Series — partial sums, convergence definition

4.3.3 · Maths › Calculus III — Sequences & Series

Series KYA hoti hai?

Partial sums se kyun define karte hain? Kyunki " $a_{1} + a_{2} + \dots$ forever" ek defined arithmetic operation nahi hai. Addition ek binary operation hai; hum ek time mein sirf do cheezein add karte hain. $s_{1}, s_{2}, s_{3}, \dots$ banake hum undefined infinite process ko ek bilkul legal infinite sequence se replace kar dete hain, aur ek sequence ya toh limit rakhti hai ya nahi.

Convergence ka matlab KYA hai?

KAISE: geometric series sum scratch se nikaalein

Yeh model example hai — derive karo, memorise mat karo.

Maano $a_{n} = a r^{n - 1}$ , toh $\sum_{n = 1}^{\infty} a r^{n - 1} = a + a r + a r^{2} + \dots$ .

Step 1 — partial sum likho. $s_{N} = a + a r + a r^{2} + \dots + a r^{N - 1} .$ Yeh step kyun? Convergence $s_{N}$ ke through define hoti hai, isliye hume $s_{N}$ ki closed form chahiye.

Step 2 — $r$ se multiply karo aur subtract karo. $r s_{N} = a r + a r^{2} + \dots + a r^{N} .$ $s_{N} - r s_{N} = a - a r^{N} .$ Yeh step kyun? $r$ se multiply karne par har term ek slot shift ho jaati hai; subtract karne par poora middle cancel ho jaata hai, sirf pehla aur aakhri piece bachta hai. Yeh classic telescoping trick hai.

Step 3 — $s_{N}$ ke liye solve karo (maano $r \neq = 1$ ). $s_{N} (1 - r) = a (1 - r^{N}) ⟹ s_{N} = \frac{a ( 1 - r ^{N} )}{1 - r} .$ Yeh step kyun? Ab $s_{N}$ $N$ mein ek explicit formula hai, toh hum limit le sakte hain.

Step 4 — limit lo. Sirf $N$ -dependent part $r^{N}$ hai.

Agar $∣ r ∣ < 1$ : $r^{N} \to 0$ , toh $s_{N} \to \frac{a}{1 - r}$ . Converges.
Agar $∣ r ∣ > 1$ : $∣ r^{N} ∣ \to \infty$ , $s_{N}$ blow up kar jaata hai. Diverges.
Agar $r = 1$ : $s_{N} = a N \to \pm \infty$ . Diverges.
Agar $r = - 1$ : $s_{N}$ alternately $a, 0, a, 0, \dots$ — koi limit nahi. Diverges.

Figure — Series — partial sums, convergence definition

KAISE: ek telescoping series scratch se

Socho $n = 1 \sum \infty \frac{1}{n ( n + 1 )}$ .

Step 1 — partial fractions. $\frac{1}{n ( n + 1 )} = \frac{1}{n} - \frac{1}{n + 1}$ . Kyun? Yeh har term ko ek difference ki tarah likhta hai taaki consecutive terms cancel ho jaayein.

Step 2 — partial sum.

