Alternating series test — Leibniz test, proof: Worked examples

4.3.10 · D3 · Maths › Calculus III — Sequences & Series › Alternating series test — Leibniz test, proof

Yeh page parent Leibniz test note ki "sab kuch ek saath try karo" companion hai. Kuch bhi compute karne se pehle, hum har tarah ke case lay out karte hain jo ek alternating-series problem tumhare saamne rakh sakti hai. Phir har worked example us cell ke saath tag ki gayi hai jise wo cover karti hai, taaki end mein tumne poora board dekh liya ho.

Har jagah use hone wale objects ka reminder:

Neeche ki picture un do knobs ko ek "flag ki taraf hopping" walk mein convert karti hai — har example ke liye ise apne dimag mein rakho.

Figure — Alternating series test — Leibniz test, proof

The scenario matrix

Cell	Decreasing?	$lim b_{n}$	Leibniz ka Verdict	Covered by
C1 Clean pass	Haan	$0$	Converges	Ex 1
C2 "To zero" fail	Haan	$\neq = 0$	Diverges (nth-term test)	Ex 2
C3 "Decreasing" fail	Nahi (wiggles)	$0$	Test apply nahi hota	Ex 3
C4 Sirf eventually decreasing	Bade $n$ ke liye haan	$0$	Converges	Ex 4
C5 Degenerate / zero terms	trivially	$0$	Converges (finite sum)	Ex 5
C6 Accuracy / error-bound problem	Haan	$0$	Converges + terms count karo	Ex 6
C7 Real-world word problem	Haan	$0$	Converges + interpret karo	Ex 7
C8 Exam twist (power series ka endpoint)	Haan	$0$	Converges conditionally	Ex 8

Do knobs → chaar logical corners (C1–C4). C5 degenerate corner hai, C6–C8 "toh phir kya" applications hain. Chalo har cell bharte hain.

Worked examples

Ex 1 — Cell C1: the clean pass

Worked example Statement

Kya $n = 1 \sum \infty \frac{( - 1 ) ^{n - 1}}{2 n - 1} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots$ converge karta hai?

Forecast: padhne se pehle guess karo — decreasing hai? zero par jaata hai? Tumhara gut verdict kya hai?

$b_{n}$ identify karo. Sign hata do: $b_{n} = \frac{1}{2 n - 1}$ . Yeh step kyun? Leibniz hamesha sirf positive sizes $b_{n}$ ki baat karta hai, signs ki kabhi nahi. Hume pehle unhe isolate karna hoga.
Decreasing check karo. $2 (n + 1) - 1 = 2 n + 1 > 2 n - 1$ , toh bada denominator chhoti fraction deta hai: $b_{n + 1} < b_{n}$ . ✓ Yeh step kyun? Knob 1. Bada bottom, chhoti value — yahi reason hai ki yeh shrink karta hai.
Limit check karo. Jab $n \to \infty$ , $2 n - 1 \to \infty$ toh $\frac{1}{2 n - 1} \to 0$ . ✓ Yeh step kyun? Knob 2. Even/odd partial sums ke merge hone ke liye sizes ka vanish hona zaroori hai.
Conclude karo. Converges. Kyun? Dono knobs pass ho gaye, toh "flag ki taraf hopping" picture literally apply hoti hai: aage $1$ , peeche $\frac{1}{3}$ , aage $\frac{1}{5}$ … har hop chhota, toh walk ek hamesha-shrinking window mein trapped hai aur ek point par settle hona hi padega. Woh settling point hi sum hai.

Verify: Yeh famous series $π /4$ ke barabar hai (Leibniz ki apni formula). Numerically $\sum_{n = 1}^{5000}$ bilkul $π /4 \approx 0.7854$ ke paas baithti hai. Sanity check: yeh $s_{1} = 1$ aur $s_{2} = \frac{2}{3}$ ke beech hai, bilkul jaisa proof predict karta hai.

Ex 2 — Cell C2: sizes vanish nahi hoti

Worked example Statement

Kya $n = 1 \sum \infty (- 1)^{n - 1} \frac{3 n}{2 n + 5}$ converge karta hai?

