Integral test — proof, p-series: Question bank — Hinglish deep dive

4.3.7 · D5 · Maths › Calculus III — Sequences & Series › Integral test — proof, p-series

Tag: #calculus3 #series/convergence

Yeh ek concept gym hai integral test aur p-series ke liye. Neeche har item ek aisi jagah ko target karta hai jahan intuition tumhe dhoka deta hai. Right side ko cover karo, puri ek sentence ki reasoning force karo, phir reveal karo. Sirf "haan/nahi" answers se kuch nahi milega — reasoning hi asli point hai.

Shuru karne se pehle, teen words jinhe hum baar baar use karte hain, bilkul zero se define kiye gaye hain:

Yeh do pictures poore time apne saamne rakho. Figure 1 dikhata hai trap kyun kaam karta hai jab $f$ decrease karta hai; Figure 2 dikhata hai yeh kyun toot jaata hai jab $f$ increase karta hai. Neeche ka har item in do pictures mein se ek hi hai, words mein.

Figure — Integral test — proof, p-series

True or false — justify

Har answer mein kyun bolo, sirf T/F nahi.

TF1.

p = 0.9999

wali ek p-series converge karti hai.

False. Convergence ke liye strictly

p > 1

chahiye;

0.9999 < 1

hai isliye

\int_{1}^{\infty} x^{- 0.9999} d x

phir bhi blow up karta hai. Cutoff razor-sharp hai, aur

0.9999

diverging side par hai.

TF2. Agar

\int_{1}^{\infty} f d x = 5

hai, toh

\sum_{n = 1}^{\infty} f (n) = 5

bhi hoga.

False. Test sirf yeh guarantee karta hai ki sum aur integral ka fate same hoga (dono finite), value same nahi. Figure 1 dekho: decreasing

f

ke liye red left-endpoint rectangles curve ke strictly upar baithte hain, isliye generally

\sum f (n) > \int_{1}^{\infty} f

hota hai (jaise

\sum 1/ n^{2} = π^{2} /6 \approx 1.645

vs integral

= 1

TF3. Harmonic series diverge karta hai, chahe uske terms

1/ n \to 0

ho jaayein.

True. Terms ka

0

ki taraf shrink karna convergence ke liye necessary hai lekin sufficient nahi; shrinking bahut slow hai, aur

\int_{1}^{\infty} d x / x = ln b \to \infty

divergence confirm karta hai.

TF4. Agar

a_{n} \to 0

hai, toh integral test se

\sum a_{n}

converge karta hai.

False. Integral test yahi nahi kehta, aur yahi classic error hai.

a_{n} \to 0

sirf tumhe turant divergence se bachata hai; tumhe phir bhi ek real test run karna hoga. Harmonic series iska counterexample hai.

TF5. Integral test ko

\sum \frac{( - 1 ) ^{n}}{n}

par use kiya ja sakta hai.

False. Test demand karta hai ki

f

positive ho; alternating terms ke mixed signs hote hain, isliye "areas can't cancel" wali assumption toot jaati hai. Iske bajaye alternating-series test use karo.

TF6. Agar

f

positive aur continuous hai lekin increasing hai, toh integral test phir bhi apply hoga.

False. Figure 2 dekho: jab

f

increase karta hai,

[n, n + 1]

par left endpoint sabse chhoti value deta hai, isliye left-endpoint rectangle ab curve ke neeche hota hai, jisse milta hai

f (n) \leq \int_{n}^{n + 1} f

— yeh decreasing case

\int_{n}^{n + 1} f \leq f (n)

ka bilkul ulta hai. Dono trap-inequalities flip ho jaati hain, isliye

S_{N}

sahi tarah se squeeze nahi hoti aur test ki hypothesis genuinely violate hoti hai. (Alag se, ek increasing positive

f

mein

a_{n} \neq \to 0

hota hai, isliye series n-th term test for divergence se waise bhi diverge ho jaati hai.)

TF7.

\sum_{n = 1}^{\infty} \frac{1}{n ^{p}}

har

p \leq 1

ke liye diverge karta hai, negative

p

bhi include karke.

True.

p \leq 1

ke liye integral

\int_{1}^{\infty} x^{- p} d x

infinite hai.

p \leq 0

ke liye terms

0

ki taraf shrink bhi nahi karti, isliye yeh turant n-th term test se diverge ho jaata hai — sab

p \leq 1

ke liye divergence.

TF8. Riemann zeta function

ζ (p) = \sum 1/ n^{p}

p = 1

ke liye defined hai.

False.

ζ

is sum se sirf

p > 1

ke liye define hoti hai, exactly wahan jahan p-series converge karti hai.

p = 1

par sum divergent harmonic series hai.

