Ratio test — proof, limitations: Worked examples

4.3.12 · D3 · Maths › Calculus III — Sequences & Series › Ratio test — proof, limitations

Intuition Yeh page kya hai

Parent note ne prove kiya tha ki ratio test kaam kyun karta hai. Yahan hum karna practice karte hain: hum ek map banate hain har tarah ki series ka jo test se mil sakti hai — har sign pattern, tricky $L = 1$ boundary, degenerate inputs, ek word problem, aur ek exam trap — phir ek-ek example har cell ke liye work out karte hain taaki koi bhi scenario tumhe surprise na kare.

Reminder of the one tool: define the shrink-ratio $ρ_{n} = \frac{a _{n + 1}}{a _{n}} L = lim_{n \to \infty} ρ_{n} .$ $ρ_{n}$ ko padhо as "agli term kitni badi hai is term se compare mein, sign ignore karke". $L$ iska long-run value hai. $L < 1$ → absolutely converge karta hai (matlab all-positive series $\sum ∣ a_{n} ∣$ khud converge karti hai, jo ordinary convergence se zyada strong hai), $L > 1$ → diverge karta hai, $L = 1$ → test kuch nahi bolta.

The scenario matrix

Har series jo ratio test ko feed kar sakte ho, in cells mein se ek mein aati hai. Hum sab cover karenge.

Cell	Kya special hai	$L$ outcome	Example #
A	Exponential ÷ factorial (clean cancellation)	$L = 0 < 1$ → converges	Ex 1
B	Factorial ÷ exponential (factorial wins)	$L = \infty > 1$ → diverges	Ex 2
C	Sign-changing / alternating terms (must use $∣ \cdot ∣$ )	$L < 1$ → abs. converges	Ex 3
D	Do exponentials, finite $0 < L < 1$	$L = r < 1$ → converges	Ex 4
E	Power-law $1/ n^{p}$ — the blind spot	$L = 1$ → inconclusive	Ex 5
F	Power series → radius of convergence, endpoints	$L = ∣ x ∣$ , phir $L = 1$ boundary	Ex 6
G	Degenerate: ek term zero hai / ratio undefined	test apply nahi hota — patch karo	Ex 7
H	Real-world word problem (drug dosing)	$L = r < 1$ → finite build-up	Ex 8
I	Exam twist: $L = 1$ but Raabe rescues it	ratio fails, sharper tool needed	Ex 9
J	Limit $L$ exist hi nahi karta (ratio oscillates)	koi single $L$ nahi → ratio test silent	Ex 10

Prerequisites jo hum use karte hain: Geometric Series, Comparison Test, Term Test (nth-term divergence), Root Test, p-Series and Integral Test, Radius of Convergence, Raabe's Test.

Cell A — exponential over factorial

Worked example Example 1 ·

$n = 1 \sum \infty \frac{3 ^{n}}{n !}$ Forecast: denominator mein factorials kisi bhi exponential se tezi se badhte hain. Guess: converges, $L = 0$ .

Step 1. Ratio likho. $ρ_{n} = \frac{a _{n + 1}}{a _{n}} = \frac{3 ^{n + 1} / ( n + 1 )!}{3 ^{n} / n !} = \frac{3 ^{n + 1}}{3 ^{n}} \cdot \frac{n !}{( n + 1 )!} .$ Yeh step kyun? Fraction ko fraction se divide karna matlab reciprocal se multiply karna; power part aur factorial part alag karna dono ko apne aap simplify hone deta hai.

Step 2. Har piece simplify karo: $\frac{3 ^{n + 1}}{3 ^{n}} = 3$ aur $\frac{n !}{( n + 1 )!} = \frac{1}{n + 1}$ (kyunki $(n + 1)! = (n + 1) \cdot n!$ , $n!$ cancel ho jaata hai). $ρ_{n} = \frac{3}{n + 1} .$ Yeh step kyun? Aage badhta factorial sirf ek naya factor $(n + 1)$ laata hai, isliye bada $n!$ cleanly cancel ho jaata hai — wahi clean cancellation kyun ratio test ko factorials pasand hain.