= \left(1-\tfrac12\right)+\left(\tfrac12-\tfrac13\right)+\cdots+\left(\tfrac1N-\tfrac1{N+1}\right).$$ Sab cancel ho jaata hai sivaaye ends ke: $s_N = 1 - \dfrac{1}{N+1}$. *Kyun?* Yahi toh telescoping ka poora point hai — inner terms annihilate ho jaate hain. **Step 3 — limit.** $s_N = 1 - \frac{1}{N+1} \to 1$. Toh series **$1$ par converge karti hai**. --- ## Ek divergent example jo tumhe zaroor jaanna chahiye > [!example] Harmonic series diverge karti hai > $\sum_{n=1}^{\infty}\frac1n$ ke terms $\to 0$ hain, phir bhi **diverges** karti hai. > Terms group karo: $1 + \tfrac12 + \underbrace{(\tfrac13+\tfrac14)}_{>\,1/2} + \underbrace{(\tfrac15+\cdots+\tfrac18)}_{>\,1/2} + \cdots$ > Har block $\tfrac12$ se zyada hai, toh $s_N$ bina bound ke badhta rehta hai. > **Kyun zaroori hai:** terms ka $0$ ki taraf jaana convergence ke liye *kaafi nahi* hai. > [!formula] $n$-th Term Test (ek necessary condition) > Agar $\sum a_n$ converge kare toh $\displaystyle\lim_{n\to\infty} a_n = 0$. > **Derivation:** $a_N = s_N - s_{N-1}$. Agar $s_N \to S$ aur $s_{N-1}\to S$, toh > $a_N \to S - S = 0$. **Contrapositive (usable test):** agar $a_n \not\to 0$, toh series diverge karti hai. > ⚠️ Converse FALSE hai — harmonic series counterexample hai. --- > [!mistake] Common errors ko steel-man karo > **Mistake 1: "$a_n \to 0$, toh series converge karti hai."** > *Kyun sahi lagta hai:* agar tum chhote se chhote tukde add kar rahe ho, toh surely total settle ho jaayega. > *Fix:* harmonic series chhote tukde forever add karta hai aur phir bhi $\infty$ tak pahunch jaata hai. Terms ki > smallness yeh guarantee nahi karti ki *accumulation* bounded rahega. $a_n\to0$ **necessary hai, sufficient nahi**. > > **Mistake 2: sequence $(a_n)$ aur series $\sum a_n$ ko confuse karna.** > *Kyun sahi lagta hai:* dono ek hi numbers se aate hain. *Fix:* $(a_n)\to 0$ *terms* ke baare mein poochh raha hai; > $\sum a_n$ ka converge karna *running totals* $(s_N)$ ke baare mein poochh raha hai. Hamesha $s_N$ se guzaro. > > **Mistake 3: $\frac{a}{1-r}$ use karna jab $|r|\ge 1$ ho.** > *Kyun sahi lagta hai:* formula itna handy hai ki tum uski validity bhool jaate ho. *Fix:* formula tab hi > exist karta hai kyunki $r^N\to 0$, jo sirf $|r|<1$ par hota hai. Usse bahar, koi finite sum nahi hai. > > **Mistake 4: yeh sochna ki divergence ka matlab "infinity par jaana" hai.** > *Kyun sahi lagta hai:* zyaatir divergent examples blow up karte hain. *Fix:* $\sum(-1)^n$ mein $s_N$ > $-1,0,-1,0,\dots$ bounce karta hai — bounded hai lekin **koi limit nahi**, isliye divergent hai. --- > [!recall]- Feynman: ek 12-saal ke bachchey ko samjhao > Socho tum apne aap ko candy dene wale ho aur har din ek piece milti hai: 1 candy, phir aadhi candy, phir > chautha hissa, phir aathwa... Har din tumhara *pile* thoda badhta hai. Ek "series" total pile hai. > Lekin tum forever wait nahi kar sakte, toh har raat pile dekho aur poochho: "Kya mera pile > kisi fixed amount, jaise 2 candies, ke karib se karib aata ja raha hai aur basically wahan rok raha hai?" > Agar haan — series **2 par converge** karti hai. Agar tumhara pile forever badhta rehta hai bina kisi ceiling ke > (jaise $1, \tfrac12, \tfrac13, \tfrac14\dots$ add karna), ya phir kehta-sunte rehta hai aur kabhi settle nahi hota, > toh **koi final pile nahi** hai — woh **diverge** karta hai. Trick yeh hai: forever add karne ki koshish mat karo. > Bas running total dekho aur dekho woh kahan ja raha hai. > [!mnemonic] Logic chain yaad karo > **"Partial sums Point the Path."** Ek series = **P**artial sums ki sequence; uski limit (agar exist kare) > sum hai. Aur term test ke liye: **"Terms not zero ⇒ no go."** --- ## Active recall > [!recall] Quick self-test > 1. $s_N$ define karo aur exactly batao ki $\sum a_n$ kab converge karta hai. > 2. Geometric series ke liye $s_N$ derive karo aur batao yeh kab converge karta hai. > 3. $a_n\to0$ convergence imply kyun NAHI karta? Example do. > 4. $n$-th term test $a_N = s_N - s_{N-1}$ se prove karo. #flashcards/maths Series kis cheez ke terms mein define hoti hai? ::: Apne sequence of partial sums $s_N=\sum_{n=1}^N a_n$ ki limit ke terms mein. $\sum a_n$ ka $S$ par converge karne ka exactly matlab kya hai? ::: $\lim_{N\to\infty} s_N = S$, ek finite number. $N$-th partial sum $s_N$ ka formula? ::: $s_N = a_1+a_2+\cdots+a_N = \sum_{n=1}^N a_n$. Geometric series $\sum_{n=1}^\infty ar^{n-1}$ kab converge karta hai, aur kismein? ::: Jab $|r|<1$, $a/(1-r)$ mein. Geometric partial sum ki closed form ($r\ne1$)? ::: $s_N = a(1-r^N)/(1-r)$. Geometric convergence ke liye $|r|<1$ kyun zaroori hai? ::: Kyunki $r^N\to0$ sirf tab hota hai jab $|r|<1$; warna $s_N$ ka koi finite limit nahi hota. $\sum 1/(n(n+1))$ ka sum aur kyun? ::: $1$ ke barabar; telescopes karke $s_N=1-1/(N+1)\to1$ ho jaata hai. $n$-th term test state karo. ::: Agar $\sum a_n$ converge kare toh $a_n\to0$; contrapositive: $a_n\not\to0\Rightarrow$ diverges. Kya $a_n\to0$ convergence ke liye sufficient hai? ::: Nahi — harmonic series $\sum 1/n$ mein $a_n\to0$ lekin diverge karti hai. Partial sums se $a_N$ derive karo. ::: $a_N = s_N - s_{N-1}$. Kya divergence ka matlab hamesha $\to\infty$ hota hai? ::: Nahi — jaise $\sum(-1)^n$ mein partial sums bounce karte hain, bounded hain lekin koi limit nahi. $s_N\to S$ ki $\varepsilon$-definition? ::: $\forall\varepsilon>0\,\exists M: N\ge M\Rightarrow |s_N-S|<\varepsilon$. --- ## Connections - [[Sequences — limits and convergence]] (ek series disguise mein ek sequence hi hai — uske partial sums) - [[Geometric series]] (prototype convergent series, upar derive ki gayi) - [[n-th Term Test for Divergence]] - [[Harmonic series]] (key divergent counterexample) - [[Telescoping series]] - [[p-series and the Integral Test]] (geometric se aage convergence decide karna) - [[Comparison Test]] · [[Ratio & Root Tests]] - [[Power series and radius of convergence]] (jahan geometric reasoning generalise hoti hai) ## 🖼️ Concept Map ```mermaid flowchart TD A[Sequence a_n] -->|add terms| B[Series sum a_n] B -->|not defined as| C[Infinite addition] C -->|replaced by| D[Partial sum s_N] A -->|finite sum| D D -->|form new sequence| E[Sequence of partial sums] E -->|take limit N to inf| F[lim s_N] F -->|equals finite S| G[Series converges to S] F -->|DNE or infinite| H[Series diverges] G -->|formalised by| I[Epsilon-M definition] B -->|model example| J[Geometric series a r^n-1] J -->|telescoping trick| K[s_N = a 1-r^N over 1-r] K -->|if abs r less than 1| G ```