Absolute vs conditional convergence: Worked examples

4.3.11 · D3 · Maths › Calculus III — Sequences & Series › Absolute vs conditional convergence

Yeh page drill hai. Parent note ne aapko theory di; yahan hum har tarah ki series ke through chalte hain jo yeh topic aap par throw kar sakta hai — ek worked example har case ke liye. Har example mein aap pehle guess karte ho, phir step by step reasoning hoti hai.

Shuru karne se pehle, woh ek sawaal yaad karo jo har jagah kaam aata hai:

Cells A, D, F, G, H ke peeche ek named result ka engine hai — aao isko clearly state kar dete hain taaki yeh kabhi mystery ki tarah invoke na ho:

Doosra engine — jab twin fail ho tab use hota hai — Alternating Series Test hai. Kyunki hum ise cells B, C aur I mein lean karte hain, aao ise poori tarah state kar dete hain taaki yeh kabhi black box na rahe:

Scenario matrix

Is topic ka har problem inhi classes mein se ek mein aata hai. Neeche ke examples un class labels ke saath hain jo wo hit karte hain, toh end tak aap ne sab dekh liye honge.

#	Case class	Tricky kya hai	Example
A	Alternating, twin ek $p$ -series hai $p > 1$ ke saath	signs irrelevant → absolute	Ex 1
B	Alternating, twin harmonic series hai ( $p = 1$ )	classic conditional trap	Ex 2
C	Alternating, twin $p$ -series hai $0 < p < 1$ ke saath	twin diverge karta hai, par AST abhi bhi save karta hai	Ex 3
D	Geometric with a sign, ratio test	jaldi absolute verdict	Ex 4
E	Terms $0$ tak jaate hi nahi	pehli hi gate pe fail → divergent	Ex 5
F	Zero / degenerate terms mixed in	kya zeros kuch break karte hain?	Ex 6
G	Non-obvious signs ( $sin n$ , pure $(- 1)^{n}$ nahi)	absolute test via Comparison Test	Ex 7
H	Word problem (oscillating physical sum)	story → series mein translate karo	Ex 8
I	Exam twist: [[Power series & radius of convergence	power series]] ke andar value $x$ , endpoint pe behaviour	absolute vs conditional at the boundary

Hum inhe order mein lete hain.

Case A — alternating, twin converge karta hai (absolute)

Worked example Ex 1 —

$n = 1 \sum \infty \frac{( - 1 ) ^{n}}{n ^{3}}$ Forecast: abhi guess karo — absolutely convergent, conditionally convergent, ya divergent?

Step 1. All-positive twin banao: $∣ a_{n} ∣ = \frac{( - 1 ) ^{n}}{n ^{3}} = \frac{1}{n ^{3}}$ . Yeh step kyun? Absolute convergence twin $\sum ∣ a_{n} ∣$ se define hoti hai, isliye hum pehle ise build karte hain.

Step 2. $\sum \frac{1}{n ^{3}}$ ko $p = 3$ wali p-series recognize karo. Yeh step kyun? $p$ -series sabse fast classifier hai jo hamare paas hai: yeh converge karta hai exactly jab $p > 1$ .

Step 3. Kyunki $p = 3 > 1$ , twin converge karta hai. Absolute Convergence Theorem (upar stated) ke zariye original bhi converge karta hai. Yeh step kyun? Ek baar twin converge kar le, toh woh theorem hume $\sum a_{n}$ ki convergence free mein deta hai — koi Alternating Series Test ki zaroorat nahi.

Verdict: absolutely convergent (cell A).

Verify: partial sum $\sum_{n = 1}^{6} \frac{( - 1 ) ^{n}}{n ^{3}} \approx - 0.8971$ , aur true value $- \frac{3}{4} ζ (3) \approx - 0.9015$ hai — partial sums pehle se limit ke kaafi paas hain, exactly wahi jo robust convergence dikhti hai. (Yahan $ζ (3) = \sum_{n \geq 1} \frac{1}{n ^{3}}$ sirf "us $p = 3$ sum ki value" ka shorthand hai; aapko zeta function ki koi bhi theory nahi chahiye, sirf itna ki yeh ek finite number $\approx 1.202$ ko name karta hai.)

