 1]$ par, height $f (n)$ ka bar use karo — left edge par ki value. Kyunki ramp dahine jaane par girta hai, $f (n)$ us strip par ramp ki sabse zyada height hai. Toh yeh bar us strip ke ramp-area ko puri tarah cover karta hai: $\int_{n}^{n + 1} f (x) d x \leq left-edge height f (n) .$

KYUN. Hum ramp ki area par ek upper bound chahte hain, apne bars se banaya hua. Ek decreasing curve par left endpoints hamesha overshoot karte hain — yahi exactly woh inequality hai jo hum chahte hain.

PICTURE. Bar bada lavender box hai; ramp-area uske andar tucked chhota region hai. Curve ke upar bar ka patchla sliver "overshoot" hai.

Ab inhe $n = 1, 2, \dots, N$ ke liye add karo. Left sides ek lambe integral mein join ho jaate hain, right sides bar-sum mein: $ramp area, 1 to N + 1 \int_{1}^{N + 1} f (x) d x \leq = S_{N}, the bar wall n = 1 \sum N f (n) .$

Step 4 — Chhote bars: ek UNDER-estimate (lower sum)

KYA. Wahi strip $[n, n + 1]$ , lekin ab height $f (n + 1)$ use karo — right edge par ki value. Girte hue ramp par yeh sabse chhoti height hai, toh yeh chhota bar puri tarah ramp ke andar fit ho jaata hai: $right-edge height f (n + 1) \leq \int_{n}^{n + 1} f (x) d x .$

KYUN. Ab hum ramp ki area par apne bars se ek lower bound chahte hain. Ek decreasing curve par right endpoints hamesha undershoot karte hain — Step 3 ka mirror image.

PICTURE. Chhota coral bar ramp-area se ghir jaata hai; bar ke upar ramp ka sliver "gap" hai.

Inhe $n = 1, 2, \dots, N - 1$ ke liye add karo. Left par hume $f (2) + f (3) + \dots + f (N)$ milta hai, jo poora sum hai apne pehle bar ke bina: $pehle bar ko chhodke saare bars S_{N} - f (1) \leq \int_{1}^{N} f (x) d x .$ $f (1)$ ko doosri side move karo: $S_{N} \leq pehla bar f (1) + \int_{1}^{N} f (x) d x .$

Step 5 — Wall ko do ramps ke beech squeeze karo

KYA. Step 3 (ek floor) ko Step 4 (ek ceiling) ke neeche stack karo. Bar-sum $S_{N}$ ab do integrals ke beech pakad liya gaya hai: $floor \int_{1}^{N + 1} f \leq S_{N} \leq ceiling f (1) + \int_{1}^{N} f .$

KYUN. Yahi toh payoff hai. $S_{N}$ floor se neeche nahi ja sakta aur ceiling se upar nahi ja sakta. Ramp ki area jo bhi fate meet kare — finite ya infinite — bar-sum usi fate par kheencha jaata hai.

PICTURE. Ek number line, beech mein bars ki wall, neeche tinted "floor" ramp aur upar tinted "ceiling" ramp. Bars gap mein wedge ho gaye hain.

Step 6 — Do verdicts padhkar nikalo

KYA. $N$ ko infinity tak jaane do aur dono edges dekho.

Convergent case: agar $\int_{1}^{\infty} f$ ek finite number $L$ hai, toh ceiling kabhi $f (1) + L$ se zyada nahi hogi. Aur $S_{N}$ sirf badhta hai (hum positive bars add karte rehte hain). Ek quantity jo hamesha upar jaati hai lekin upar se cap hai zaroor settle hogi — toh $\sum a_{n}$ converge karta hai.

Divergent case: agar $\int_{1}^{\infty} f = \infty$ , toh floor $\int_{1}^{N + 1} f$ infinity tak chadh jaata hai, aur yeh $S_{N}$ ke neeche hai, toh yeh $S_{N}$ ko bhi apne saath upar dhakelta hai — $\sum a_{n}$ diverge karta hai. $■$

KYUN. Ramp sirf do hi cheezein kar sakta hai (hamesha badhna, ya level off hona). Squeeze har ek ko sum ke matching verdict mein convert karta hai.

PICTURE. Do panels: baayein, ek levelling ramp jisme $S_{N}$ ek dashed cap par flat ho raha hai; dahine, ek rising floor jo $S_{N}$ ko hamesha ke liye upar kheeench raha hai.

