4.3.8 · Maths › Calculus III — Sequences & Series
Agar tum ek choti seedhi pe khade ho jo roof tak nahi pahunchi, toh tum bhi roof tak nahi pahunchoge.
Agar tum ek lambi seedhi pe khade ho jo roof tak pahunchi, toh tum bhi pahunch jaoge.
Direct Comparison Test (DCT) bilkul yahi hai: jo series ek convergent series ke neeche dabi ho woh converge karti hai, aur jo series ek divergent series ke upar push ho woh diverge karti hai . Hum sum compute nahi karte — bas use known behaviours ke beech trap karte hain.
Definition Direct Comparison Test
Maano 0 ≤ a n ≤ b n sab n ≥ N ke liye (eventually). Tab:
Agar ∑ b n converge kare, toh ∑ a n converge karega. (choti wali ek finite roof ke neeche trap hai)
Agar ∑ a n diverge kare, toh ∑ b n diverge karega. (badi wali ek infinite floor ke upar push hai)
Zaroori baat: is test ko non-negative terms chahiye (a n , b n ≥ 0 ). Yeh term-by-term compare karta hai.
Hum ise Monotone Convergence Theorem se banate hain: ek sequence jo increasing aur upar se bounded ho, converge karti hai.
Maano a n ≥ 0 . Partial sums define karo:
S n = a 1 + a 2 + ⋯ + a n .
Step 1 — Partial sums increasing hain.
S n + 1 − S n = a n + 1 ≥ 0 , toh S n monotone increasing hai.
Yeh step kyun? Terms ki non-negativity hi guarantee karti hai ki partial sums kabhi neeche nahi jaate — isliye DCT ko a n ≥ 0 chahiye.
Step 2 — Partial sums ko bound karo.
Kyunki a n ≤ b n hai,
S n = ∑ k = 1 n a k ≤ ∑ k = 1 n b k ≤ ∑ k = 1 ∞ b k =: B .
Yeh step kyun? Agar ∑ b n converge kare, toh uske partial sums finite total B se bounded hain. Term-by-term domination woh bound S n tak pass kar deti hai.
Step 3 — Monotone Convergence lagao.
S n increasing aur B se upar bounded hai. Isliye S n ek finite limit par converge karta hai. Hence ∑ a n converge karta hai. ∎
Contrapositive se divergence wala half milta hai.
Agar ∑ a n diverge kare, toh S n → ∞ . Kyunki ∑ k = 1 n b k ≥ S n , bade partial sums bhi → ∞ hain, toh ∑ b n diverge karta hai. ∎
Intuition Sirf yahi do directions kyun?
Choti series ka converge karna tumhe badi ke baare mein kuch nahi batata (badi phir bhi blow up ho sakti hai). Badi series ka diverge karna tumhe choti ke baare mein kuch nahi batata (woh phir bhi converge ho sakti hai). Sirf "sahi direction" wale comparisons information dete hain — dono arrows yaad karo.
Ensure karo ki terms eventually ≥ 0 hain.
Convergence ya divergence guess karo (dominant power dekho → p -series / geometric socho).
Inequality sahi direction mein banao:
Convergence prove karne ke liye, ek bada convergent b n dhundho.
Divergence prove karne ke liye, ek chota divergent b n dhundho.
Inequality ko algebraically justify karo.
Known benchmark result batao aur conclude karo.
"SCALE" — S maller C onverges A bove, L arger diE s below.
Ya simply: "Convergence neeche flow karta hai, divergence upar."
Worked example Example 1 — Convergence:
n = 1 ∑ ∞ n 2 + 5 1
Guess: 1/ n 2 jaisa behave karta hai → converge karega.
Inequality: n 2 + 5 > n 2 ⇒ n 2 + 5 1 < n 2 1 .
Yeh step kyun? Hume ek bada convergent partner chahiye jo hamare series ko finite roof ke neeche trap kare.
Benchmark: ∑ 1/ n 2 ek p -series hai jisme p = 2 > 1 → converge karta hai.
Kyunki 0 ≤ n 2 + 5 1 ≤ n 2 1 aur upar wala converge karta hai, hamar series converge karti hai. ✅
Worked example Example 2 — Divergence:
n = 2 ∑ ∞ n − 1 1
Guess: 1/ n jaisa → diverge karega (harmonic).
Inequality: n − 1 < n ⇒ n − 1 1 > n 1 .
Yeh step kyun? Divergence prove karne ke liye hume ek chota divergent partner chahiye jo hume infinite floor se upar push kare.
Benchmark: ∑ 1/ n diverge karta hai (harmonic, p = 1 ).
Kyunki n − 1 1 ≥ n 1 ≥ 0 aur chota wala diverge karta hai, hamar series diverge karti hai. ✅
Worked example Example 3 — Sahi side chunna:
n = 1 ∑ ∞ n ln n (n ≥ 3 ke liye)
Guess: ln n grow karta hai, toh terms 1/ n se zyada hain → divergence.
Inequality: n ≥ 3 ke liye, ln n > 1 , toh n ln n > n 1 .
Yeh step kyun? Chota divergent partner 1/ n → hamare series ko diverge karne par push karta hai.
