Telescoping series: Question bank — Hinglish deep dive

4.3.5 · D5 · Maths › Calculus III — Sequences & Series › Telescoping series

Shuru karne se pehle, ek shared vocabulary reminder taaki neeche har symbol samajh aaye:

Sahi hai ya galat — justify karo

Ek series tabhi telescope karti hai jab use partial fractions se likha ja sake.

Galat — partial fractions sirf sabse common tool hai

a_{n} = b_{n} - b_{n + 1}

banane ka; logs, trig differences, aur factorial ratios bhi telescope karte hain. Defining property difference form hai, na woh method jo use reveal karta hai.

Agar saare interior terms cancel ho jaayein, toh sum

b_{1}

hona chahiye.

Galat — cancellation ke baad

b_{1} - b_{N + 1}

bachta hai, aur sum hai

b_{1} - lim_{n \to \infty} b_{n}

. Sirf tab jab

b_{n} \to 0

hota hai, leftover gayab hota hai aur sum

b_{1}

ho jaata hai.

Ek telescoping series hamesha converge karti hai.

Galat — yeh converge karti hai tabhi jab

lim_{n \to \infty} b_{n}

exist kare (finite ho); agar

b_{n}

diverge kare ya badhta rahe, toh

S_{N}

bhi aisa hi karta hai. Telescoping

S_{N}

ka ek closed form guarantee karta hai, convergence nahi.

Agar

a_{n} = b_{n} - b_{n + 2}

hai toh exactly do terms har end pe survive karti hain.

Sahi — gap

2

ke saath, cancellation sirf har doosri helper value se match karti hai, isliye do leading (

b_{1}, b_{2}

) aur do trailing terms bachti hain. General rule: gap

k

leaves

k

leading aur

k

trailing survivors.

a_{n} = b_{n + 1} - b_{n}

likhna (helper increasing) telescoping tod deta hai.

Galat — yeh phir bhi telescope karta hai; bas

S_{N} = b_{N + 1} - b_{1}

milta hai, standard result ka sign flip. Difference ki direction sirf swap karti hai ki kaunsi boundary positive hai.

\sum_{n = 1}^{\infty} (b_{n} - b_{n + 1})

mein starting index

1

3

badalne par answer badal jaata hai.

Sahi — surviving lead term

b_{1}

ki jagah

b_{3}

ban jaata hai, toh sum hai

b_{3} - lim b_{n}

. Jo pehle do terms tumne drop kiye woh real contributions the, cancelled nahi the.

Agar

b_{n} \to 0

ho toh telescoping sum guaranteed positive hota hai.

Galat — sum hai

b_{1} - 0 = b_{1}

, jiska sign sirf

b_{1}

ka sign hai. Koi cheez force nahi karti ki

b_{1} > 0

ho; jaise ek helper jisme

b_{1} = - ln 2

hai negative sum deta hai.

Do alag helper sequences

b_{n}

ek hi series

a_{n}

produce kar sakti hain.

Sahi — har

b_{n}

mein koi bhi constant

c

add karne se

b_{n} - b_{n + 1}

unchanged rehta hai, isliye

b_{n}

sirf ek constant tak hi determined hota hai. Sum

b_{1} - lim b_{n}

unaffected rehta hai kyunki

c

cancel ho jaata hai.

Why questions

Partial fraction decomposition itni baar telescoping kyun unlock karti hai?

Yeh ek fraction ko simpler pieces mein todti hai jinke denominators ek shift se alag hote hain (jaise

\frac{1}{n}

aur

\frac{1}{n + 1}

), jo bilkul wahi

b_{n}

–

b_{n + 1}

pattern hai jo neighbour-to-neighbour cancel karta hai. Dekho Partial Fraction Decomposition.

Sirf boundary terms kyun bachti hain, koi interior wali nahi?

Har interior helper value

b_{k}

ek baar

+

ke saath appear karta hai (term

k

se) aur ek baar

-

ke saath (term

k - 1

se), toh woh pairs mein annihilate ho jaate hain. Sirf

b_{1}

(koi earlier partner nahi) aur

b_{N + 1}

(koi later partner nahi) unpaired hain.

Telescoping series methods mein "rare aur precious" kyun maana jaata hai?

Zyaadatar tools (dekho Convergence Tests for Series) sirf decide karte hain ki series converge karti hai ya nahi; telescoping actually exact value deliver karta hai, kyunki sum reduce ho jaata hai ek finite boundary expression mein.