TF9. Agar

\sum f (n)

converge karta hai, toh

\int_{1}^{\infty} f d x

bhi converge karna chahiye.

True — provided $f$ positive, continuous aur decreasing ho (integral-test ke hypotheses; inke bina equivalence guarantee nahi hai). Un conditions ke under equivalence dono taraf chalti hai: sum ka convergence integral ka convergence force karta hai lower-sum inequality

\int_{1}^{N + 1} f \leq S_{N}

ke through, jiska left side bounded rehta hai. Yeh dono saath jeete aur maarte hain.

Spot the error

Har line ek flawed argument batati hai; reveal uss exact broken step ka naam leta hai.

SE1. "

\sum \frac{1}{n l n n}

converge karta hai kyunki

\frac{1}{n l n n}

\frac{1}{n}

se zyada tezi se shrink karta hai."

Error hai "shrinks faster" par trust karna bina test ke. Substitution

u = ln x

deta hai

\int_{2}^{\infty} \frac{d x}{x l n x} = \int_{l n 2}^{\infty} \frac{d u}{u} = \infty

, isliye yeh diverge karta hai — extra

ln n

isse bachane ke liye kaafi nahi hai.

SE2. "

f (x) = \frac{s i n ^{2} x}{x ^{2}}

positive aur continuous hai, isliye main integral test apply karunga."

Missing hypothesis:

f

decreasing nahi hai (yeh

sin^{2} x

ki wajah se oscillate karta hai). Integral test ke liye eventually-decreasing

f

chahiye; yahan

1/ x^{2}

ke saath comparison ki zaroorat hogi.

SE3. "

\sum 1/ n^{2}

ke liye, kyunki

\int_{1}^{\infty} x^{- 2} d x = 1

hai, sum bhi

1

ke barabar hoga."

Error sum aur integral ko equal karna hai. Test sirf bound karta hai:

1 = \int_{1}^{\infty} x^{- 2} d x < \sum 1/ n^{2} = π^{2} /6 \approx 1.645

. Convergence share karna

\neq =

value share karna.

SE4. "

\int_{1}^{\infty} x^{- p} d x = \frac{x ^{1 - p}}{1 - p}_{1}^{\infty}

, aur

p = 1

plug in karne par

\frac{1}{0}

milta hai, jo undefined hai lekin finite-ish hai, isliye harmonic borderline convergent hai."

Error

p = 1

par

p \neq = 1

wale antiderivative ko use karna hai. Jab

p = 1

hota hai toh antiderivative

ln x

hota hai,

x^{1 - p} / (1 - p)

nahi; aur

ln b \to \infty

, isliye harmonic diverge karta hai outright — "borderline convergent" nahi.

SE5. "Rectangles curve ki area ke barabar hain, isliye integral test exact hai."

Decreasing

f

ke rectangles kabhi curve ki area ke barabar nahi hote — left-endpoint rectangles overshoot karte hain, right-endpoint wale undershoot karte hain (Figure 1 dekho). Yeh gap exactly wahi hai jo trapping inequality

\int_{1}^{N + 1} f \leq S_{N} \leq f (1) + \int_{1}^{N} f

measure karti hai.

SE6. "

\sum_{n = 1}^{\infty} \frac{1}{n l n n}

— test

n = 1

se apply karo."

n = 1

par,

ln 1 = 0

hai isliye

f (1) = \frac{1}{1 \cdot 0}

undefined hai. Test

n = 2

se shuru karo; test kisi bhi

n = N

se start karne mein khush hai kyunki finitely many terms convergence kabhi nahi badlti.

Why questions

WHY1.

f

decreasing kyun hona chahiye, sirf positive aur continuous nahi?

Decreasing guarantee karta hai ki left-endpoint height

f (n)

[n, n + 1]

par max hai aur right-endpoint height

f (n + 1)

min hai (Figure 1). Yeh dono one-directional inequalities deta hai jo

S_{N}

ko trap karti hain; monotonicity ke bina, ek rectangle curve ke partly upar aur partly neeche ho sakta hai aur squeeze collapse ho jaata hai.

WHY2. Test convergence kyun batata hai lekin sum batane se kyun mana karta hai?

Kyunki yeh bounds ki ek pair deliver karta hai

\int_{1}^{N + 1} f \leq S_{N} \leq f (1) + \int_{1}^{N} f

, equation nahi. Bounds yeh pin down karte hain ki

S_{N}

finite hai ya nahi, lekin unke beech ek genuine gap chhod dete hain — sum kahin andar hota hai, aur test kabhi exactly nahi batata kahan.

WHY3.

\sum \frac{1}{n l n n}

handle karte waqt substitution

u = ln x

kyun aata hai?