Step 3. Limit lo. $L = lim_{n \to \infty} \frac{3}{n + 1} = 0 < 1 \Rightarrow converges absolutely.$ Yeh step kyun? Fixed numerator divided by growing denominator → $0$ ho jaata hai.

Verify: Yeh exponential series hai minus $n = 0$ term: $\sum_{n = 1}^{\infty} 3^{n} / n! = e^{3} - 1 \approx 19.09$ — ek finite number, convergence ke saath consistent.

Cell B — factorial over exponential

Worked example Example 2 ·

$n = 1 \sum \infty \frac{n !}{5 ^{n}}$ Forecast: ab factorial upar hai. Guess: diverges, $L = \infty$ .

Step 1. Ratio: $ρ_{n} = \frac{( n + 1 )!}{5 ^{n + 1}} \cdot \frac{5 ^{n}}{n !} = \frac{( n + 1 )!}{n !} \cdot \frac{5 ^{n}}{5 ^{n + 1}} = (n + 1) \cdot \frac{1}{5} .$ Yeh step kyun? Pehle jaisi split; yahan $\frac{( n + 1 )!}{n !} = n + 1$ badhta hai, aur $\frac{5 ^{n}}{5 ^{n + 1}} = \frac{1}{5}$ ek constant brake hai — lekin ek constant growing factor ko rok nahi sakta.

Step 2. Limit: $L = lim_{n \to \infty} \frac{n + 1}{5} = \infty > 1 \Rightarrow diverges.$ Yeh step kyun? $L = \infty$ divergence case mein aata hai; terms eventually blow up kar jaati hain.

Verify: Growth argue karo ratio se khud, kisi ek lucky term se nahi. $n \geq 5$ ke liye humein milta hai $ρ_{n} = \frac{n + 1}{5} \geq \frac{6}{5} > 1$ , toh har baad ki term strictly pehle se badi hai: $a_{n + 1} > a_{n}$ for all $n \geq 5$ . Strictly increasing positive sequence $0$ ki taraf tend nahi kar sakti, toh $a_{n} \neq \to 0$ aur Term Test (nth-term divergence) divergence force karta hai. (Ek single value jaise $a_{20} \approx 2.5 \times 1 0^{4}$ sirf illustrate karta hai; reason hai ratio ka $1$ se upar rehna.) ✓

Cell C — sign-changing terms (absolute values matter)

Worked example Example 3 ·

$n = 1 \sum \infty \frac{( - 1 ) ^{n} 4 ^{n}}{( 2 n )!}$ Forecast: $(- 1)^{n}$ signs flip karta hai, lekin ratio test $∣ \cdot ∣$ use karta hai jo sign erase karta hai. $(2 n)!$ $4^{n}$ ko crush kar deta hai. Guess: absolutely converges, $L = 0$ .

Step 1. Absolute ratio (sign $∣ \cdot ∣$ ke andar vanish ho jaata hai): $ρ_{n} = \frac{( - 1 ) ^{n + 1} 4 ^{n + 1} / ( 2 n + 2 )!}{( - 1 ) ^{n} 4 ^{n} / ( 2 n )!} = \frac{4 ^{n + 1}}{4 ^{n}} \cdot \frac{( 2 n )!}{( 2 n + 2 )!} .$ Yeh step kyun? $∣(- 1)^{n + 1} / (- 1)^{n} ∣ = 1$ , toh absolute values rakhne se hum series ko positive series ki tarah treat kar sakte hain. Sign-changing series par $∣ \cdot ∣$ kabhi mat chhodo: bina iske raw ratio $a_{n + 1} / a_{n}$ negative hoga, aur negative "ratio" ko geometric bound se compare nahi kar sakte (jiske liye positive $r < 1$ chahiye). Poora geometric-domination argument silently break ho jaata hai. Isliye definition insist karta hai $∣ a_{n + 1} / a_{n} ∣$ par.

Step 2. Simplify: $\frac{4 ^{n + 1}}{4 ^{n}} = 4$ , aur $(2 n + 2)! = (2 n + 2) (2 n + 1) (2 n)!$ toh $\frac{( 2 n )!}{( 2 n + 2 )!} = \frac{1}{( 2 n + 2 ) ( 2 n + 1 )}$ . $ρ_{n} = \frac{4}{( 2 n + 2 ) ( 2 n + 1 )} .$ Yeh step kyun? Factorial ko do steps neeche expand karne se exactly pata chalta hai kaunse factors cancel hote hain.