Pura infinite sum sirf

b_{1}

aur

lim b_{n}

par kyun depend karta hai, middle ko ignore karke?

Kyunki har middle contribution construction se cancel ho jaati hai, total poori tarah ek "boundary" quantity hai — jaise ek definite integral sirf apne endpoints par depend karta hai. Interior genuinely value ke liye irrelevant hai.

Hume

lim_{n \to \infty} b_{n}

kyun compute karna chahiye, na ki sirf

lim a_{n}

lim a_{n} = 0

convergence ke liye required hai lekin sum ke baare mein kuch nahi batata. Value

b_{1} - lim b_{n}

mein rehti hai, toh helper ka limit — Limits of Sequences ke zariye — answer fix karta hai.

Telescoping integration ke peeche fundamental idea ka discrete cousin kyun hai?

Ek difference

b_{n} - b_{n + 1}

ka sum endpoints tak collapse ho jaata hai, ek derivative

F^{'}

ka integral karne se

F (b) - F (a)

milne ki tarah. Yahi Method of Differences hai, finite/discrete analogue.

Edge cases

Ek series jisme

b_{n} \to 3

ho (nonzero limit): kya yeh phir bhi telescope karti hai, aur sum kya hai?

Haan, yeh phir bhi telescope karti hai — telescoping difference form ke baare mein hai, limit ke nahi. Sum hai

b_{1} - 3

; machinery unchanged hai, sirf surviving limit nonzero hai.

b_{n}

oscillate kare, jaise

b_{n} = (- 1)^{n}

: kya series convergent hai?

Nahi —

S_{N} = b_{1} - b_{N + 1}

do values ke beech flipping karta rehta hai, toh

lim b_{N + 1}

exist nahi karta. Series oscillation se diverge karti hai chahe algebraically yeh "telescope" kare.

Ek single-term "series"

\sum_{n = 1}^{1} (b_{n} - b_{n + 1})

: kya hota hai?

Yeh sirf

b_{1} - b_{2}

hai aur cancel karne ke liye kuch nahi — telescoping gracefully degenerate ho jaata hai ek difference mein. Formula

S_{N} = b_{1} - b_{N + 1}

phir bhi

N = 1

ke saath hold karta hai.

Gap puri length ke barabar ho, jaise

\sum_{n = 1}^{3} (b_{n} - b_{n + 3})

: kitna cancel hoga?

Kuch cancel nahi hoga — gap

3

ke saath sirf

3

terms mein, koi

- b_{n + 3}

apne matching

+ b_{n + 3}

se range ke andar nahi milta. Saare chhe pieces survive karte hain:

b_{1} + b_{2} + b_{3} - b_{4} - b_{5} - b_{6}

b_{n} \to \infty

ho (helper unbounded badhta rahe): sum kya hai?

Series diverge karti hai —

S_{N} = b_{1} - b_{N + 1} \to - \infty

. Telescoping phir bhi exact partial sum deta hai, lekin limit finitely exist nahi karti, toh koi sum nahi hai.

Do pieces equal hoon,

a_{n} = b_{n} - b_{n} = 0

: kya yeh ek telescoping series hai?

Trivially haan, sum

0

ke saath — har term zero hai. Yeh degenerate boundary case hai jahan "first minus last"

b_{1} - b_{1}

deta hai jab tum note karo ki

b_{n + 1} = b_{n}

sab kuch nothing mein collapse kar deta hai.

Recall Ek-line self-test

$\sum_{n = 1}^{\infty} (\frac{n}{n + 1} - \frac{n + 1}{n + 2})$ ka sum? ::: $\frac{1}{2} - 1 = - \frac{1}{2}$ , kyunki $lim b_{n} = 1 \neq = 0$ .

Connections

Telescoping series (Hinglish) — woh parent topic jinhe ye traps drill karte hain.
Partial Fraction Decomposition — woh engine jise zyaadatar traps abuse karti hain.
Sequence of Partial Sums — jahan collapse define hota hai.
Limits of Sequences — woh limit jo har edge case decide karti hai.
Convergence Tests for Series — kyun "converges" ≠ "is value ka matlab".
Method of Differences — discrete-analogue lens.
Geometric Series — doosri exact-sum family, contrast ke liye.

Question bank — Telescoping series