Telescoping series: Worked examples — Hinglish deep dive

4.3.5 · D3 · Maths › Calculus III — Sequences & Series › Telescoping series

Shuru karne se pehle, ek word jo hum baar baar use karte hain: ek partial sum $S_{N}$ ka matlab sirf itna hai ki "pehle $N$ terms jodo aur ruk jao." $S_{N}$ ko $N$ badhne ke saath dekhna hi decide karta hai ki infinite sum kya hai — Sequence of Partial Sums dekho.

Scenario matrix

Har telescoping problem in cells mein se ek (ya ek blend) hai:

Cell	Kya badalta hai	Danger	Example
C1 Gap 1	$a_{n} = b_{n} - b_{n + 1}$ , clean	kuch nahin — textbook case	Ex 1
C2 Gap 2	$a_{n} = b_{n} - b_{n + 2}$	2 survivors har end pe	Ex 2
C3 Shifted start	series $n = 3$ se shuru, $1$ se nahin	lead term $b_{3}$ hai, $b_{1}$ nahin	Ex 3
C4 Non-vanishing tail	$b_{n} \to L \neq = 0$	answer hai $b_{1} - L$ , nahin $b_{1}$	Ex 4
C5 Root / conjugate form	difference ke peeche chhupa hua	dekhne ke liye rationalise karna padega	Ex 5
C6 Product/log form	$\prod$ ya $ln$ , second difference	ratios ke roop mein telescopes hota hai	Ex 6
C7 Word problem	real quantities (medicine, tiles)	$b_{n} - b_{n + 1}$ mein translate karo	Ex 7
C8 Exam twist	diverges / degenerate	pehchano ki yeh finite number par collapse nahin hota	Ex 8

Ab hum har cell ko hit karte hain.

C1 — clean gap-1 case

Worked example Ex 1 (cell C1):

$n = 1 \sum \infty \frac{1}{( 2 n - 1 ) ( 2 n + 1 )}$ Forecast: andaaza lagao — kya sum $1$ se bada hai ya chhota? (Yeh $\frac{1}{2}$ hai; yeh yaad rakho.)

Step 1. Partial Fraction Decomposition se split karo: $\frac{1}{( 2 n - 1 ) ( 2 n + 1 )} = \frac{A}{2 n - 1} + \frac{B}{2 n + 1}$ . Yeh step kyun? Hum $b_{n} - b_{n + 1}$ ki shape chahte hain; partial fractions hi woh tool hai jo denominator mein product se woh difference banata hai.

Step 2. Denominators clear karo: $1 = A (2 n + 1) + B (2 n - 1)$ . $n = \frac{1}{2}$ rakho $\Rightarrow 1 = 2 A \Rightarrow A = \frac{1}{2}$ . $n = - \frac{1}{2}$ rakho $\Rightarrow 1 = - 2 B \Rightarrow B = - \frac{1}{2}$ . $\frac{1}{( 2 n - 1 ) ( 2 n + 1 )} = \frac{1}{2} (\frac{1}{2 n - 1} - \frac{1}{2 n + 1}) .$ Yeh step kyun? Ab $b_{n} = \frac{1}{2 n - 1}$ aur $b_{n + 1} = \frac{1}{2 n + 1}$ — term bilkul $\frac{1}{2} (b_{n} - b_{n + 1})$ hai, ek gap-1 telescope.

Step 3. Partial sum: $S_{N} = \frac{1}{2} (b_{1} - b_{N + 1}) = \frac{1}{2} (1 - \frac{1}{2 N + 1})$ . Yeh step kyun? "First minus last" — parent note ka master result.

Step 4. Limit: $\frac{1}{2 N + 1} \to 0$ , toh $\sum = \frac{1}{2} (1 - 0) = \frac{1}{2}$ . Limits of Sequences dekho.

Verify: pehle 3 terms jodo: $\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} = 0.333 + 0.0667 + 0.0286 = 0.4286$ , aur $\frac{1}{2} (1 - \frac{1}{7}) = 0.4286$ ✓. Match karta hai, aur jab $N$ finite hai toh $\frac{1}{2}$ se neeche hai, jaisa hona chahiye.

C2 — gap 2 (do survivors har end pe)

Worked example Ex 2 (cell C2):

$n = 1 \sum \infty \frac{2}{n ( n + 2 )}$ Forecast: kyunki har end pe do terms bachte hain, expect karo ki answer ek "sum of two boundary pieces" hoga, ek nahin.