Kyunki

d u = d x / x

\frac{d x}{x l n x}

ban jaata hai

\frac{d u}{u}

— disguise mein harmonic integral. Substitution exactly isliye choose ki jaati hai taaki reveal ho sake ki

ln

factor bahut slowly grow karta hai divergence ko fix karne ke liye.

WHY4. Exact boundary

p = 1

kyun hai aur

p = 1.5

p = 0.5

kyun nahi?

Improper integral ko limit ke roop mein evaluate karo

\int_{1}^{\infty} x^{- p} d x = lim_{b \to \infty} \int_{1}^{b} x^{- p} d x

, jahan

b

sirf woh moving upper cutoff hai jise hum infinity ki taraf push karte hain.

p \neq = 1

ke liye yeh equal hota hai

lim_{b \to \infty} \frac{b ^{1 - p} - 1}{1 - p}

se, jiska fate

1 - p

ke sign par hinge karta hai:

p > 1

ke liye exponent

1 - p < 0

hai isliye

b^{1 - p} \to 0

(finite area),

p < 1

ke liye yeh

> 0

hai isliye

b^{1 - p} \to \infty

. Exactly

p = 1

par antiderivative

ln b

mein switch ho jaata hai, aur

lim_{b \to \infty} ln b = \infty

. Behaviour bilkul

1

par character change karta hai.

WHY5. Integral test

n = 1

ki jagah

n = N

se kyun shuru kar sakte hain?

Convergence ek tail property hai — finitely many terms chop karne se sum finite amount se change hota hai lekin uska finite/infinite fate kabhi nahi. Isliye agar

f

sirf eventually positive-decreasing hai (kisi

N

se aage), tail test karo; head harmless hai.

WHY6. Comparison test p-series ko apna favourite yardstick kyun use karta hai?

Kyunki p-series mein ek known, clean convergence rule hai (

p > 1

), isliye ek unknown series ko

1/ n^{p}

se compare karna use turant settle kar deta hai. P-series ruler hai kyunki integral test ne ise already exactly measure kar liya hai.

Edge cases

EC1.

p = 1

exactly par

\sum 1/ n^{p}

ka kya hota hai?

Yeh harmonic series hai, jo diverge karta hai — boundary diverging side se belong karti hai. "Push past 1" ka matlab strictly

1

se zyada hai.

EC2.

p < 0

ke baare mein kya, jaise

\sum n^{2}

Turant diverge karta hai:

p < 0

ke saath terms

n^{∣ p ∣}

bound ke bina grow karte hain, isliye

a_{n} \neq \to 0

aur n-th term test for divergence ise integral test ki zaroorat se pehle hi khatam kar deta hai.

EC3.

p = 0

ke baare mein kya, yaani

\sum 1/ n^{0} = \sum 1

Diverge karta hai: har term

1

hai, partial sums

S_{N} = N \to \infty

hain. Terms

0

ki taraf nahi jaate, isliye n-th term test ise turant rule out kar deta hai.

EC4. Agar

f

decreasing hai lekin sirf eventually (maan lo pehle increase karta hai phir decrease), kya hum phir bhi test kar sakte hain?

Haan — test ko tail

[N, \infty)

par apply karo jahan

f

decreasing hai. Pehle ke finitely many terms ek finite constant add karte hain aur convergence nahi badal sakte.

EC5. Kya

\int_{1}^{\infty} f d x

ke converge karne ke liye

f (x) \to 0

zaroori hai?

Nahi — lekin ek positive decreasing

f

ke liye, agar integral converge karta hai toh

f (x)

\to 0

hona chahiye; warna tail area infinite hoti. "Integral converges

\Rightarrow f \to 0

" direction humari hypotheses ke under hold karta hai.

EC6.

\sum \frac{1}{n ( l n n ) ^{2}}

ka fate

\sum \frac{1}{n l n n}

se compare karke kya hai?

Extra power result flip kar deta hai:

u = ln x

ke saath,

\int_{2}^{\infty} \frac{d x}{x ( l n x ) ^{2}} = \int_{l n 2}^{\infty} \frac{d u}{u ^{2}}

converge karta hai (yeh

p = 2

wala p-integral hai), isliye yeh series converge karti hai jabki

\sum \frac{1}{n l n n}

diverge karta hai.

ln

ek second-order p-series ki tarah act karta hai.

Connections

Comparison test — jahan yeh p-series yardsticks use hote hain.
Limit comparison test — ek unknown series ko $1/ n^{p}$ ke against compare karta hai.
Improper integrals — woh machinery jo upar har "why" ke andar evaluate hoti hai.
Harmonic series — star boundary case $p = 1$ .
Riemann zeta function — $ζ (p) = \sum 1/ n^{p}$ , sirf wahan defined jahan p-series converge karti hai.
n-th term test for divergence — TF4, EC2, EC3 ke peeche necessary-not-sufficient trap.