Step 3. Limit: $L = lim_{n \to \infty} \frac{4}{( 2 n + 2 ) ( 2 n + 1 )} = 0 < 1 \Rightarrow converges absolutely.$

Verify (do checks). (i) Signed sum hai $cos (2) - 1$ ( $cos x = \sum (- 1)^{n} x^{2 n} / (2 n)!$ at $x = 2$ ): $cos 2 - 1 \approx - 1.416$ , finite → series converge karti hai. (ii) Absolute sum $\sum \frac{4 ^{n}}{( 2 n )!} = cosh (2) - 1 \approx 2.762$ (kyunki $cosh x = \sum x^{2 n} / (2 n)!$ at $x = 2$ ) bhi finite hai — yeh woh check hai jo actually absolute convergence confirm karta hai, exactly wahi claim jo ratio test ne ki thi. ✓

Cell D — do exponentials, finite ratio

Worked example Example 4 ·

$n = 1 \sum \infty \frac{3 ^{n}}{4 ^{n} + n}$ (geometric figure) Forecast: large $n$ ke liye $+ n$ , $4^{n}$ ke next mein negligible hai, toh terms $(3/4)^{n}$ jaisi behave karti hain. Guess: converges with $L = 3/4$ .

Step 1. Ratio: $ρ_{n} = \frac{3 ^{n + 1}}{4 ^{n + 1} + ( n + 1 )} \cdot \frac{4 ^{n} + n}{3 ^{n}} = 3 \cdot \frac{4 ^{n} + n}{4 ^{n + 1} + n + 1} .$ Yeh step kyun? Standard split; messy denominators abhi ke liye rehte hain — hum inhe divide karke clean karte hain.

Step 2. Fraction ke top aur bottom ko $4^{n}$ se divide karo: $ρ_{n} = 3 \cdot \frac{1 + n / 4 ^{n}}{4 + ( n + 1 ) / 4 ^{n}} .$ Yeh step kyun? Dominant term $4^{n}$ se divide karne par sub-dominant $n / 4^{n} \to 0$ visible ho jaata hai, true long-run ratio isolate ho jaata hai.

Step 3. Limit: $n / 4^{n} \to 0$ , toh $L = 3 \cdot \frac{1 + 0}{4 + 0} = \frac{3}{4} < 1 \Rightarrow converges.$

Figure — Cell D (log-scale term plot). Horizontal axis term index

n = 1 \dots 12

hai; vertical axis (logarithmic) term value

a_{n}

hai. Blue curve pure geometric sequence

(3/4)^{n}

hai; orange curve hamare actual terms

3^{n} / (4^{n} + n)

hain. Dono curves almost ek doosre ke upar hain aur dono log scale par ek straight line mein girte hain (geometric decay ki pehchaan), ek thin green band ke saath jo woh choti gap mark karta hai jahan geometric curve hamare se thodi upar hai. Kyunki hamare terms ek convergent Geometric Series of ratio

3/4 < 1

se dominate hain, Comparison Test proof khatam karta hai — algebraic result

L = 3/4

se match karta hai.

Verify: $ρ_{n}$ numerically compute karo $n = 15$ par: yeh $0.75$ ke bahut close hona chahiye. ✓

Cell F — power series → radius of convergence & endpoints

Worked example Example 6 ·

$n = 1 \sum \infty \frac{x ^{n}}{n ^{2}}$ Forecast: power series par ratio test $L = ∣ x ∣$ produce karta hai, ek Radius of Convergence deta hai. Phir dono endpoints $x = \pm 1$ par $L = 1$ hit karta hai aur alag checks chahiye.

Step 1. Ratio (kyunki $x$ negative ho sakta hai isliye $∣ \cdot ∣$ rakho): $ρ_{n} = \frac{∣ x ∣ ^{n + 1}}{( n + 1 ) ^{2}} \cdot \frac{n ^{2}}{∣ x ∣ ^{n}} = ∣ x ∣ \cdot \frac{n ^{2}}{( n + 1 ) ^{2}} .$ Yeh step kyun? $x$ -powers $∣ x ∣$ dete hain; polynomial part ek $\to 1$ factor hai.

