Divergence test (necessary but not sufficient): Worked examples

4.3.6 · D3 · Maths › Calculus III — Sequences & Series › Divergence test (necessary but not sufficient)

Kuch bhi karne se pehle, yaad karo ki yeh test sirf do verdicts de sakta hai:

Recall Rule ek saanss mein

$L = n \to \infty lim a_{n}$ compute karo (woh term jo tum add kar rahe ho, jab $n$ bahut bada ho jaata hai). Agar $L \neq = 0$ hai ya limit exist nahi karti (DNE), toh series diverges karti hai. Agar $L = 0$ hai, toh test inconclusive hai — yeh kuch nahi kehta, aur tumhe koi doosra tool use karna hoga.

Yahan $a_{n}$ ka matlab sirf itna hai ki "woh $n$ -va number jise tum sum kar rahe ho", aur $lim_{n \to \infty} a_{n}$ poochh raha hai "jab $n$ infinity ki taraf jaata hai, kya woh number kisi ek value par settle hota hai, aur agar haan, toh kaunsi?" Dekho Limits of Sequences — yahi woh engine hai jo neeche sab kuch chalata hai.

Scenario matrix

Har series jo tum Divergence Test mein daal sakte ho, exactly inhi cells mein se ek mein aati hai. Right column woh verdict hai jo sirf Divergence Test deta hai.

Cell	$a_{n}$ $n \to \infty$ par kya karta hai	Test verdict	Example
A	Nonzero number par settle karta hai (jaise rational function, equal degrees)	DIVERGES	Ex 1
B	Bina bound ke badhta hai ( $a_{n} \to \pm \infty$ )	DIVERGES	Ex 2
C	Oscillate karta hai, kabhi settle nahi hota (limit DNE)	DIVERGES	Ex 3
D	Ek nonzero constant ki taraf famous limit se jaata hai ( $\to e$ )	DIVERGES	Ex 4
E	Dhire dhire zero ki taraf jaata hai → phir bhi diverges (test blind)	INCONCLUSIVE	Ex 5
F	Tezi se zero ki taraf jaata hai → actually converges (test blind)	INCONCLUSIVE	Ex 6
G	Sign flip karta hai lekin zero tak shrink hota hai (alternating)	INCONCLUSIVE	Ex 7
H	Word problem — pebbles/medicine doses, pehle $a_{n}$ extract karna hoga	depends	Ex 8
I	Exam twist — lagtaa hai decay kar raha hai lekin ek hidden constant survive karta hai	DIVERGES	Ex 9

Figure — Divergence test (necessary but not sufficient)

Left side wali red curve dekho: uske terms zero ke upar level off ho rahe hain — test fire karta hai. Right side par terms sab zero ki taraf rengti hain, aur test dono ko alag nahi kar sakta chahe ek series finite number mein add ho aur doosri blow up kare.

Cell A — term nonzero number par settle karta hai

Worked example Example 1 (Cell A) — equal-degree rational terms

$\sum_{n = 1}^{\infty} \frac{4 n ^{3} - 2 n}{7 n ^{3} + n ^{2} + 5} .$ Forecast: answer chhupao aur guess karo. Upar aur neeche dono ka fastest-growing piece $n^{3}$ hai. Kisi aisi cheez ka ratio jo same rate se badhti hai, kya settle hoga? Zero nahi — leading coefficients ka seedha ratio. Toh predict karo: diverges.

Step 1. Numerator aur denominator dono ko $n^{3}$ se divide karo, jo highest power present hai. $a_{n} = \frac{4 n ^{3} - 2 n}{7 n ^{3} + n ^{2} + 5} = \frac{4 - \frac{2}{n ^{2}}}{7 + \frac{1}{n} + \frac{5}{n ^{3}}} .$ Yeh step kyun? Fastest-growing term se divide karne par har slower term mein $n$ denominator mein aa jaata hai, jise hum phir zero par bhej sakte hain. Yeh dikhata hai ki koun dominate karta hai.