Step 1. $\frac{2}{n ( n + 2 )} = \frac{A}{n} + \frac{B}{n + 2}$ . Clear karo: $2 = A (n + 2) + B n$ . $n = 0 \Rightarrow A = 1$ ; $n = - 2 \Rightarrow B = - 1$ . Toh term $= \frac{1}{n} - \frac{1}{n + 2}$ . Yeh step kyun? Same partial-fraction engine, lekin note karo ki denominators 2 se differ karte hain, 1 se nahin.

Step 2. Yeh $b_{n} - b_{n + 2}$ hai jahan $b_{n} = \frac{1}{n}$ — ek gap-2 telescope. 5 terms likho: $S_{N} = (1 - \frac{1}{3}) + (\frac{1}{2} - \frac{1}{4}) + (\frac{1}{3} - \frac{1}{5}) + (\frac{1}{4} - \frac{1}{6}) + \dots$ Yeh step kyun? Parent note ka Mistake 2: gap 2 ke saath tumhe physically dekhna padega ki kaun se pieces milte hain. $- \frac{1}{3}$ teesre term ke $+ \frac{1}{3}$ ka wait karta hai, agle ka nahin.

Step 3. Cancellation ke baad survivors hain pehle do " $+$ " pieces ( $1, \frac{1}{2}$ ) aur aakhri do " $-$ " pieces ( $- \frac{1}{N + 1}, - \frac{1}{N + 2}$ ): $S_{N} = 1 + \frac{1}{2} - \frac{1}{N + 1} - \frac{1}{N + 2} .$ Yeh step kyun? Gap $k$ ⟹ $k$ survivors har side. Yahaan $k = 2$ .

Step 4. Limit: dono tails $\to 0$ , toh $\sum = 1 + \frac{1}{2} = \frac{3}{2}$ .

Verify: pehle 4 terms $= 0.6667 + 0.25 + 0.1333 + 0.0833 = 1.1333$ ; formula $S_{4} = 1.5 - \frac{1}{5} - \frac{1}{6} = 1.1333$ ✓.

C3 — shifted starting index

Worked example Ex 3 (cell C3):

$n = 3 \sum \infty \frac{1}{n ( n + 1 )}$ Forecast: parent ne prove kiya tha ki $\sum_{n = 1}^{\infty} = 1$ . Kyunki hum $n = 1, 2$ drop kar rahe hain, yeh 1 se kam hona chahiye.

Step 1. Term $= \frac{1}{n} - \frac{1}{n + 1}$ (parent se). Gap-1, $b_{n} = \frac{1}{n}$ . Yeh step kyun? Jaana-pehchana split reuse karo — algebra ko parwah nahin ki hum kahaan se shuru kar rahe hain.

Step 2. Starting index $3$ hai, toh surviving lead term $b_{3} = \frac{1}{3}$ hai, $b_{1}$ nahin: $S_{N} = b_{3} - b_{N + 1} = \frac{1}{3} - \frac{1}{N + 1} .$ Yeh step kyun? Parent mein Mistake 3: "first minus last" matlab tumhara first. Bounds track karo.

Step 3. Limit: $n = 3 \sum \infty = \frac{1}{3} - 0 = \frac{1}{3}$ .

Verify (do tarike se): directly $\frac{1}{3 \cdot 4} + \frac{1}{4 \cdot 5} + \frac{1}{5 \cdot 6} + \dots$ ; ya missing terms ko full sum se ghataao: $1 - \frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3} = 1 - \frac{1}{2} - \frac{1}{6} = \frac{1}{3}$ ✓. Dono $\frac{1}{3} < 1$ dete hain. ✓

C4 — tail vanish nahin karta

Worked example Ex 4 (cell C4):

$n = 1 \sum \infty (\frac{n}{n + 1} - \frac{n + 1}{n + 2})$ Forecast: yeh tempting lagta hai ki answer pehla " $+$ " piece $\frac{1}{2}$ hai. Ruko — yahaan tail zero nahin jaata!

Step 1. Yeh already ek difference hai: $b_{n} = \frac{n}{n + 1}$ , aur $a_{n} = b_{n} - b_{n + 1}$ . Gap-1. Yeh step kyun? Koi partial fractions nahin chahiye — telescoping form hamare haath mein hai; bas $b_{n}$ identify karo.