Step 2. Limit: $n^{2} / (n + 1)^{2} \to 1$ , toh $L = ∣ x ∣$ . $∣ x ∣ < 1$ ke liye converge karta hai, $∣ x ∣ > 1$ ke liye diverge karta hai → radius $R = 1$ . Yeh step kyun? $L < 1$ exactly $∣ x ∣ < 1$ hai; boundary $∣ x ∣ = 1$ open chhodi hai.

Step 3 — endpoints ( $L = 1$ , test blind).

$x = 1$ : $\sum 1/ n^{2} = ζ (2) = π^{2} /6 \approx 1.645$ → converges (p-series, $p = 2$ ).
$x = - 1$ : $\sum (- 1)^{n} / n^{2}$ → absolutely convergent (iska absolute version wahi $\sum 1/ n^{2}$ hai) → converges. Yeh step kyun? $L = 1$ par ratio test chhod do aur har endpoint directly inspect karo.

Interval of convergence: $[- 1, 1]$ .

Verify: $x = 1$ par value $π^{2} /6 \approx 1.645$ hai; $x = - 1$ par sum $- π^{2} /12 \approx - 0.822$ hai — dono finite. ✓

Cell G — degenerate input (a zero term)

Worked example Example 7 ·

$n = 1 \sum \infty a_{n}, a_{n} = \frac{1 - cos ( nπ )}{2} \cdot \frac{1}{2 ^{n}}$ Forecast: $1 - cos (nπ)$ even $n$ ke liye $0$ hai aur odd $n$ ke liye $2$ . Toh har even term zero hai — ratio $a_{n + 1} / a_{n}$ $0$ se divide karta hai aur undefined hai. Ratio test ko eventually-nonzero terms chahiye; yahan woh kabhi settle nahi karti. Hume patch karna hoga.

Step 1. Zeros spot karo. $cos (nπ) = (- 1)^{n}$ , toh $\frac{1 - ( - 1 ) ^{n}}{2}$ hai $1$ (odd $n$ ke liye) ya $0$ (even $n$ ke liye). Is tarah $a_{n} = {1/ 2^{n}, 0, n odd n even.$ Yeh step kyun? Blindly ratio form karne se pehle, hamesha check karo ki koi term vanish toh nahi kar rahi — zero denominator formula kill kar deta hai.

Step 2. Sirf nonzero terms rakho. Nonzero terms exactly odd indices hain, jinhe hum $n = 1, 3, 5, \dots$ list kar sakte hain. Ek naya counter $k$ introduce karo $n = 2 k + 1$ set karke; jaise $k = 0, 1, 2, \dots$ yeh har odd $n$ se guzarta hai ( $k = 0 \to n = 1$ , $k = 1 \to n = 3$ , …). Phir $2^{n} = 2^{2 k + 1}$ aur surviving series hai $\sum_{k = 0}^{\infty} \frac{1}{2 ^{2 k + 1}} = \sum_{k = 0}^{\infty} \frac{1}{2} \cdot \frac{1}{4 ^{k}} .$ Yeh step kyun? Nonzero terms par re-indexing karna (explicit substitution $n = 2 k + 1$ ke saath) ek degenerate case ko clean Geometric Series mein badal deta hai.

Step 3. Yeh geometric hai first term $1/2$ aur ratio $1/4 < 1$ ke saath → converges to $\frac{1/2}{1 - 1/4} = \frac{1/2}{3/4} = \frac{2}{3} .$ Yeh step kyun? Geometric sum $\frac{a}{1 - r}$ ; ratio $< 1$ convergence guarantee karta hai — ratio test ki zaroorat nahi.

Verify: $\frac{2}{3} \approx 0.6667$ ; partial sum $1/2 + 1/8 + 1/32 + \dots$ iske taraf badhta hai. ✓

Takeaway: jab terms zero hit karti hain, ratio test ka formula undefined hai — zeros strip karo, ya Comparison Test/Root Test use karo (root test zero terms tolerate karta hai kyunki woh $n ∣ a_{n} ∣$ use karta hai, koi division nahi).

