4.3.5
Ek telescoping series woh hoti hai jisme har term ko ek single sequence ke do pieces ke difference ke roop mein likha ja sake, taaki jab aap terms add karo, interior pieces cancel ho jayein aur sirf kuch hi bachein — bilkul waise jaise ek folding telescope apne andar collapse ho jaati hai.
Zyaadatar series ko exactly sum karna impossible hai — hume sirf convergence tests milte hain. Telescoping un rare cases mein se ek hai jahan hume EXACT sum milta hai , sirf "yeh converge karta hai" nahi. Trick yeh hai: sum ko ek cancellation ke problem mein badlo.
Definition Telescoping seriesEk series ∑ n = 1 ∞ a n telescoping hai agar koi sequence b n a n = b n − b n + 1 . S N = ∑ n = 1 N a n = b 1 − b N + 1 . ⟺ lim N → ∞ b N + 1 ∑ a n = b 1 − lim N → ∞ b N + 1
WHAT chahiye: S N = ∑ n = 1 N ( b n − b n + 1 )
Har term likh do — yahi poora game hai:
S N = ( b 1 − b 2 ) + ( b 2 − b 3 ) + ( b 3 − b 4 ) + ⋯ + ( b N − b N + 1 ) Yeh step kyun? Main expand karta hoon taaki physically dekh sakoon ki kya kisse touch kar raha hai. Term k − 1 − b k k + b k
Survivors aur cancellers ko group karo:
S N = b 1 + 0 ( − b 2 + b 2 ) + 0 ( − b 3 + b 3 ) + ⋯ + 0 ( − b N + b N ) − b N + 1 Yeh step kyun? 2 ≤ k ≤ N b k + − pehla b 1 aakhri − b N + 1
S N = b 1 − b N + 1 Limit lo: agar b N + 1 → L n = 1 ∑ ∞ a n = b 1 − L
Series rarely pehle se b n − b n + 1 manufacture karna padta hai, usually partial fractions ke zariye.
n = 1 ∑ ∞ n ( n + 1 ) 1 Step 1 — difference mein split karo. Likho n ( n + 1 ) 1 = n A + n + 1 B Kyun? Mujhe b n − b n + 1
Denominators clear karo: 1 = A ( n + 1 ) + B n n = 0 A = 1 n = − 1 B = − 1 n ( n + 1 ) 1 = n 1 − n + 1 1 Step 2 — b n Yahan b n = n 1 a n = b n − b n + 1 Kyun? Confirm karta hai ki yeh telescopes karta hai — term n − n + 1 1 n + 1 + n + 1 1
Step 3 — partial sum. S N = b 1 − b N + 1 = 1 − N + 1 1
Step 4 — limit. N → ∞ lim ( 1 − N + 1 1 ) = 1. n = 1 ∑ ∞ n ( n + 1 ) 1 = 1
Worked example Two-step gap:n = 1 ∑ ∞ n ( n + 2 ) 1 Step 1. n ( n + 2 ) 1 = n 1/2 − n + 2 1/2 = 2 1 ( n 1 − n + 2 1 ) Kyun? A = 2 1 , B = − 2 1 1 = A ( n + 2 ) + B n
Step 2 — gap of 2 se savdhan raho. Yeh b n − b n + 1 b n − b n + 2 do terms bachti hain.