Forecast: terms sign flip karte rehte hain — kya sirf flipping se yeh bach jaayega?

$b_{n} = \frac{3 n}{2 n + 5}$ identify karo. Kyun? Testing se pehle sizes isolate karo.
Limit compute karo. Upar aur neeche $n$ se divide karo: $\frac{3}{2 + 5/ n} \to \frac{3}{2} = 1.5 \neq = 0$ . Yeh step kyun? Jab trouble ka shak ho toh Knob 2 pehle check karo — agar size die nahi karta, toh baaki kuch bhi matter nahi karta.
nth-Term Test for Divergence invoke karo. Kyunki term size $0$ par nahi jaati, actual terms $(- 1)^{n - 1} b_{n}$ bhi $0$ par nahi jaate (yeh $\pm 1.5$ ke around oscillate karte hain). Jo series ke terms vanish nahi hote, woh converge nahi kar sakti. Yeh step kyun? Leibniz ek convergence checker hai; jab uski condition 2 fail ho, hum case nth-term test ko dete hain, jo hard diverges deta hai.
Conclude karo. Diverges. Kyun? Hopping picture mein hops kabhi shrink nahi karte — yeh hamesha roughly $1.5$ se jump karte rehte hain, toh walker kabhi chhoti window mein trapped nahi hota aur kabhi settle nahi kar sakta. Koi settling point nahi matlab koi sum nahi. Leibniz apply hi nahi hota.

Verify: $b_{100} = \frac{300}{205} \approx 1.463$ , $b_{1000} = \frac{3000}{2005} \approx 1.496$ — $1.5$ ki taraf chadh raha hai, $0$ ke paas bilkul nahi. ✓

Ex 3 — Cell C3: yeh wiggle karta hai

Worked example Statement

Us alternating series $\sum (- 1)^{n - 1} b_{n}$ ko consider karo jiske sizes yeh sequence hain $b_{n} : 1, \frac{1}{3}, \frac{1}{2}, \frac{1}{5}, \frac{1}{4}, \frac{1}{7}, \frac{1}{6}, \dots$ (parent note ka "wiggle": odd-position aur even-position values interleaved hain toh list upar-neeche jaati hai). Kya Leibniz apply hota hai?

Forecast: sizes clearly zero ki taraf ja rahi hain — kya Leibniz ke liye yeh kaafi hai?

Limit check karo. Har value growing $k$ ke liye $\frac{1}{k}$ mein se koi ek hai, toh $b_{n} \to 0$ . ✓ Kyun? Knob 2 pass ho gaya — victory declare karne ka temptation hoga.
Decreasing check karo. List in order padho: $b_{1} = 1 > b_{2} = \frac{1}{3}$ (neeche), lekin $b_{2} = \frac{1}{3} < b_{3} = \frac{1}{2}$ (upar). Toh yeh neeche phir upar jaata hai — monotone nahi. Yeh step kyun? Knob 1 fail ho gaya. Proof ne terms ko brackets $(b_{2 k - 1} - b_{2 k})$ mein group kiya tha aur zaroori tha ki har ek $\geq 0$ ho. Agar sizes kabhi-kabhi badhti hain, toh ek bracket negative ho sakta hai aur "monotone increasing even sums" argument collapse ho jaata hai.
Conclude karo. Leibniz koi conclusion nahi deta. Kyun? "Shrinking window mein trapped" ki guarantee is par rely karti thi ki har hop pichle se chhota ho. Yahan ek hop pichle wale se bada ho sakta hai, toh window phir se wide ho sakti hai — squeeze jo settling force karta hai woh toot jaata hai, aur test chup reh jaata hai. (Yeh doosre tools se phir bhi converge kar sakta hai, lekin yeh test kuch nahi keh sakta.)

Verify: Isi sequence ke liye, step 3 par rise hai $b_{3} - b_{2} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} > 0$ — ek genuine increase, toh monotonicity exactly jaisi claim ki gayi waise tooti hai. ✓

Ex 4 — Cell C4: sirf eventually decreasing

Worked example Statement

Kya $n = 1 \sum \infty (- 1)^{n - 1} \frac{ln n}{n}$ converge karta hai? (Note: $b_{1} = \frac{l n 1}{1} = 0$ , $b_{2} = \frac{l n 2}{2} \approx 0.347$ , $b_{3} \approx 0.366$ — shuru mein yeh badhta hai!)