Cell H — real-world word problem (drug dosing)

Worked example Example 8 · Steady-state medicine

Ek patient roz ek pill leta hai. Body itna clear karti hai ki sirf $70%$ har dose ka ek din baad bachta hai. Agar har pill $200$ mg deliver karta hai, toh long run mein (just after $n$ -th dose, as $n \to \infty$ ) body mein kitna drug hai? Kya total build-up finite rehta hai?

Forecast: doses stack hoti hain lekin har purani dose $\times 0.7$ per day decay karti hai — ek geometric pile-up. Ratio $0.7 < 1$ → finite.

Step 1. $k$ din pehle li gayi dose ki amount model karo: yeh $200 \cdot (0.7)^{k}$ mg tak decay ho gayi hai. Aaj ki dose ke baad, total hai $A = k = 0 \sum \infty 200 (0.7)^{k}$ . Yeh step kyun? Har dose independently age hoti hai; saari past doses ko sum karne se current load milta hai.

Step 2. Series $\sum 200 (0.7)^{k}$ par ratio test karo: $ρ_{k} = \frac{200 ( 0.7 ) ^{k + 1}}{200 ( 0.7 ) ^{k}} = 0.7 = L < 1 \Rightarrow converges.$ Yeh step kyun? Ek pure geometric series ka constant ratio uske common ratio ke equal hota hai — sabse clean possible ratio test.

Step 3. Sum karo: $A = \frac{200}{1 - 0.7} = \frac{200}{0.3} \approx 666.7 mg (steady state).$ Yeh step kyun? Convergence (ratio $< 1$ ) humein geometric closed form use karne deta hai.

Verify: units poore mein mg hain; $200/0.3 = 666. \overline{6}$ mg — ek bounded steady-state, toh drug nahi accumulate karta dangerous infinity tak. ✓

Cell I — exam twist: $L = 1$ , but Raabe rescues

Worked example Example 9 ·

$n = 1 \sum \infty \frac{1 \cdot 3 \cdot 5 \dots ( 2 n - 1 )}{2 \cdot 4 \cdot 6 \dots ( 2 n )} \cdot \frac{1}{n}$ Forecast: double-product ratio $\to 1$ simplify ho jaata hai, toh ratio test $L = 1$ dega (inconclusive). Yeh classic exam trap hai — expect kiya jaata hai ki tum Raabe tak escalate karo.

Step 1. Double product ko $c_{n} = \frac{1 \cdot 3 \dots ( 2 n - 1 )}{2 \cdot 4 \dots ( 2 n )}$ kaho, toh $a_{n} = c_{n} / n$ . Iska ratio: $\frac{c _{n + 1}}{c _{n}} = \frac{2 n + 1}{2 n + 2}, \frac{a _{n + 1}}{a _{n}} = \frac{2 n + 1}{2 n + 2} \cdot \frac{n}{n + 1} .$ Yeh step kyun? Product ko ek step aage badhana naye numerator/denominator factor $(2 n + 1) / (2 n + 2)$ se multiply karta hai; extra $1/ n$ $n / (n + 1)$ contribute karta hai.

Step 2. Limit: $L = lim_{n \to \infty} \frac{2 n + 1}{2 n + 2} \cdot \frac{n}{n + 1} = 1 \cdot 1 = 1 \Rightarrow ratio test inconclusive.$ Yeh step kyun? Dono factors $1$ approach karte hain — crowded boundary phir se aa gayi.

Step 3 — Raabe tak escalate karo. Terms positive hain, toh Raabe ki hypothesis hold karti hai. Upar di gayi definition use karte hue, $\frac{a _{n}}{a _{n + 1}} = \frac{( 2 n + 2 ) ( n + 1 )}{( 2 n + 1 ) n} = \frac{2 n ^{2} + 4 n + 2}{2 n ^{2} + n},$ $\frac{a _{n}}{a _{n + 1}} - 1 = \frac{3 n + 2}{2 n ^{2} + n}, R = lim_{n \to \infty} n \cdot \frac{3 n + 2}{2 n ^{2} + n} = \frac{3}{2} .$ Kyunki $R = \frac{3}{2} > 1$ → converges. Yeh step kyun? Raabe us rate mein zoom in karta hai jis par ratio $1$ approach karta hai — woh finer ruler jo parent note ne promise kiya tha jab ratio test blind hota hai.

