Alternating series test — Leibniz test, proof

4.3.10 · Maths › Calculus III — Sequences & Series

Alternating series KYA hota hai?

Isko derive/prove KAISE karein (scratch se)

Hum proof banate hain partial sums ko squeeze karke. Maano $s_{N} = \sum_{n = 1}^{N} (- 1)^{n - 1} b_{n}$ .

Step 1 — Even partial sums $s_{2 m}$ dekho. $s_{2 m} = (b_{1} - b_{2}) + (b_{3} - b_{4}) + \dots + (b_{2 m - 1} - b_{2 m}) .$ Yeh grouping kyun? Kyunki $b_{n}$ decrease ho raha hai, har bracket $\geq 0$ hai. Aur non-negative brackets add karne se $s_{2 m}$ badhta hai jaise $m$ badhta hai. Toh ${s_{2 m}}$ increasing hai.

Step 2 — Dikhaao ki $s_{2 m}$ upar se bounded hai. Alag tarike se regroup karo: $s_{2 m} = b_{1} - (b_{2} - b_{3}) - (b_{4} - b_{5}) - \dots - b_{2 m} .$ Regroup kyun? Yahan har bracket $\geq 0$ hai, aur hum unhe subtract kar rahe hain, saath mein $b_{2 m} \geq 0$ bhi subtract karte hain. Toh $s_{2 m} \leq b_{1}$ .

Step 3 — Conclude karo ki even sums converge karte hain. Ek increasing sequence jo upar se bounded ho, converge karti hai (Monotone Convergence Theorem). Limit ko $s$ bulao: $lim_{m \to \infty} s_{2 m} = s, s \leq b_{1} .$

Step 4 — Odd sums ko joodo. Note karo ki $s_{2 m + 1} = s_{2 m} + b_{2 m + 1}$ . Limits lo: $lim_{m \to \infty} s_{2 m + 1} = s + lim_{m \to \infty} b_{2 m + 1} = s + 0 = s .$ Yeh khatam kyun karta hai? Even aur odd partial sums same limit $s$ share karte hain. Agar partial sums ki dono subsequences $s$ pe jaati hain, toh poori sequence $s_{N} \to s$ hoti hai. Isliye series converge karti hai. $■$

Figure — Alternating series test — Leibniz test, proof

Bonus: error bound (proof se free gift)

Kyunki partial sums $s$ ke aas-paas oscillate karte hain, har baar kam jump karke:

Worked examples

Common mistake Classic errors ko steel-man karna

Mistake A: "Terms 0 pe jaate hain, toh converge karta hai." Kyun sahi lagta hai: Alternating series ke liye humne abhi $b_{n} \to 0$ use kiya — sufficient lagta hai. Fix: $b_{n} \to 0$ akela kabhi bhi kaafi nahi hota in general (harmonic series $\sum 1/ n$ diverge karti hai!). Leibniz decreasing + alternation bhi maangta hai.

Mistake B: "Leibniz mujhe bata raha hai sum $b_{1}$ ke barabar hai." Kyun sahi lagta hai: Humne prove kiya $s \leq b_{1}$ . Fix: $b_{1}$ sirf ek upper bound hai, value nahi. True $s$ , $s_{1}$ aur $s_{2}$ ke beech rehta hai.

Mistake C: "Converges ⟹ absolutely converges." Kyun sahi lagta hai: Convergence usually matlab "well-behaved." Fix: Leibniz aksar sirf conditional convergence deta hai (alternating harmonic). $\sum b_{n}$ ko absolute convergence ke liye hamesha alag se check karo.

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho tum ek jhande ki taraf hop kar rahe ho. Pehle ek bada hop aage, phir ek chhota hop peechhe, phir aur chhota hop aage, phir aur chhota peechhe... Har hop pehle se chhhota hota hai, aur woh bahut tiny hote jaate hain. Tum ek chhoti si jagah mein hamesha uchchhal nahi sakte — tumhe ek jagah par settle karna hi padega. Wohi jagah sum hai! Aur tum jahan bhi ho, jhanda (true answer) tumhari agli hop ki size se zyaada door nahi hai.

Active-recall flashcards

Leibniz (alternating series) test ki do conditions batao.

(1)

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

(decreasing) aur (2)

b_{n} \to 0

, jahan

b_{n} = ∣ a_{n} ∣ > 0

Proof mein

b_{n}

ka decreasing hona kyun zaroori hai?

Yeh grouped brackets

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

banata hai, jisse bounded partial sums monotone milte hain.

Proof mein condition

b_{n} \to 0

kahaan use hoti hai?

Yeh dikhane ke liye ki even aur odd partial sums same limit share karte hain:

s_{2 m + 1} = s_{2 m} + b_{2 m + 1} \to s + 0

Alternating series error bound kya hai?

∣ s - s_{N} ∣ \leq b_{N + 1}

— error zyaada se zyaada pehle omit hue term ke barabar hai.

Kya

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

converge karta hai? Absolutely?

Converge karta hai (

ln 2

tak) lekin sirf conditionally, kyunki

\sum 1/ n

diverge karta hai.

Agar

b_{n} \to 0

hai lekin monotone nahi hai, toh kya Leibniz se convergence conclude kar sakte ho?

Nahi — monotonicity zaroori hai; test apply nahi hota.

Agar

b_{n} \neq \to 0

ho, toh kya keh sakte ho?

Series

n

th-term test se diverge karti hai; Leibniz irrelevant hai.

True sum do consecutive partial sums ke beech kyun hota hai?

Partial sums limit ko alternately overshoot/undershoot karte hain, shrinking jumps

b_{N + 1}

ke saath.

Connections

Convergence of Series — overview
Absolute vs Conditional Convergence
Monotone Convergence Theorem
nth-Term Test for Divergence
Harmonic Series
Power Series — radius of convergence (endpoint behaviour mein aksar Leibniz use hota hai)