Alternating series test — Leibniz test, proof: Question bank

4.3.10 · D5 · Maths › Calculus III — Sequences & Series › Alternating series test — Leibniz test, proof

Notation reminder taaki kuch bhi bina jaane use na ho: ek alternating series woh sum hai jiske terms har step pe sign flip karte hain. $n$ -ve term ko $a_{n}$ kahenge (to $a_{1}$ pehla term hai, $a_{2}$ doosra, aur aise aage — signs samet). Iska size hai $b_{n} = ∣ a_{n} ∣ > 0$ , yaani absolute value jisme se sign hata diya. Tab hum likhte hain $\sum_{n = 1}^{\infty} (- 1)^{n - 1} b_{n}$ , jahan $(- 1)^{n - 1}$ sirf sign-flipper hai aur $b_{n}$ mein saari real content hai. $s_{N}$ hai $N$ -va partial sum ( $N$ terms ke baad ka running total), aur $s$ hai woh true infinite sum jab woh exist karta hai.

Bracket-squeeze se hum matlab rakhte hain proof ka core move: terms ko brackets mein group karna jaise $(b_{1} - b_{2}) + (b_{3} - b_{4}) + \dots$ taaki har bracket $\geq 0$ ho, jo partial sums ko ek increasing floor aur ek fixed ceiling $b_{1}$ ke beech trap kar leta hai — wahi squeeze convergence force karta hai.

True or false — justify

Agar

b_{n} \to 0

, to

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

converge karta hai.

False. Terms ka vanish hona sirf condition 2 hai; bina monotone decrease ke proof ka bracket-squeeze fail ho sakta hai, isliye convergence guaranteed nahi hai.

Agar

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

converge karta hai, to

b_{n} \to 0

True. Koi bhi convergent series (alternating ho ya na ho) ke terms zero ki taraf tend karne chahiye — yeh nth-term test ka ulta padhna hai.

Leibniz test yeh prove kar sakta hai ki ek series diverge karti hai.

False. Yeh ek one-way test hai: isse pass karna convergence prove karta hai, lekin fail karna (jaise

b_{n}

monotone nahi) apne aap mein kuch prove nahi karta.

Agar

b_{n} \neq \to 0

, to alternating series diverge karti hai.

True, lekin nth-term test se, Leibniz se nahi — jo terms vanish nahi hote woh kabhi finite limit mein sum nahi ho sakte.

Alternating harmonic series aur harmonic series ka convergence behaviour same hai.

False.

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

converge karta hai (

ln 2

ko); plain harmonic series

\sum 1/ n

diverge karta hai. Alternation usse bacha leta hai.

Har convergent alternating series sirf conditionally convergent hoti hai.

False. Kuch absolutely converge karti hain, jaise

\sum (- 1)^{n - 1} / n^{2}

, kyunki

\sum 1/ n^{2}

converge karta hai. Alternation ka present hona conditional convergence force nahi karta.

True sum pehle term

b_{1}

ke barabar hota hai.

False. Proof dikhata hai ki

s \leq b_{1}

; yeh ek upper bound hai, value nahi. Actual

s

s_{1} = b_{1}

aur

s_{2} = b_{1} - b_{2}

ke strictly beech hota hai.

Ek Leibniz series ke liye, partial sums limit ke dono taraf hote hain: consecutive

s_{N}

values

s

ke opposite sides par hote hain.

True. Odd partial sums overshoot karte hain aur even ones undershoot karte hain (ya ulta), har jump shrink hota hai, aur exactly yahi reason hai ki

s

unke beech trap rehta hai.

Error bound

∣ s - s_{N} ∣ \leq b_{N + 1}

ke liye terms ka decreasing hona zaruri hai, sirf

\to 0

kaafi nahi.

True. Jo straddling yeh bound deta hai woh monotonicity par rely karta hai; bina iske sum consecutive partial sums ke beech lie karne ki zarurat nahi hai.

Monotone decrease sirf "eventually" (bade

n

ke liye) hold karni chahiye,

n = 1

se zaruri nahi.

True. Finitely many early terms convergence ko affect nahi kar sakte, isliye kisi

n = N

se aage decrease kaafi hai — bas tail par test apply karo.

Spot the error

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

Leibniz se converge karta hai kyunki yeh alternate karta hai aur har term 1 se chhoti hai."

Error yeh hai: condition 2 violated hai —

b_{n} = \frac{n}{n + 1} \to 1 \neq = 0

. 1 se bounded hona irrelevant hai; series nth-term test se diverge karti hai.

b_{n} = 1, \frac{1}{3}, \frac{1}{2}, \frac{1}{5}, \frac{1}{4}, \dots \to 0

aur alternate karta hai, isliye Leibniz convergence deta hai."

Error yeh hai:

b_{n}

monotone decreasing nahi hai (yeh wiggle karta hai). Bracket argument

(b_{2 k - 1} - b_{2 k}) \geq 0

break ho jaata hai, isliye Leibniz simply apply nahi hota.

"Kyunki Leibniz ne prove kiya

s \leq b_{1}

, hum conclude karte hain

s = b_{1}

jab pehle ke baad ke terms tiny hain."

Error yeh hai: ek inequality equality nahi hoti.

s

s_{1}

aur

s_{2} = b_{1} - b_{2}

ke beech pin hota hai, isliye

s < b_{1}

jab bhi

b_{2} > 0

ho.

"Series converge karti hai, isliye main freely terms rearrange kar sakta hoon bina sum badlaaye."

Error yeh hai: conditional convergence (jaise alternating harmonic) free rearrangement allow nahi karta — Riemann ka theorem kehta hai tum ise kuch bhi sum kar sakte ho. Sirf absolutely convergent series safely rearrange hoti hain (dekho Absolute vs Conditional Convergence).