Case B — alternating, twin harmonic hai (conditional trap)

Worked example Ex 2 —

$n = 1 \sum \infty \frac{( - 1 ) ^{n + 1}}{n}$ Forecast: absolute, conditional, ya divergent? (Yeh model conditional series hai — guess commit karo.)

Step 1. Twin: $∣ a_{n} ∣ = \frac{1}{n}$ , toh $\sum ∣ a_{n} ∣ = \sum \frac{1}{n}$ — Harmonic series. Yeh step kyun? Hamesha pehle twin test karo.

Step 2. Harmonic series $p = 1$ wali $p$ -series hai, jo diverge karti hai. Toh series absolutely convergent nahi hai. Yeh step kyun? $p = 1$ exactly woh borderline hai jahan $p$ -series converge karna band kar deti hai.

Step 3. Ab original ko Alternating Series Test se test karo. $b_{n} = \frac{1}{n}$ likhne par: (i) positive, (ii) decreasing ( $\frac{1}{n + 1} < \frac{1}{n}$ ), (iii) $\to 0$ . Teeno AST conditions hold karti hain, toh $\sum a_{n}$ converge karta hai. Yeh step kyun? Twin diverge ho gaya, toh hum recipe ke AST branch par aa jaate hain.

Verdict: conditionally convergent (cell B). Yeh $ln 2$ ke barabar hai.

Verify: $\sum_{n = 1}^{8} \frac{( - 1 ) ^{n + 1}}{n} \approx 0.6345$ vs $ln 2 \approx 0.6931$ — converge ho raha hai, lekin slowly (partial sums dhheere dhheere chalte hain), fragile convergence ki pehchaan.

Case C — alternating, twin $p < 1$ (phir bhi conditional)

Worked example Ex 3 —

$n = 1 \sum \infty \frac{( - 1 ) ^{n + 1}}{n}$ Forecast: yahan twin harmonic series se bhi zyada hard diverge karta hai — kya original phir bhi converge karta hai?

Step 1. Twin: $∣ a_{n} ∣ = \frac{1}{n} = \frac{1}{n ^{1/2}}$ , $p = \frac{1}{2}$ wali p-series. Yeh step kyun? Twin banao, $p$ se classify karo.

Step 2. $p = \frac{1}{2} < 1$ , toh twin diverge karta hai. Absolutely convergent nahi. Yeh step kyun? Koi bhi $p \leq 1$ $p$ -series ko diverge kara deta hai.

Step 3. Original par Alternating Series Test apply karo $b_{n} = \frac{1}{n}$ ke saath: positive ✓, decreasing ✓ (bada $n$ → bada $n$ → chhota reciprocal), $\to 0$ ✓. Toh $\sum a_{n}$ converge karta hai. Yeh step kyun? Twin fail ho gaya; AST fallback hai aur teeno conditions pass hoti hain.

Verdict: conditionally convergent (cell C).

Verify: $\sum_{n = 1}^{10} \frac{( - 1 ) ^{n + 1}}{n} \approx 0.6068$ ; true value $\approx 0.6049$ hai. AST error bound (upar wale theorem se) kehta hai ki error $\leq$ pehla omitted term $\frac{1}{11} \approx 0.3015$ hai — aur sach mein $∣0.6068 - 0.6049∣ \approx 0.0019$ iske andar hai.

Case D — geometric, ratio test (absolute)

Worked example Ex 4 —

$n = 1 \sum \infty \frac{( - 3 ) ^{n}}{4 ^{n}}$ Forecast: ratio $\frac{- 3}{4} = \frac{3}{4} < 1$ — yeh aapko kya batata hai?

Step 1. Twin: $∣ a_{n} ∣ = \frac{3 ^{n}}{4 ^{n}} = (\frac{3}{4})^{n}$ . Yeh step kyun? Hum absolute convergence chahte hain, isliye pehle sign strip karo.

Step 2. Twin par Ratio Test apply karo: $L = lim_{n \to \infty} \frac{∣ a _{n + 1} ∣}{∣ a _{n} ∣} = lim_{n \to \infty} \frac{( 3/4 ) ^{n + 1}}{( 3/4 ) ^{n}} = \frac{3}{4} .$ Yeh step kyun? $∣ a_{n} ∣$ par Ratio Test directly absolute convergence test karta hai — aur yeh natural tool hai jab terms powers hote hain.