Step 2. $n \to \infty$ bhejo. $n$ downstairs wala har fraction khatam ho jaata hai: $L = \frac{4 - 0}{7 + 0 + 0} = \frac{4}{7} .$ Yeh step kyun? $\frac{2}{n ^{2}} \to 0$ , $\frac{1}{n} \to 0$ , $\frac{5}{n ^{3}} \to 0$ — growing denominator ke upar ek constant vanish ho jaata hai.

Step 3. Kyunki $L = \frac{4}{7} \neq = 0$ , Divergence Test fire karta hai: series diverges.

Verify: bade $n = 1000$ se sanity-check: $\frac{4 ( 1 0 ^{9} ) - 2000}{7 ( 1 0 ^{9} ) + 1 0 ^{6} + 5} \approx 0.5714 \approx \frac{4}{7}$ . Terms hamesha $0.57$ ke aaspaas rehte hain, toh inhe infinitely baar add karne se finite number nahi nikal sakta. ✔

Cell B — term bina bound ke badhta hai

Worked example Example 2 (Cell B) — numerator jeet jaata hai

$\sum_{n = 1}^{\infty} \frac{n ^{2} + 1}{3 n + 4} .$ Forecast: upar ki degree 2 hai, neeche ki degree 1. Upar wala out-grow karta hai neeche wale ko, toh term khud hi blow up karta hai. Predict karo: diverges (Cell A se bhi zyada strongly — terms sirf bade nahi rehte, explode hote hain).

Step 1. $n$ se divide karo (denominator mein highest power, taaki growth read kar sakein): $a_{n} = \frac{n ^{2} + 1}{3 n + 4} = \frac{n + \frac{1}{n}}{3 + \frac{4}{n}} .$ Yeh step kyun? Yeh upar surviving $n$ ko isolate karta hai, dikhata hai ki term $\frac{n}{3}$ ki tarah grow karta hai.

Step 2. Jab $n \to \infty$ : numerator $\to \infty$ , denominator $\to 3$ , toh $L = lim_{n \to \infty} \frac{n + \frac{1}{n}}{3 + \frac{4}{n}} = + \infty.$ Yeh step kyun? Ek finite constant se divide ki gayi infinity-bound quantity phir bhi infinity ki taraf jaati hai.

Step 3. $L = + \infty \neq = 0$ , toh series diverges.

Verify: $n = 100$ par, $a_{100} = \frac{10001}{304} \approx 32.9$ ; $n = 1000$ par, $a_{1000} \approx 333.1$ — terms badhte rehte hain, $L = \infty$ confirm karta hai. ✔

Red points plot ke top se upar march karte hain — ek term jo grow karta hai woh running total ko kabhi settle nahi hone de sakta.

Cell C — term oscillate karta hai (limit DNE)

Worked example Example 3 (Cell C) — bouncing sign

$\sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n} n}{n + 1} .$ Forecast: ek second ke liye sign ignore karo — $\frac{n}{n + 1} \to 1$ . Lekin $(- 1)^{n}$ term ko $+ 1$ aur $- 1$ ke beech flip karta hai. Kya yeh ek value par settle hota hai? Nahi. Predict karo: diverges kyunki limit exist nahi karti.

Step 1. Size ko sign se alag karo. Size hai $\frac{n}{n + 1} = \frac{1}{1 + \frac{1}{n}} \to 1$ . Yeh step kyun? Jaanna ki size $1$ ki taraf jaati hai (na $0$ ki taraf) batata hai ki flips $- 1$ aur $+ 1$ ke aaspaas numbers ke beech hain, na shrinking numbers ke beech.

Step 2. Even aur odd $n$ ko alag track karo:

even $n$ : $a_{n} \to + 1$ ,
odd $n$ : $a_{n} \to - 1$ .

Yeh step kyun? Agar do subsequences do alag values ki taraf aim karti hain, toh poori sequence ka koi single limit nahi hota — limit exist nahi karti.

Step 3. DNE is test ke liye " $\neq = 0$ " count hoti hai, toh series diverges.

