Alternating series test — Leibniz test, proof: Visual walkthrough

4.3.10 · D2 · Maths › Calculus III — Sequences & Series › Alternating series test — Leibniz test, proof

Step 0 — Hum dekh kya rahe hain

Toh $s_{N} = b_{1} - b_{2} + b_{3} - b_{4} + \dots$ , $N$ hops ke baad rukna hai. Har $\pm$ us hop ki direction hai. Picture dekho: hum ek horizontal number line par rehte hain, aur har hop ek aisa arrow hai jiski length $b_{n}$ hai aur jiski direction alternate karti hai.

Figure — Alternating series test — Leibniz test, proof

Hum exactly wahi assume karte hain jo test demand karta hai:

Decreasing: $b_{n + 1} \leq b_{n}$ — har hop pichle se lamba nahi hota.
Vanishing: $b_{n} \to 0$ — hops bilkul kuch nahi tak shrink ho jaate hain.

Step 1 — Partial sums ko landing spots ki tarah plot karo

KYA: Arrows ki jagah, har hop ke baad sirf landing spot mark karo: $s_{1}, s_{2}, s_{3}, \dots$

KYUN: Sawal "kya series converge karti hai?" bilkul waisa hi sawal hai jaise "kya landing spots $s_{N}$ ek point par settle hote hain?" Hops ko spots mein convert karna ek walking problem ko ek sequence-of-points problem mein badal deta hai, jise hum test karna jaante hain.

PICTURE: Dekho kaise spots right, left, right, left bounce karte hain — lekin bounces chhoti hoti jaati hain. $s_{1}$ kaafi right par hai, $s_{2}$ left jump karta hai (by $b_{2}$ ), $s_{3}$ right nudge karta hai (by $b_{3} < b_{2}$ ), aur aise hi.

Step 2 — Spots ko do teams mein baanto: even aur odd

KYA: Even spots ko ek tarah colour karo, odd spots ko doosre tarah.

KYUN: Har team monotone nikally hai (steadily ek direction mein move karti hai), aur monotone sequences woh hain jinhe hum Monotone Convergence Theorem se pin down kar sakte hain. Mixed zig-zag monotone nahi hai, isliye hum us theorem ko directly uspar use nahi kar sakte — isliye split kiya.

PICTURE: Even team rightward creep karti hai (number line upar); odd team leftward creep karti hai. Woh ek doosre ki taraf squeeze ho rahi hain.

Step 3 — Even team sirf right move karti hai (yeh increasing hai)

KYA: Terms ko ek baar mein do-do group karo. Har bracket hai (ek bada hop) minus (ek chhota-ya-equal hop).

KYUN: Kyunki $b_{n}$ decreasing hai, $b_{2 k - 1} \geq b_{2 k}$ , isliye har bracket $\geq 0$ hai. $s_{2 m}$ se $s_{2 m + 2}$ jaane par sirf ek aur non-negative bracket add hota hai. Kuch $\geq 0$ add karna tumhe sirf right push kar sakta hai. Isliye even spots kabhi left nahi jaate: $s_{2} \leq s_{4} \leq s_{6} \leq \dots$

PICTURE: Hops ka har naya pair ek green block hai non-negative width ka; blocks stack karna even spot ko steadily rightward march karta hai.

Step 4 — Lekin even team $b_{1}$ ko kabhi pass nahi kar sakti (yeh bounded hai)

KYA: Usi sum ko $b_{1}$ se shuru karke re-bracket karo, $b_{2}$ ko $b_{3}$ ke saath pair karo, wagera.

Worked example Regrouping, term-by-term (kuch add ya remove nahi kiya)

Original signed sum se shuru karo aur bracket boundaries ko ek step right shift karo. Dekho ki har $b_{k}$ apna sign rakhta hai — hum sirf choose karte hain ki brackets kahan khulein aur bandh hoon: $s_{2 m} = b_{1} - b_{2} + b_{3} - b_{4} + b_{5} - \dots + b_{2 m - 1} - b_{2 m}$ Front se $b_{1}$ aur back se $- b_{2 m}$ nikalo, aur interior terms ko $(- b_{2} + b_{3}), (- b_{4} + b_{5}), \dots$ pair karo: $s_{2 m} = b_{1} + (- b_{2} + b_{3}) + (- b_{4} + b_{5}) + \dots + (- b_{2 m - 2} + b_{2 m - 1}) - b_{2 m}$ Ab har interior bracket se minus factor out karo, $(- b_{2 k} + b_{2 k + 1})$ ko $- (b_{2 k} - b_{2 k + 1})$ mein badlo: $s_{2 m} = b_{1} - (b_{2} - b_{3}) - (b_{4} - b_{5}) - \dots - (b_{2 m - 2} - b_{2 m - 1}) - b_{2 m}$ Terms count karo: $b_{1}$ ek baar (+), $b_{2 m}$ ek baar (−), aur har middle $b_{k}$ apne original sign ke saath ek baar — shuruat se identical, sirf re-bracketed.