Forecast: pehle couple of terms ke liye yeh increase karta hai — kya isse yeh kill ho jaata hai?

$b_{n} = \frac{ln n}{n}$ identify karo, limit. Growth rates se, $ln n$ rengta hai jabki $n$ dauraata hai, toh $\frac{ln n}{n} \to 0$ . ✓ Kyun? Pehle Knob 2.
Derivative se decreasing test karo. Maano $f (x) = \frac{ln x}{x}$ . Derivative kyun? Kyunki "kya yeh shrink kar raha hai?" bilkul wahi sawaal hai "kya slope negative hai?" — derivative woh tool hai jo measure karta hai ki koi smooth function upar jaata hai ya neeche. Quotient rule: $f^{'} (x) = \frac{1 - l n x}{x ^{2}} .$ Yeh exactly tab negative hai jab $ln x > 1$ , yaani $x > e \approx 2.718$ . Toh $b_{n}$ saare $n \geq 3$ ke liye decrease karta hai. Neeche ki figure yeh dikhati hai: curve chadhti hai, $x = e$ ke paas peak karti hai, phir hamesha ke liye girti hai. Yeh step kyun? Knob 1 ko sirf eventual decrease chahiye — Leibniz proof kehta hai "saare bade $n$ ke liye." Early terms ka ek finite pile sirf ek fixed number hai jo add ho jaata hai; yeh convergence nahi badal sakta.
Conclude karo. Converges (conditionally). Kyun? Pehle do rising terms sirf starting point ko ek fixed amount se shift karte hain; $n = 3$ se aage hops monotonically $0$ tak shrink karte hain, jo exactly woh shrinking-window setup hai jo settling force karta hai. Convergent tail mein add kiya ek fixed head-start phir bhi convergent hai. (Conditionally, kyunki $\sum \frac{l n n}{n}$ diverge karta hai.)

Verify: $f^{'} (4) = \frac{1 - ln 4}{16} = \frac{1 - 1.386}{16} \approx - 0.0242 < 0$ , confirm karta hai ki $e$ ke baad decrease hota hai. ✓

Ex 5 — Cell C5: degenerate / zero terms

Worked example Statement

$b_{n} = 0$ for all $n \geq 4$ aur $b_{1} = 5, b_{2} = 3, b_{3} = 1$ wale $n = 1 \sum \infty (- 1)^{n - 1} b_{n}$ ke baare mein Leibniz machinery kya kehti hai?

Forecast: ek "series" jo secretly finite hai — kya test zeros par choke karta hai?

Terms list karo. $5 - 3 + 1 - 0 + 0 - 0 + \dots = 3$ . $n = 3$ ke baad har term kuch contribute nahi karta. Yeh step kyun? Exact zeros ki tail matlab partial sums move karna band kar dete hain — degenerate lekin perfectly legal.
Do knobs check karo. $b_{n}$ : $5, 3, 1, 0, 0, \dots$ (weakly) decreasing hai kyunki poore mein $b_{n + 1} \leq b_{n}$ (equalities allowed hain — test $\leq$ use karta hai, $<$ nahi). ✓ Aur $b_{n} \to 0$ (literally $0$ hai). ✓ Yeh step kyun? Yeh woh corner case hai jo dikhata hai ki definition mein $\leq$ deliberate hai — flat/zero pieces theek hain.
Conclude karo. Converges, aur yahan hum sum exactly padhte hain: $s = 3$ . Kyun? Jab hops exactly $0$ ho jaate hain, walker literally ruk jaata hai — settle hone ke liye kuch bacha hi nahi, woh already apni final spot par baitha hai. Jo walk ruk jaati hai woh "settles" ka most extreme case hai.