Step 3. Kyunki $L = \frac{3}{4} < 1$ , twin converge karta hai → series absolutely convergent hai. Yeh step kyun? Ratio test rule: $L < 1 \Rightarrow$ $\sum ∣ a_{n} ∣$ convergence.

Verdict: absolutely convergent (cell D). Pehle term $- \frac{3}{4}$ aur ratio $- \frac{3}{4}$ wali geometric series ke roop mein, sum $\frac{- 3/4}{1 - ( - 3/4 )} = - \frac{3}{7}$ hai.

Verify: $\frac{- 3/4}{1 + 3/4} = \frac{- 3/4}{7/4} = - \frac{3}{7} \approx - 0.4286$ , aur $\sum_{n = 1}^{12} \approx - 0.4286$ already. ✓

Case E — terms zero tak nahi pahunchte (divergent)

Worked example Ex 5 —

$n = 1 \sum \infty (- 1)^{n} \frac{n}{n + 1}$ Forecast: signs nicely alternate karte hain — "conditional" kehne ka temptation hoga. Kya yeh trap hai?

Step 1. Magnitude dekho: $∣ a_{n} ∣ = \frac{n}{n + 1}$ . Jaise $n \to \infty$ , yeh $\to 1$ jaata hai, $0$ nahi. Yeh step kyun? Kisi bhi convergence test se pehle, kisi bhi convergent series ke terms $0$ tak jaane chahiye — term test / divergence test. Pehle yeh gate check karo.

Step 2. Kyunki $a_{n}$ $+ 1$ aur $- 1$ ke paas values ke beech oscillate karta hai aur $0$ ke paas nahi jaata, series diverge karti hai. Yeh step kyun? Agar $a_{n} \neq \to 0$ , toh partial sums settle nahi ho sakte. Yeh series ko AST ya twin ke relevant hone se pehle hi maar deta hai.

Step 3. Yahan AST try mat karo — AST ko magnitudes $\to 0$ chahiye (condition 3), jo fail ho rahi hai, isliye AST apply nahi hota. Divergence test ne pehle hi verdict de diya. Yeh step kyun? AST reach karna jab uski hypothesis fail ho rahi ho, ek classic error hai; divergence test ne pehle hi verdict de diya tha.

Verdict: divergent (cell E) — recipe mein teesra outcome: na absolutely na conditionally convergent, simply koi finite sum nahi hai.

Verify: odd partial sums $- 0.5$ ke paas hain aur even wale $0$ ke paas; $a_{100} = \frac{100}{101} \approx 0.990 \neq = 0$ , confirm karta hai ki terms vanish nahi hote.

Case F — degenerate / zero terms (kya convergence survive karta hai?)

Worked example Ex 6 —

$n = 1 \sum \infty a_{n}, a_{n} = ⎩ ⎨ ⎧ \frac{( - 1 ) ^{n + 1}}{n ^{2}} 0 n odd n even$ Forecast: aadhe terms literally $0$ hain. Kya zeros insert karne se convergence ya type change hoti hai?

Step 1. Zeros kisi bhi partial sum mein contribute nahi karte. Toh $\sum a_{n}$ ke same partial sums hain (constant padding ke saath) jaise $\sum_{k odd} \frac{1}{k ^{2}}$ , positive terms ka ek sub-sum. Yeh step kyun? $0$ ke barabar ek term add karna running total kabhi nahi badalta — degenerate terms invisible hote hain.

Step 2. Twin: $\sum ∣ a_{n} ∣ = \sum_{k odd} \frac{1}{k ^{2}} \leq \sum_{n = 1}^{\infty} \frac{1}{n ^{2}}$ , jo converge karta hai ( $p = 2$ p-series). Comparison Test ke zariye, twin converge karta hai. Yeh step kyun? Ek positive series ko ek known convergent series se bound karna prove karta hai ki twin converge karta hai.

Step 3. Twin converges ⇒ absolutely convergent (Absolute Convergence Theorem). Zeros na help karte hain na hurt. Yeh step kyun? Absolute convergence poori tarah twin se decide hoti hai, aur $0$ ka twin $0$ hai.

Verdict: absolutely convergent (cell F).