Step 2. Partial sum: $S_{N} = b_{1} - b_{N + 1} = \frac{1}{2} - \frac{N + 1}{N + 2}$ . Yeh step kyun? Master result phir se — lekin ab doosra piece dhyan se dekho.

Step 3. Tail ka limit: $N \to \infty lim \frac{N + 1}{N + 2} = 1$ , nahin $0$ . Yeh step kyun? $\frac{N + 1}{N + 2} = \frac{1 + 1/ N}{1 + 2/ N} \to \frac{1}{1} = 1$ . Yahi cell C4 ka poora point hai.

Step 4. Toh $\sum = \frac{1}{2} - 1 = - \frac{1}{2}$ — ek negative sum, aur $b_{1} = \frac{1}{2}$ nahin.

Verify: $S_{3} = \frac{1}{2} - \frac{4}{5} = - 0.3$ ; aur $\frac{1}{2} - \frac{N + 1}{N + 2}$ at $N = 3$ hai $\frac{1}{2} - 0.8 = - 0.3$ ✓. $N \to \infty$ ke saath yeh $- 0.5$ ki taraf jaata hai. ✓

C5 — root ke peeche chhupa hua difference

Worked example Ex 5 (cell C5):

$n = 1 \sum \infty \frac{1}{n + n + 1}$ Forecast: yahaan koi obvious difference nahin hai. Guess karo ki kya yeh converge bhi karta hai — terms $\frac{1}{2 n}$ jaisi shrink karti hain, jo... dhyan se dekho.

Step 1. Rationalise karo: upar aur neeche $n + 1 - n$ se multiply karo: $\frac{1}{n + n + 1} \cdot \frac{n + 1 - n}{n + 1 - n} = \frac{n + 1 - n}{( n + 1 ) - n} = n + 1 - n .$ Yeh step kyun? Conjugate trick hamara tool hai: $(n + 1 + n) (n + 1 - n) = (n + 1) - n = 1$ ugly fraction ko ek clean difference $b_{n + 1} - b_{n}$ mein badal deta hai jahan $b_{n} = n$ .

Step 2. Toh $a_{n} = n + 1 - n = b_{n + 1} - b_{n}$ (ek rising telescope). Partial sum: $S_{N} = N + 1 - 1 = N + 1 - 1.$ Yeh step kyun? Likho: $(2 - 1) + (3 - 2) + \dots + (N + 1 - N)$ — middle cancel hota hai, last minus first bachta hai.

Step 3. Limit: $N + 1 - 1 \to \infty$ . Series diverge karta hai. Yeh step kyun? Yahaan $b_{n} = n \to \infty$ , toh koi finite sum exist nahin karta — telescope extend karta rehta hai, kabhi fold shut nahin hota. Telescoping dono cases mein exact behaviour batata hai (dekho Convergence Tests for Series).

Verify: $S_{3} = (2 - 1) + (3 - 2) + (4 - 3) = 4 - 1 = 1$ ; formula $4 - 1 = 1$ ✓. Bina bound ke badh raha hai. ✓ $diverges$

C6 — product / log form (second difference)

Worked example Ex 6 (cell C6):

$n = 2 \prod \infty (1 - \frac{1}{n ^{2}})$ Forecast: yeh ek product hai, lekin " $1 - tiny$ " ke products finite limit ki taraf jaate hain. Guess: $\frac{1}{2}$ ? $1$ ? $0$ ?

Step 1. Har term ko factor karo: $1 - \frac{1}{n ^{2}} = \frac{( n - 1 ) ( n + 1 )}{n ^{2}}$ . Yeh step kyun? Difference of squares wo ratio structure expose karta hai jo product mein telescope karega.

Step 2. $N$ tak partial product: $P_{N} = \prod_{n = 2}^{N} \frac{( n - 1 ) ( n + 1 )}{n ^{2}} = neeche shift karta hai (n = 2 \prod N \frac{n - 1}{n}) \cdot upar shift karta hai (n = 2 \prod N \frac{n + 1}{n}) .$ Yeh step kyun? Do telescoping products mein split karna $b_{n} - b_{n + 1}$ ka multiplicative version hai: consecutive ratios cancel ho jaate hain.