"Monotonicity check karne ke liye maine

b_{5} < b_{4}

confirm kiya, isliye

b_{n}

decreasing hai."

Error yeh hai: ek comparison ek pattern nahi hai. Tumhe sabhi (bade)

n

ke liye

b_{n + 1} \leq b_{n}

chahiye, jo usually

f^{'} (x) \leq 0

se show karte hain jahan

f (x) = b_{x}

ho, ya algebraic ratio/difference argument se.

b_{n} \to 0

yahan converge karne ke liye kaafi tha, aur

\sum 1/ n

mein

1/ n \to 0

hai, isliye harmonic series bhi converge karti hai."

Error yeh hai: alternation plus monotonicity ne kaam kiya, sirf

b_{n} \to 0

ne nahi. Plain harmonic series mein partial sums trap karne ke liye koi sign-flipping nahi hai, aur woh diverge karta hai.

Why questions

Hum

(b_{1} - b_{2}) + (b_{3} - b_{4}) + \dots

group kyun karte hain instead of term by term sum karne ke?

Grouping expose karta hai ki har bracket

\geq 0

hai (decrease se), jo instantly dikhata hai ki even partial sums increasing hain — Monotone Convergence setup ka pehla half.

Hum dobaara

b_{1} - (b_{2} - b_{3}) - \dots - b_{2 m}

regroup kyun karte hain?

Upper bound paane ke liye: har naya bracket subtract hota hai aur non-negative hai, jo force karta hai

s_{2 m} \leq b_{1}

. Increasing plus bounded above phir Monotone Convergence Theorem se convergence deta hai.

Even partial sums ka converge karna proof ko akela kyun finish nahi karta?

Ek subsequence ka converge karna poori sequence ko converge force nahi karta; odd sums kahin aur ja sakte hain. Hume dikhana hoga ki dono same

s

ki taraf jaate hain.

lim b_{n} = 0

exactly woh ingredient kyun hai jo odd aur even limits ko merge karta hai?

Kyunki

s_{2 m + 1} = s_{2 m} + b_{2 m + 1}

, aur

b_{2 m + 1} \to 0

odd limit ko even limit ke barabar bana deta hai

s + 0 = s

. Bina iske dono limits alag honge aur

s_{N}

ka koi ek limit nahi hoga.

Error bound "first omitted term" free mein kyun milta hai?

Kyunki

s

s_{N}

aur

s_{N + 1}

ke beech hota hai, aur unka difference exactly

b_{N + 1}

hai, isliye

s_{N}

s

tak ki distance woh gap exceed nahi kar sakti.

Leibniz succeed karne ke baad bhi hum

\sum b_{n}

alag se kyun check karte hain?

Leibniz sirf convergence certify karta hai, absolute convergence nahi.

\sum b_{n}

test karne se pata chalta hai ki yeh absolute hai ya merely conditional — rearrangement aur further manipulation ke liye crucial.

Leibniz power series endpoints ke liye itna useful kyun hai?

Ek endpoint par series often alternating ho jaati hai decreasing terms ke saath, jahan absolute-convergence tests inconclusive hote hain lekin Leibniz cleanly convergence settle kar deta hai.

Edge cases

Agar

b_{n}

eventually constant ho (maan lo

b_{n} = c > 0

sabhi bade

n

ke liye), kya Leibniz apply hota hai?

Nahi. Yeh weakly decreasing hai lekin

b_{n} \to c \neq = 0

, isliye condition 2 fail hoti hai aur series nth-term test se diverge karti hai.

Agar

b_{n} = 0

infinitely many

n

ke liye ho, kya series abhi bhi "alternating" hai?

Borderline: definition chahti hai

b_{n} > 0

. Zeros strict sign-flip pattern todh dete hain; pehle zero terms drop karo aur reindex karo test apply karne se pehle.

Kya hoga agar kuch

n

ke liye

b_{n + 1} = b_{n}

ho (decrease mein ties)?

Theek hai — test allow karta hai

b_{n + 1} \leq b_{n}

, isliye ties allowed hain. Har bracket abhi bhi

\geq 0

hai (possibly

= 0

), aur argument kaam karta hai.

Ek series alternating hai aur

b_{n} \to 0

, lekin

b_{n}

finite stretch mein increase karta hai phir hamesha ke liye decrease karta hai. Converge karega?

Haan. Convergence sirf tail par depend karta hai; Leibniz wahan se apply karo jahan monotonicity shuru hoti hai, kyunki finite front sum change karta hai lekin convergence nahi.

Kya ek alternating series converge kar sakti hai even though woh DONO Leibniz conditions fail kare?

Haan — Leibniz sufficient hai, necessary nahi. Ek cleverly arranged non-monotone, phir bhi vanishing series converge ho sakti hai; Leibniz bas woh tool nahi hoga jo ise prove kare.

Agar har

b_{n}

kisi convergent absolute series ke terms jaise ho, kya conditional/absolute distinction abhi bhi zaruri hai?

Haan — agar

\sum b_{n}

converge karta hai to alternating version absolutely converge karta hai, isliye "conditional" apply nahi hoga. Distinction hamesha

\sum b_{n}

par hinge karti hai, signs par kabhi nahi.

Recall Har trap ki one-line summary

Leibniz ko Down to Zero chahiye (decreasing aur vanishing); yeh sirf convergence prove karta hai, value nahi upper bound deta hai, aur kabhi absolute vs conditional nahi batata — uske liye $\sum b_{n}$ check karo.

Connections

4.3.10 Alternating series test — Leibniz test, proof (Hinglish)
Convergence of Series — overview
Absolute vs Conditional Convergence
Monotone Convergence Theorem
nth-Term Test for Divergence
Harmonic Series
Power Series — radius of convergence