Verify: $a_{100} = \frac{100}{101} \approx + 0.990$ , $a_{101} = - \frac{101}{102} \approx - 0.990$ . Terms hamesha $\pm 1$ ke aaspaas rehte hain — koi settling nahi, koi finite total nahi. ✔

Cell D — famous nonzero limit

Worked example Example 5 (Cell E) — harmonic blind spot

$\sum_{n = 1}^{\infty} \frac{1}{n} .$ Forecast: $\frac{1}{n} \to 0$ . Kya isse convergence guarantee hoti hai? Parent note se predict karo: test kuch nahi kehta, aur waqai mein yeh famous series diverge karti hai.

Step 1. $L = lim_{n \to \infty} \frac{1}{n} = 0$ . Yeh step kyun? Ek ko hamesha bade hote number se divide karne par result zero ki taraf shrink hota hai.

Step 2. $L = 0$ , toh Divergence Test inconclusive hai — yeh dono taraf rule nahi kar sakta.

Step 3. Ek stronger tool lao. p-Series Test $p = 1$ ke saath (ya parent note mein grouping argument) dikhata hai ki $\sum \frac{1}{n}$ diverges hai chahe uske terms vanish ho jaayein.

Verify: partial sums $ln N$ ki tarah badhte hain: $S_{100} \approx 5.19$ , $S_{10000} \approx 9.79$ — bina limit ke creep karte rehte hain, $a_{n} \to 0$ hone ke bawajood divergence confirm karta hai. ✔

Dono red term-curves zero par flatten ho jaati hain (top panel), test ke liye indistinguishable. Bottom panel running totals dikhata hai: ek badhta rehta hai (diverges), doosra $\frac{π ^{2}}{6}$ par flatten ho jaata hai (converges). Test sirf top panel dekhta hai — isliye woh dono ko alag nahi kar sakta.

Cell G — alternating, zero tak shrinks

Worked example Example 7 (Cell G) — sign flip karta hai lekin size khatam ho jaati hai

$\sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n + 1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots .$ Forecast: magnitude $\frac{1}{n} \to 0$ , toh term (sign ke saath) bhi $0$ ki taraf jaati hai. Predict karo: Divergence Test inconclusive hai.

Step 1. $∣ a_{n} ∣ = \frac{1}{n} \to 0$ , aur squeezing se $a_{n} = \frac{( - 1 ) ^{n + 1}}{n} \to 0$ milta hai. Yeh step kyun? Ek quantity jo $- \frac{1}{n}$ aur $+ \frac{1}{n}$ ke beech trapped hai, jahan dono $0$ ki taraf jaati hain, use bhi $0$ tak jaane par force kiya jaata hai.

Step 2. $L = 0$ , toh Divergence Test kuch nahi kehta.

Step 3. Ek alag tool (alternating-series argument) dikhata hai ki yeh actually converges karti hai — $ln 2$ tak. Cell E ke $\sum \frac{1}{n}$ se compare karo: same-sized terms, lekin alternating signs cancellation se isse bacha lete hain.

Verify: $ln 2 \approx 0.6931$ ; partial sum $S_{1000} \approx 0.6926$ , $ln 2$ ki taraf aa raha hai. Zero-limit terms ⇒ test silent ⇒ humein doosra tool chahiye tha. ✔

Cell H — word problem (pehle $a_{n}$ extract karo)

Worked example Example 8 (Cell H) — never-shrinking dose

Ek patient pill leta hai. Har ghante unka body ek fixed fraction keep karta hai toh $n$ -vi dose se jo drug baad mein present hai woh "total exposure" tally mein $a_{n} = 5 \cdot \frac{n}{n + 3}$ mg contribute karta hai. Kya total exposure $\sum a_{n}$ ek finite number par settle hoti hai? Forecast: har contribution $5 \cdot \frac{n}{n + 3} \to 5$ mg ki taraf jaati hai — yeh shrink nahi karti. Predict karo: total exposure diverges (bina bound ke badhti hai).