Likh ke dikhaao:
S N = 2 1 [ ( 1 − 3 1 ) + ( 2 1 − 4 1 ) + ( 3 1 − 5 1 ) + ⋯ + ( N 1 − N + 2 1 ) ] + 3 1 − 3 1 Start mein survivors: 1 2 1 End mein: − N + 1 1 , − N + 2 1 S N = 2 1 ( 1 + 2 1 − N + 1 1 − N + 2 1 ) Step 3 — limit. ∑ = 2 1 ( 1 + 2 1 ) = 4 3
Worked example Logs bhi telescope karte hain:n = 2 ∑ ∞ ln ( 1 − n 2 1 ) Step 1. 1 − n 2 1 = n 2 ( n − 1 ) ( n + 1 ) = ln ( n − 1 ) + ln ( n + 1 ) − 2 ln n Kyun? Logs products ko sums mein badal dete hain; yeh ln n Step 2 — telescoping regroup. Term ko consecutive ratios ke difference ke roop mein likho:
ln ( 1 − n 2 1 ) = [ ln n n + 1 ] − [ ln n − 1 n ] . Kyun? Expand karo: ln n n + 1 − ln n − 1 n = ln ( n + 1 ) + ln ( n − 1 ) − 2 ln n b n + 1 − b n b n = ln n − 1 n Step 3 — n = 2 N ln n − 1 n − ln 1 2 + ln N N + 1 S N = ln N N + 1 − ln 1 2 = ln 2 N N + 1 . Step 4 — limit. S N = ln 2 N N + 1 N → ∞ ln 2 1 = − ln 2.
Common mistake Classic errors ko steel-man karnaMistake 1 — "interior terms hamesha fully cancel ho jaate hain, toh sum bas b 1
Kyun sahi lagta hai: n ( n + 1 ) 1 N + 1 1 → 0 = b 1 Fix: Sum hai b 1 − lim b N + 1 b n → 0 ∑ ( n + 1 n − n + 2 n + 1 ) b 1 = 2 1 lim b n = 1 = 2 1 − 1 = − 2 1 2 1
Mistake 2 — gap > 1 hone par extra survivors bhool jaana.
Kyun sahi lagta hai: "Cancellation hoti hai, toh bas pehla aur aakhri rakh lo." Lekin b n − b n + 2 do lead aur do trailing terms bachte hain. Pattern trust karne se pehle hamesha 4–5 terms likh ke dekho.
Mistake 3 — galat starting index. Agar ek series n = 3 b 3 b 1
Recall Feynman: ek 12-saal ke bachche ko explain karoSocho ek line mein bache hain jinmein se har ek ek card pakde hua hai. Pehla baccha kehta hai "+1", aur saath mein "− 2 1 2 1 − 3 1 3 1 − 4 1 pehla number (+1, isko cancel karne wala koi pehle nahi tha) aur bilkul aakhri chota sa number bachta hai. Jaise-jaise line infinitely lambi hoti jaati hai, aakhri number zero tak shrink ho jaata hai — toh poora shorgul wala sum bas 1 hai. Woh ends-tak-collapse-ho-jaana bilkul waise hai jaise ek telescope band ho jaati hai.
"FIRST minus LAST, the MIDDLE goes past."
S N = b first − b last+1
Telescoping series kya define karta hai? Ek series ∑ a n b n a n = b n − b n + 1
∑ n = 1 N ( b n − b n + 1 ) S N S N = b 1 − b N + 1
∑ n = 1 ∞ n ( n + 1 ) 1 1 S N = 1 − N + 1 1 → 1
Telescoping difference create karne ke liye kaun sa tool use hota hai? Partial fraction decomposition.
∑ n = 1 ∞ n ( n + 2 ) 1 4 3
Infinite sum hamesha sirf b 1 Kyunki ∑ a n = b 1 − lim n → ∞ b n b n → 0
Gap-k a n = b n − b n + k k k
Telescoping series ke converge karne ki condition? lim n → ∞ b n
Partial Fraction Decomposition — woh engine jo b n − b n + 1 Geometric Series — exact-sum ki doosri family; telescoping se compare karo.Sequence of Partial Sums — telescoping is sequence ke collapse hone se define hoti hai.Convergence Tests for Series — telescoping value deta hai, sirf convergence nahi.Limits of Sequences — bacha hua lim b n Method of Differences — discrete analogue / finite sums mein yahi idea.
Exact sum, not just convergence
1 over n n+1 = 1/n - 1/n+1