Verify: $s_{3} = 5 - 3 + 1 = 3$ , aur $s_{N} = 3$ for all $N \geq 3$ kyunki baad ke terms $0$ hain. Error bound $∣ s - s_{3} ∣ \leq b_{4} = 0$ — perfectly tight. ✓

Ex 6 — Cell C6: accuracy / error-bound problem

Worked example Statement

$n = 1 \sum \infty \frac{( - 1 ) ^{n - 1}}{n ^{3}}$ ke liye, kitne terms $1 0^{- 3}$ se kam error guarantee karte hain?

Forecast: denominator mein cube bahut tezi se shrink karta hai — guess karo ki tumhe dozens chahiye ya sirf kuch handful terms.

Convergence jaldi confirm karo. $b_{n} = \frac{1}{n ^{3}}$ decrease karta hai aur $\to 0$ ⟹ converges. ✓ Kyun? Error bound tab hi use kar sakte ho jab test ki conditions hold karein.
Error bound likho. $∣ s - s_{N} ∣ \leq b_{N + 1} = \frac{1}{( N + 1 ) ^{3}}$ . Yeh step kyun? Proof ne yeh free diya — true sum consecutive partial sums ke beech trapped hai, toh miss zyada se zyada agla hop hai.
Inequality solve karo. Chahiye $\frac{1}{( N + 1 ) ^{3}} < 1 0^{- 3}$ , yaani $(N + 1)^{3} > 1000$ , toh $N + 1 > 10$ , jisse $N \geq 10$ milta hai. Yeh step kyun? Hum bound ko invert karte hain smallest $N$ dhundhne ke liye jo error ko target ke neeche squeeze kare.
Conclude karo. Pehle $10$ terms ka sum kaafi hai. Exactly 10 kyun? Kyunki $b_{11}$ pehla unused hop hai, aur humne use tolerance ke neeche force kiya; trapped true sum us hop se zyada door nahi ho sakta, toh 10 terms already answer ko required precision tak pin kar dete hain. (Ex 2 of parent se compare karo, jahan $1/ n$ ko 200 terms chahiye the — cubing kahin zyada tezi se shrink karta hai.)

Verify: $b_{11} = \frac{1}{1 1 ^{3}} = \frac{1}{1331} \approx 7.51 \times 1 0^{- 4} < 1 0^{- 3}$ ✓, jabki $b_{10} = \frac{1}{1000} = 1 0^{- 3}$ strictly below nahi hai — toh $N = 10$ exactly cutoff hai.

Ex 7 — Cell C7: real-world word problem

Worked example Statement

Ek ball drop ki jaati hai aur bounce karti hai. Woh energy lose karti hai toh har bounce ki height chhoti hoti jaati hai. Ek physicist damped oscillation ke signed displacement contributions ko $\sum_{n = 1}^{\infty} (- 1)^{n - 1} \frac{1}{n n}$ metres (upar = $+$ , neeche = $-$ ) se model karta hai. Kya net displacement ek finite value par settle hota hai, aur 8 bounces ke baad kitne error mein?

Forecast: damped, shrinking, sign-flipping — kya ball ek finite spot par settle hoti hai?

$b_{n} = \frac{1}{n n} = n^{- 3/2}$ identify karo. Kyun? Physical " $n$ -th nudge ki size", hamesha positive.
Knobs check karo. Jab $n$ badhta hai, $n^{3/2}$ badhta hai, toh $b_{n}$ decrease karta hai aur $\to 0$ . ✓✓ Kyun? Settling physical answer ko meaningful banane wale "zeroing in" behaviour ke liye dono zaroori hain.
Convergence conclude karo. Displacement ek finite $s$ par settle hota hai. Kyun? Shrinking, sign-flipping nudges exactly wahi flag-ki-taraf-hopping walk hai; har nudge pichle se chhota hai, toh ball hamesha-tightening interval mein trapped hai aur ek resting displacement approach karna hi padega — physically "settles" ka yahi matlab hai.
8 terms ke baad error. $∣ s - s_{8} ∣ \leq b_{9} = 9^{- 3/2} = \frac{1}{27} \approx 0.037$ m. Kyun? Remainder bound "mujhe kitna yakeen hai?" ko actual distance metres mein turn karta hai — note karo ki units survive karte hain: $b_{9}$ metres mein hai, toh error metres mein hai.