Step 1. Term extract karo: $a_{n} = 5 \cdot \frac{n}{n + 3}$ . $n$ se divide karo: $a_{n} = 5 \cdot \frac{1}{1 + \frac{3}{n}}$ . Yeh step kyun? Word problem mein ek ordinary sequence chhupa hai; koi bhi test apply karne se pehle humein $a_{n}$ naam dena hoga.

Step 2. $L = 5 \cdot \frac{1}{1 + 0} = 5 \neq = 0$ (units: mg).

Step 3. $L = 5 mg \neq = 0$ , toh Divergence Test se total exposure diverges — physically, contributions kabhi taper off nahi karti, toh cumulative exposure bina bound ke badhti hai.

Verify: $a_{1000} = 5 \cdot \frac{1000}{1003} \approx 4.985$ mg — abhi bhi $5$ ke paas, nonzero limit aur unbounded total confirm karta hai. ✔

Cell I — exam twist (hidden surviving constant)

Worked example Example 9 (Cell I) — lagtaa hai decay kar raha hai, lekin…

$\sum_{n = 1}^{\infty} (n^{2} + n - n) .$ Forecast: pehli nazar mein $n^{2} + n \approx n$ , toh term vanish ho sakti hai. Lekin kya leftover sach mein zero ho jaata hai? Carefully predict karo — yeh hi trap hai.

Step 1. Do bade nearly-equal roots ke subtraction ko tame karne ke liye conjugate se multiply karo: $a_{n} = \frac{( n ^{2} + n - n ) ( n ^{2} + n + n )}{n ^{2} + n + n} = \frac{n ^{2} + n - n ^{2}}{n ^{2} + n + n} = \frac{n}{n ^{2} + n + n} .$ Yeh step kyun? Do bade nearly-equal numbers subtract karne se saari precision kho jaati hai; conjugate trick ise ek clean quotient mein convert karta hai jiska hum limit le sakte hain.

Step 2. Upar aur neeche dono ko $n$ se divide karo: $a_{n} = \frac{1}{1 + \frac{1}{n} + 1} ⟶ \frac{1}{1 + 1} = \frac{1}{2} .$ Yeh step kyun? Root se $n$ bahar nikalna $n^{2} + n = n 1 + \frac{1}{n}$ use karta hai; phir $\frac{1}{n} \to 0$ .

Step 3. $L = \frac{1}{2} \neq = 0$ — "decay" ek illusion tha. Series diverges.

Verify: $n = 1000$ par, $1000000 + 1000 - 1000 \approx 1000.4998 - 1000 = 0.4998 \approx \frac{1}{2}$ . Gap $\frac{1}{2}$ par park karta hai, kabhi zero nahi. ✔

Har cell ka recap

Recall Self-test: har cell ke liye verdict batao

Cell A (nonzero settle) ::: DIVERGES Cell B (infinity tak badhta hai) ::: DIVERGES Cell C (oscillate karta hai, limit DNE) ::: DIVERGES Cell D (famous $\to e^{x}$ ) ::: DIVERGES Cell E ( $\to 0$ slowly, jaise $1/ n$ ) ::: INCONCLUSIVE (aur actually diverges) Cell F ( $\to 0$ fast, jaise $1/ n^{2}$ ) ::: INCONCLUSIVE (aur actually converges) Cell G (alternating $\to 0$ ) ::: INCONCLUSIVE (aur actually converges to $ln 2$ ) Cell I (hidden surviving constant) ::: DIVERGES

Connections

Divergence test — Hinglish parent — woh rule jise yeh examples exercise karte hain.
Limits of Sequences — har "compute $L$ " step yahan rehta hai.
Partial Sums and Series Convergence — Cell F ke figure mein running-total picture.
Harmonic Series — Cell E ka star blind-spot example.
p-Series Test — Cells E aur F resolve karta hai jo test nahi kar sakta.
Integral Test, Comparison Test — "next tools" jab $L = 0$ ho.
Telescoping Series — jahan $a_{N} = S_{N} - S_{N - 1}$ idea wapas aata hai.

Worked examples — Divergence test (necessary but not sufficient)