Verify: $R = 3/2 = 1.5 > 1$ , aur leading behaviour $a_{n} \sim \frac{C}{n ^{3/2}}$ (ek $p = 3/2 > 1$ series) independently convergence confirm karta hai p-Series and Integral Test se. ✓

Cell J — jab limit $L$ exist hi nahi karta

Worked example Example 10 ·

$n = 1 \sum \infty a_{n}$ with $a_{n} = {1/ 2^{n}, 1/ 3^{n}, n even n odd$ Forecast: ratios do bahut alag values ke beech alternate karenge, toh koi single $L$ exist nahi karega — ratio test silent ho jaayega aur hum doosre tarike se finish karenge.

Step 1. Consecutive ratios dekho. Odd $n$ se even $n + 1$ aur vice versa jaane par wildly different sizes milte hain, e.g. $\frac{a _{2}}{a _{1}} = \frac{1/4}{1/3} = \frac{3}{4}, \frac{a _{3}}{a _{2}} = \frac{1/27}{1/4} = \frac{4}{27} \approx 0.148,$ aur do alternating values ka yeh pattern hamesha ke liye repeat hota hai. Yeh step kyun? Hum kuch $ρ_{n}$ compute karte hain taaki dikhe ki woh ek single number approach nahi kar rahe.

Step 2. Conclude karo ki ratio test inapplicable hai: $lim_{n \to \infty} ρ_{n}$ exist nahi karta, toh $L < 1/ L > 1/ L = 1$ trichotomy ke paas kuch feed karne ko nahi hai. Yeh step kyun? Koi limit nahi ⇒ koi verdict nahi; yeh recognize karna tumhe ek meaningless " $L$ " likhne se bachata hai.

Step 3 — comparison se finish karo. Har term satisfy karta hai $0 \leq a_{n} \leq 1/ 2^{n}$ , aur $\sum 1/ 2^{n}$ ek convergent Geometric Series hai. Comparison Test ke according series converges. Yeh step kyun? Jab ratio test silent ho, ek direct geometric bound phir bhi settle kar deta hai.

Verify: parity se split karo aur do geometric series sum karo — even part $\sum_{m \geq 1} 1/ 4^{m} = 1/3$ , odd part $\sum_{m \geq 0} 1/ 3^{2 m + 1} = 3/8$ ; total $= 1/3 + 3/8 = 17/24 \approx 0.708$ , finite. ✓

Recall Kaun se cell ko kaun sa tool chahiye?

Exponential-vs-factorial (Cells A–D) → ::: ratio test cleanly jeet jaata hai. Pure power law $1/ n^{p}$ (Cell E) → ::: p-series / integral test. Power series endpoints jahan $L = 1$ ho (Cell F) → ::: har endpoint haath se check karo. Koi term zero ho, ratio undefined ho (Cell G) → ::: zeros strip karo / root test use karo. Real-world geometric pile-up (Cell H) → ::: ratio common ratio ke equal hai; geometric sum use karo. $L = 1$ factorial-product ratio se (Cell I) → ::: Raabe's test. Ratio oscillate kare aur $L$ exist na kare (Cell J) → ::: ratio test silent hai; comparison use karo (ya limsup root test).

Connections

Ratio test — proof, limitations (index 4.3.12) — parent: proof aur theory.
Geometric Series — Cells D, G, H, J dil se geometric hain.
Comparison Test — Cells D, G aur J finish karta hai.
Term Test (nth-term divergence) — Cell B mein sanity check.
p-Series and Integral Test — Cells E aur F endpoints ke liye sahi tool.
Radius of Convergence — Cell F yahi compute karta hai.
Root Test — Cell G ke liye zero-tolerant sibling aur Cell J ke liye limsup rescue.
Raabe's Test — Cell I ke liye sharper ruler.

Worked examples — Ratio test — proof, limitations