Verify: $\sum_{k odd, k \leq 9} \frac{1}{k ^{2}} = 1 + \frac{1}{9} + \frac{1}{25} + \frac{1}{49} + \frac{1}{81} \approx 1.1838$ ; poora odd-square sum $\frac{π ^{2}}{8} \approx 1.2337$ hai — bounded aur converging. ✓

Case G — non-obvious signs (absolute ke liye comparison)

Worked example Ex 7 —

$n = 1 \sum \infty \frac{sin n}{n ^{2}}$ Forecast: $sin n$ ka sign unpredictably jump karta hai — clean $(- 1)^{n}$ nahi. Kya hum phir bhi decide kar sakte hain?

Step 1. Twin: $∣ a_{n} ∣ = \frac{∣ sin n ∣}{n ^{2}}$ . Bound $∣ sin n ∣ \leq 1$ use karo sabhi $n$ ke liye. Yeh step kyun? $sin$ $[- 1, 1]$ mein trapped hai, isliye uski absolute value kabhi $1$ se exceed nahi karta — yeh ek clean over-estimate deta hai.

Step 2. Isliye $\frac{∣ sin n ∣}{n ^{2}} \leq \frac{1}{n ^{2}}$ . Kyunki $\sum \frac{1}{n ^{2}}$ converge karta hai ( $p = 2$ ), Comparison Test $\sum ∣ a_{n} ∣$ ko converge hone ke liye force karta hai. Yeh step kyun? Jab signs messy hoon, AST useless hota hai — lekin twin ek nice convergent series se bounded hai, toh hum seedha absolute convergence ke liye jaate hain.

Step 3. Twin converges ⇒ original absolutely converge karta hai. Yeh step kyun? Absolute Convergence Theorem apply hota hai ek baar twin settle ho jaaye.

Verdict: absolutely convergent (cell G).

Verify: $\sum_{n = 1}^{20} \frac{s i n n}{n ^{2}} \approx 1.0121$ ; har partial sum bound $\sum \frac{1}{n ^{2}} = \frac{π ^{2}}{6} \approx 1.6449$ ke andar rehta hai. ✓

Case H — word problem (oscillating physical sum)

Ek number line par robot $0$ se start karta hai. Step $n$ par woh $\frac{1}{2 ^{n}}$ metres ki distance move karta hai, lekin har step direction flip karta hai: right, left, right, left, … Woh kahan pahunchega, aur kya woh final position "robust" hai?

Figure — Absolute vs conditional convergence

Upar ki figure yahi derivation draw karke dikhati hai: steps ke saath left-to-right padho. Har coloured arrow ek term $\frac{( - 1 ) ^{n + 1}}{2 ^{n}}$ hai — magenta arrow step 1 hai (right, length $\frac{1}{2}$ ), violet arrow step 2 (left, length $\frac{1}{4}$ ), orange arrow step 3 (right, length $\frac{1}{8}$ ), aur aise aage. Picture do cheezein obvious banati hai: (a) har arrow pehle wale se exactly aadhi length ka hai, aur (b) tips $x = \frac{1}{3}$ wali dashed navy line (red dot) ke paas paas march karte hain bina zyada door cross kiye. Woh "closing in" hi convergence hai visually.

Worked example Ex 8 — total displacement

$n = 1 \sum \infty \frac{( - 1 ) ^{n + 1}}{2 ^{n}}$ Forecast: robot zig-zag karta hai ever-smaller steps ke saath. Absolute, conditional, ya divergent displacement?

Step 1. Model: rightward $+$ hai, leftward $-$ hai, step size $\frac{1}{2 ^{n}}$ . Total displacement $= \sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n + 1}}{2 ^{n}}$ . Yeh exactly figure mein arrow chain hai — magenta → violet → orange arrows trace karo aur dekho kaise red dot ki taraf shrink hote hain. Yeh step kyun? "Halving steps ka zig-zag" ko signs aur magnitudes mein translate karna poora modelling job hai.

Step 2. Twin: $∣ a_{n} ∣ = \frac{1}{2 ^{n}} = (\frac{1}{2})^{n}$ , ratio $\frac{1}{2} < 1$ wali geometric series → converge karti hai (total path length finite hai, $\sum \frac{1}{2 ^{n}} = 1$ ). Yeh step kyun? Agar direction ignore karke bhi total distance travelled finite ho, toh position robustly pin down hai.