Diverge karti hai. ✅
Worked example Example 4 — Jab bound shape karni pade:
n = 1 ∑ ∞ n 2 2 + sin n
Numerator wiggle karta hai lekin − 1 ≤ sin n ≤ 1 , toh 1 ≤ 2 + sin n ≤ 3 .
Inequality: n 2 2 + sin n ≤ n 2 3 .
Yeh step kyun? Messy numerator ko uske maximum se replace karo taaki ek clean bada convergent series mile.
∑ 3/ n 2 = 3 ∑ 1/ n 2 converge karta hai → hamar series converge karti hai. ✅
Common mistake "Maine galat direction mein bound kiya aur phir bhi conclude kar liya."
Kyun sahi lagta hai: tumne ek inequality dhundh li — surely koi bhi inequality kaam aati hai!
Trap: Convergence prove karne ke liye tumhe ek bada convergent series chahiye, lekin students aksar a n ≥ n 2 1 (ek chota convergent series) dikhate hain. Woh useless direction hai — a n phir bhi diverge ho sakta hai.
Fix: Arrow ko goal se match karo : convergence ⇒ upar se trap karo; divergence ⇒ neeche se push karo.
Common mistake "Maine DCT negative terms wali series par use ki."
Kyun sahi lagta hai: inequality phir bhi sahi lag rahi thi.
Trap: Proof ka Step 1 toot jaata hai — agar terms negative ho sakti hain toh partial sums monotone nahi rehte, isliye Monotone Convergence fail ho jaata hai.
Fix: Sign-changing terms ke liye, absolute values par Comparison use karo (phir yeh absolute convergence ka statement hai) ya Alternating Series Test use karo.
n 2 + 5 1 < n 2 1 — lekin chote n ka kya?"
Kyun sahi lagta hai: chinta hoti hai ki inequality shuruaat mein fail ho jaaye.
Trap: yahan koi issue nahi — lekin general mein test ko sirf n ≥ N ke liye inequality chahiye. Finite terms convergence ko kabhi affect nahi karte.
Fix: "Eventually" kaafi hai; yeh state karo.
Recall Feynman: ek 12-saal ke bachche ko samjhao
Socho tum hamesha ke liye chhoti-chhoti candy ki dher jod rahe ho. Sab khud se nahi jod sakte, toh tum cheat karte ho: tum ek alag candy ki dher dhundho jise tum pehle se samajhte ho.
Agar tumhari dher hamesha choti ho ek aisi se jo ek finite jar mein add ho jaati hai, toh tumhari dher bhi ek jar mein fit ho jaayegi — woh badhna band ho jaati hai. (converges)
Agar tumhari dher hamesha badi ho ek aisi se jo infinity tak grow karti hai, toh tumhari dher bhi infinity tak grow karegi. (diverges)
Tum apni khud ki candies kabhi count nahi karte — bas pile-to-pile kisi trusted dost se compare karte ho!
Direct Comparison Test terms par kya do conditions chahiye? Woh non-negative honi chahiye (0 ≤ a n ≤ b n ), kam se kam eventually (for n ≥ N ).
Agar 0 ≤ a n ≤ b n aur ∑ b n converge kare, toh ∑ a n ka kya hoga? Woh converge karega (finite roof ke neeche trap hai).
Agar 0 ≤ a n ≤ b n aur ∑ a n diverge kare, toh ∑ b n ka kya hoga? Woh diverge karega (infinite floor ke upar push hai).
DCT ki proof ke peeche kaunsa theorem hai? Monotone Convergence Theorem (increasing + bounded above ⇒ converges).
DCT ke liye terms non-negative kyun honi chahiye? Taaki partial sums monotone increasing rahen, jisse Monotone Convergence apply ho sake.
DCT se ek series ko CONVERGE prove karne ke liye, tum use ek ___ series se compare karte ho jo ___ ho. bigger; convergent.
DCT se ek series ko DIVERGE prove karne ke liye, tum use ek ___ series se compare karte ho jo ___ ho. smaller; divergent.
Kya a n ≤ n 2 1 dikhana (smaller, convergent) prove karta hai ki ∑ a n converge karta hai? Nahi — woh useless direction hai; a n phir bhi diverge ho sakta hai.
∑ n 2 2 + s i n n ke liye, kaunsa bound convergence deta hai?n 2 2 + s i n n ≤ n 2 3 , aur ∑ 3/ n 2 converge karta hai.
Jab terms ka sign badle toh DCT ki jagah kaunsa test use karte hain? Absolute values compare karo (absolute convergence) ya Alternating Series Test use karo.
p-series test — sabse common benchmark (p > 1 converge karte hain, p ≤ 1 diverge).
Geometric series — ek aur common comparison partner.
Limit comparison test — DCT ka gentle cousin jab inequality awkward ho.
Monotone convergence theorem — proof ke peeche ka engine.
Harmonic series — canonical divergent benchmark.
Absolute convergence — jab terms sab positive nahi hote tab kya karo.
Integral test — benchmark series classify karne ka alternate tarika.
increasing + bounded converges
Monotone Convergence Theorem
Non-negative terms an,bn >= 0
Term-by-term 0 <= an <= bn
Partial sums Sn increasing
Sum bn conv => Sum an conv
Benchmarks p-series / geometric