Verify: $9^{3/2} = (9)^{3} = 3^{3} = 27$ , toh $b_{9} = 1/27 \approx 0.0370$ m. ✓ Units check: metres in, metres out. ✓

Ex 8 — Cell C8: exam twist (power-series endpoint)

Worked example Statement

Power series $n = 1 \sum \infty \frac{x ^{n}}{n}$ ka radius of convergence $1$ hai. Endpoint $x = - 1$ par, kya yeh converge karta hai?

Forecast: ratio test endpoints decide nahi kar sakta — yahan kaun sa test humari madad karta hai?

$x = - 1$ substitute karo. Har term $\frac{( - 1 ) ^{n}}{n}$ ban jaata hai, toh series hai $n = 1 \sum \infty \frac{( - 1 ) ^{n}}{n} = - 1 + \frac{1}{2} - \frac{1}{3} + \dots$ . Yeh step kyun? Endpoints haath se test karne padte hain — Power Series — radius of convergence dekho; radius test exactly boundary par chup ho jaata hai.
Alternating shape pehchaano. Sign factor out karo: yeh $- \sum (- 1)^{n - 1} \frac{1}{n}$ hai, $b_{n} = \frac{1}{n}$ wali ek alternating series. Kyun? Overall minus sign convergence affect nahi karta; hume sirf alternating skeleton chahiye.
Leibniz apply karo. $b_{n} = \frac{1}{n}$ decrease karta hai aur $\to 0$ ⟹ converges. Yeh step kyun? Leibniz conditionally convergent endpoint decide karne ka standard tool hai — compare karo $x = + 1$ par Harmonic Series se, jo diverge karta hai.
Conclude karo. $x = - 1$ par converges (conditionally), $x = + 1$ par diverges, toh interval of convergence $[- 1, 1)$ hai. Asymmetry kyun? $x = - 1$ par signs alternate karte hain aur shrinking-hop squeeze kick karta hai; $x = + 1$ par saare terms positive $\frac{1}{n}$ hain — koi alternation nahi, koi squeeze nahi, aur harmonic series infinity ki taraf march karti hai.

Verify: $x = - 1$ par sum $- ln 2 \approx - 0.6931$ hai; partial sums ise bracket karte hain, jaise $s_{2} = - 1 + 0.5 = - 0.5$ aur $s_{3} \approx - 0.833$ , $- 0.6931$ ke doono taraf. ✓ $x = + 1$ par yeh divergent harmonic series hai. ✓

Recall Matrix par quick self-test

Har Leibniz case decide karne wale do knobs kaun se hain? ::: Kya $b_{n}$ (eventually) decreasing hai, aur kya $b_{n} \to 0$ hai. Agar $b_{n} \to 0$ hai lekin wiggle karta hai, toh verdict kya hai? ::: Leibniz chup hai — koi aur method try karo. Agar $b_{n} \neq \to 0$ hai, toh verdict kya hai? ::: nth-term test se Diverges; Leibniz apply hi nahi hota. "Sirf large $n$ ke liye decreasing" phir bhi kyun kaam karta hai? ::: Terms ka finite head ek fixed constant hai; convergence sirf tail par depend karti hai. Accuracy demand ko term count mein kaise convert karte hain? ::: Remainder bound se $b_{N + 1} < tolerance$ solve karo.

Connections

Parent: Leibniz test & proof
Convergence of Series — overview
Absolute vs Conditional Convergence
Monotone Convergence Theorem
nth-Term Test for Divergence
Harmonic Series
Power Series — radius of convergence

Worked examples — Alternating series test — Leibniz test, proof

The scenario matrix

Worked examples

Ex 1 — Cell C1: the clean pass

Ex 2 — Cell C2: sizes vanish nahi hoti

Ex 3 — Cell C3: yeh wiggle karta hai

Ex 4 — Cell C4: sirf eventually decreasing

Ex 5 — Cell C5: degenerate / zero terms

Ex 6 — Cell C6: accuracy / error-bound problem

Ex 7 — Cell C7: real-world word problem

Ex 8 — Cell C8: exam twist (power-series endpoint)

Connections

Case Map