Step 3. Twin converges ⇒ absolutely convergent (Absolute Convergence Theorem). Sum $= \frac{1/2}{1 - ( - 1/2 )} = \frac{1/2}{3/2} = \frac{1}{3}$ . Yeh step kyun? Geometric-series formula with first term $\frac{1}{2}$ , ratio $- \frac{1}{2}$ .

Verdict: absolutely convergent (cell H). Robot $x = \frac{1}{3}$ m par settle ho jaata hai — exactly figure mein red dot — aur yeh position order-independent hai: aap steps kisi bhi order mein le sakte ho aur $\frac{1}{3}$ par land karoge.

Verify: $\sum_{n = 1}^{10} \frac{( - 1 ) ^{n + 1}}{2 ^{n}} \approx 0.3330 \to \frac{1}{3}$ , aur total distance $\sum_{n = 1}^{10} \frac{1}{2 ^{n}} \approx 0.999 \to 1$ . ✓ (Units: metres throughout.)

Case I — exam twist: power series ka endpoint

Worked example Ex 9 —

$n = 1 \sum \infty \frac{x ^{n}}{n}$ ke liye, do endpoints $x = 1$ aur $x = - 1$ par convergence classify karo. Forecast: power series ka radius $1$ hai. Interesting drama hamesha endpoints par hota hai — har end par type guess karo.

Step 1. $x = 1$ par: series $\sum \frac{1 ^{n}}{n} = \sum \frac{1}{n}$ ban jaati hai, Harmonic series, jo diverge karti hai. Yeh step kyun? Endpoint value plug in karo; yahan yeh pure $p = 1$ series mein collapse ho jaata hai.

Step 2. Toh $x = 1$ par series divergent hai — na absolute na conditional. (Right endpoint excluded hai.) Yeh step kyun? Ek endpoint par divergence matlab woh endpoint interval of convergence ka part nahi hai.

Step 3. $x = - 1$ par: series $\sum \frac{( - 1 ) ^{n}}{n}$ ban jaati hai. Twin $= \sum \frac{1}{n}$ diverge karta hai (absolute nahi), lekin Alternating Series Test ke zariye ( $b_{n} = \frac{1}{n}$ decreasing $\to 0$ ke saath) series converge karti hai. Yeh step kyun? Same endpoint machinery, lekin ab alternating signs ise bacha leti hain.

Verdict: $x = 1$ par divergent; $x = - 1$ par conditionally convergent (cell I). Bilkul same magnitude series divergent ya conditionally convergent ho sakti hai — purely depend karta hai us sign par jo endpoint supply karta hai — yahi is poore topic ka essence hai.

Verify: $x = - 1$ par, $\sum_{n = 1}^{8} \frac{( - 1 ) ^{n}}{n} \approx - 0.6345 \to - ln 2 \approx - 0.6931$ ; $x = 1$ par partials $\sum_{n = 1}^{8} \frac{1}{n} \approx 2.7179$ chadhte rehte hain (koi limit nahi). ✓

Recall Kaun sa cell kaun sa tha? (self-quiz)

Twin $p > 1$ wali $p$ -series hai ::: absolutely convergent (cells A, F, G) Twin harmonic hai, original alternate karta hai ::: conditionally convergent (cells B, I at $x = - 1$ ) Twin $0 < p < 1$ wali $p$ -series hai, original alternate karta hai ::: conditionally convergent (cell C) Terms $0$ tak fail ho jaate hain ::: divergent — turant ruk jao (cell E) Geometric with ratio magnitude $< 1$ ::: absolutely convergent (cells D, H) $\sum x^{n} / n$ ka endpoint $x = 1$ ::: divergent (cell I)

Connections

4.3.11 Absolute vs conditional convergence — parent theory
Alternating Series Test — cells B, C, I ko bachaya
Comparison Test — cells F, G decide kiye
Ratio Test — cell D decide kiya
p-series — universal classifier
Harmonic series — conditional villain
Power series & radius of convergence — cell I mein endpoint drama
Riemann Rearrangement Theorem — isliye absolute/conditional split matter karta hai