KYUN: Har bracket $(b_{2 k} - b_{2 k + 1})$ $\geq 0$ hai (decreasing $b_{n}$ ), leftover $b_{2 m} \geq 0$ hai, aur inhe sab ko $b_{1}$ se subtract kiya gaya hai. $b_{1}$ se non-negative cheezein subtract karna tumhe sirf $b_{1}$ par ya neeche chhod sakta hai: $s_{2 m} \leq b_{1} har m ke liye.$

Toh even spots fenced in hain: woh right march kar sakte hain (Step 3) lekin $b_{1}$ par ek wall unhe rokti hai.

PICTURE: $b_{1}$ par amber wall; even spots left se uski taraf approach karte hain lekin kabhi cross nahi karte.

Step 5 — Ek increasing, fenced-in team zaroor settle karegi (Monotone Convergence)

KYA: Step 3 (increasing) aur Step 4 (bounded above by $b_{1}$ ) combine karo.

KYUN: Increasing + bounded above ⟹ ek limit exist karta hai. Use $s$ naam do: $lim_{m \to \infty} s_{2 m} = s, s \leq b_{1} .$ Yahan $s$ even team ka settling-point hai, aur yeh wall par ya neeche baith hai.

PICTURE: Even dots ek invisible ceiling $s$ ( $b_{1}$ ke thoda neeche) ke against pile up hote hain, arbitrarily close hote jaate hain.

Step 6 — Odd team usi spot par land karti hai (yahan $b_{n} \to 0$ enter karta hai)

KYA: Even limit lo aur bilkul agla hop add karo.

KYUN — sirf woh jagah jahan hum vanishing use karte hain: Agar hops zero tak shrink nahi hote, toh odd team $s$ se ek fixed gap $b_{2 m + 1}$ right par settle hoti — do alag limits, no convergence. Lekin $b_{n} \to 0$ woh gap band kar deta hai: $lim_{m \to \infty} s_{2 m + 1} = s + 0 = s .$ Even team → $s$ . Odd team → $s$ . Same point.

PICTURE: Left (even) aur right (odd) dots ke beech shrinking amber bridge $b_{2 m + 1}$ zero length tak collapse hota hai.

Step 7 — Free bonus: error bound, ek nazar mein

KYA: Notice karo ki $s$ hamesha do consecutive spots $s_{N}$ aur $s_{N + 1}$ ke beech rehta hai.

KYUN: Har parity mein true sum $s$ do consecutive spots $s_{N}$ aur $s_{N + 1}$ ke beech pakda jaata hai, aur un neighbours ke beech distance exactly agle hop ki length hai, $∣ s_{N + 1} - s_{N} ∣ = b_{N + 1}$ . Ek interval ke andar trapped point dono ends se zyada door nahi ho sakta interval ki width se — toh tumhari $s$ se distance $b_{N + 1}$ ko beat nahi kar sakti.

PICTURE: True sum $s$ shrinking interval $[s_{2}, s_{1}], [s_{2}, s_{3}], \dots$ ke andar pinned hai; har interval ki width agla $b$ hai.

Step 8 — Degenerate cases (kya toot ta hai agar koi hypothesis mar jaaye)

PICTURE: Left panel — near-constant length ke hops ( $b_{n} \to 1$ ), spots hamesha bounce karte rehte hain, kabhi settle nahi karte. Right panel — non-monotone hops ek spot ko "galat" direction mein jump karate hain, fence todke.

Ek-picture summary

Upar sab kuch ek single frame mein: do monotone teams (even ↑, odd ↓) ek shared limit $s \leq b_{1}$ ki taraf squeeze ho rahi hain, collapsing bridge $b_{2 m + 1}$ , aur error bound $b_{N + 1}$ final trap ki width ke roop mein.

Recall Feynman: poora walkthrough plain words mein

Tum ek line par hop karte ho: ek bada step aage, ek chhota step peeche, ek aur chhota step aage, phir peeche — har step pichle se chhota. Har baar jahan land karo, mark karo. "Peeche" landings (even) steadily right creep karte hain lekin tumhara bahut pehla landing $b_{1}$ kabhi pass nahi karte — woh ek wall hai. Jo cheez sirf right jaaye lekin wall se takraaye woh zaroor kahi rok jaayegi; use $s$ kaho. "Aage" landings (odd) hamesha ek tiny step se right par hote hain ek peeche wale landing ke, aur kyunki steps kuch nahi tak shrink ho jaate hain, woh same $s$ par khatam hote hain. Dono tarah ke landings ek point par pile up hote hain — woh point sum hai. Aur jahan bhi ruko, true answer tumhara agla step size $b_{N + 1}$ se zyada door nahi hai. Yahi poora test hai: hops zero tak jaayein ⟹ tum kahi land karo, truth se ek hop ke andar. Alternating harmonic series $1 - \frac{1}{2} + \frac{1}{3} - \dots$ ke liye woh landing spot $ln 2$ hota hai.

Connections

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