Step 7 — p-series feed karo aur cases mein split karo

KYA. $a_{n} = \frac{1}{n ^{p}}$ lo, toh $f (x) = x^{- p}$ . Yeh har $p > 0$ ke liye positive, continuous, aur decreasing hai — hypotheses met. Step 6 se, sum ki fate = ramp ki fate. Ramp area compute karo.

Case $p \neq = 1$ : $\int_{1}^{\infty} x^{- p} d x = lim_{b \to \infty} \frac{b ^{1 - p} - 1}{1 - p} .$

$p > 1$ : exponent $1 - p < 0$ , toh $b^{1 - p} \to 0$ . Area $= \frac{1}{p - 1}$ (finite) $\Rightarrow$ converges.
$0 < p < 1$ : exponent $1 - p > 0$ , toh $b^{1 - p} \to \infty$ . Area infinite $\Rightarrow$ diverges.

Case $p = 1$ (Harmonic series): power rule kaam nahi karta, integral logarithm ban jaata hai: $\int_{1}^{\infty} \frac{d x}{x} = lim_{b \to \infty} ln b = \infty \Rightarrow diverges .$

KYUN. Exponent $1 - p$ ki sign decide karti hai ki ramp ki tail vanish hogi ya explode karegi — woh single sign hi $p = 1$ par razor's edge hai.

PICTURE. Teen ramps overlaid: ek steep ( $p = 2$ , finite tail area, tinted aur closed off), borderline ( $p = 1$ , tail area hamesha ke liye creep karta hua), aur ek gentle ( $p = \frac{1}{2}$ , tail infinity tak flood karti hui).

Ek picture mein summary

Sab kuch ek frame mein: bars ki wall ( $S_{N}$ ), ramp ( $y = f (x)$ ), upar niklte hue tall left-endpoint bars (upper sum), neeche chhupte hue short right-endpoint bars (lower sum), aur squeeze arrows jo $S_{N}$ ko do integrals ke beech pin karte hain.

Recall Feynman: poora walkthrough plain words mein dobara sunao

Main jaanna chahta hoon ki $a_{1} + a_{2} + a_{3} + \dots$ ko hamesha ke liye add karne par kuch milta hai ya yeh infinity tak bhaag jaata hai. Main hamesha ke liye add nahi kar sakta, toh main har term ko width $1$ aur height $a_{n}$ ke ek block ki tarah draw karta hoon — ab sum sirf blocks ki wall ki shaded area hai. Blocks ke paas mein main ek smooth slide $y = f (x)$ rakhta hoon jiska area main ek integral se calculate kar sakta hoon. Is downhill slide par, agar main har strip ka left edge use karoon toh block thoda zyada tall hota hai — yeh slide se zyada cover karta hai (upper sum). Agar main right edge use karoon toh block thoda chhota hota hai — slide block se zyada cover karti hai (lower sum). Toh block-wall do integrals ke beech squeeze ho jaati hai: "slide ki area" aur "slide ki area plus ek extra pehla block". Kyunki wall sirf badhti hai (saare blocks upar point karte hain), agar slide ki area finite hai toh wall cap ho jaati hai aur settle ho jaani chahiye; agar slide ki area infinite hai toh wall bhi infinity tak shove ho jaati hai. Hamesha same fate. Phir main $1/ n^{p}$ try karta hoon: uski slide $x^{- p}$ hai, jiska tail area sirf tabhi kuch nahi rehta jab $p > 1$ ho, aur warna flood kar deta hai. Exactly $p = 1$ par integral $ln b$ ban jaata hai, jo hamesha ke liye upar creep karta hai — toh harmonic series diverge karta hai. Woh razor-sharp cutoff at $p = 1$ hi punchline hai.

Connections

Integral test — proof, p-series (index 4.3.7) — parent note jisko yeh walkthrough visualise karta hai.
Improper integrals — woh machinery jo ramp ki area infinity tak compute karti hai.
Harmonic series — exact borderline $p = 1$ jo humne barely diverge hote dekhi.
Comparison test — jab p-series verdict par bharosa ho, toh ise yardstick ki tarah use karo.
Limit comparison test — same yardstick, softer matching rule.
Riemann zeta function — woh true value jahan sum land karta hai (jo test kabhi reveal nahi karta).
n-th term test for divergence — kyun $a_{n} \to 0$ akela kabhi enough nahi hota.

#calculus3 #series/convergence