Step 3. Har product telescope karta hai: $\prod_{n = 2}^{N} \frac{n - 1}{n} = \frac{1}{2} \cdot \frac{2}{3} \dots \frac{N - 1}{N} = \frac{1}{N}, \prod_{n = 2}^{N} \frac{n + 1}{n} = \frac{3}{2} \cdot \frac{4}{3} \dots \frac{N + 1}{N} = \frac{N + 1}{2} .$ Yeh step kyun? Ek telescoping product mein, ek factor ka numerator agle ka denominator cancel karta hai — middle collapses bilkul waise hi jaisa sum mein hota hai, sirf multiply hoke.

Step 4. Multiply karo: $P_{N} = \frac{1}{N} \cdot \frac{N + 1}{2} = \frac{N + 1}{2 N} \to \frac{1}{2}$ . Yeh step kyun? $\frac{N + 1}{2 N} = \frac{1 + 1/ N}{2} \to \frac{1}{2}$ . ( $ln$ lo aur parent ka $\sum ln (1 - \frac{1}{n ^{2}}) = - ln 2$ recover ho jaata hai, kyunki $ln \frac{1}{2} = - ln 2$ ✓.)

Verify: $P_{4} = (1 - \frac{1}{4}) (1 - \frac{1}{9}) (1 - \frac{1}{16}) = 0.75 \cdot 0.8889 \cdot 0.9375 = 0.625$ ; aur $\frac{5}{8} = 0.625$ ✓. $\frac{1}{2}$ ki taraf ja raha hai.

C7 — real-world word problem

Worked example Ex 7 (cell C7): drug clearance

Ek patient ke kidneys ek drug ko process karte hain aisa ki day $n$ pe nikali gayi drug ki maatra hai $R_{n} = \frac{100}{n} - \frac{100}{n + 1}$ milligrams (ek manufactured telescoping model). Kitne mg total milaakar hamesha ke liye nikale jaate hain? Forecast: kyunki day-1 removal mein already sabse bada piece hai, total ka andaaza $50$ mg ke aas paas lagao.

Step 1. Pehchano ki $R_{n} = b_{n} - b_{n + 1}$ hai jahan $b_{n} = \frac{100}{n}$ mg. Gap-1 telescope. Yeh step kyun? Ek word problem usi waqt solve ho jaata hai jab tum units mein chhupa hua $b_{n} - b_{n + 1}$ shape dhundh lete ho.

Step 2. $N$ din mein total: $S_{N} = b_{1} - b_{N + 1} = 100 - \frac{100}{N + 1}$ mg. Yeh step kyun? First minus last — beech ke dinon ke contributions running total mein cancel ho jaate hain.

Step 3. $N \to \infty$ ke saath: kul nikala gaya $= 100 - 0 = 100 mg$ . Yeh step kyun? $\frac{100}{N + 1} \to 0$ , toh body eventually $100$ mg kul clear kar deti hai.

Verify (units + numbers): har $R_{n}$ hai mg $-$ mg $=$ mg ✓. 4 din baad: $100 - \frac{100}{5} = 80$ mg nikala; direct sum $R_{1} + R_{2} + R_{3} + R_{4} = 50 + 16. \overline{6} + 8. \overline{3} + 5 = 80$ mg ✓. Mera " $\approx 50$ " ka forecast galat tha — tail bahut contribute karta hai; total $100$ mg hai.

C8 — exam twist: woh trap jo diverge karta hai

Worked example Ex 8 (cell C8):

$n = 1 \sum \infty (\frac{1}{n} - \frac{1}{n + 1} + \frac{1}{n})$ — trick pakdo Forecast: yeh lagta hai telescoping jaisa. Kya yeh converge karta hai?

Step 1. Pehle summand simplify karo: $\frac{1}{n} - \frac{1}{n + 1} + \frac{1}{n} = \frac{2}{n} - \frac{1}{n + 1}$ . Yeh step kyun? Kabhi blindly telescope mat karo — like terms combine karo taaki true term dikhey. Extra $+ \frac{1}{n}$ clean difference ko tod deta hai.

Step 2. $(\frac{1}{n} - \frac{1}{n + 1}) + \frac{1}{n}$ ke roop mein likhao. Pehla bracket telescope karke $1$ deta hai; bacha hua $\sum \frac{1}{n}$ harmonic series hai. Yeh step kyun? Genuinely-telescoping part ko non-telescoping part se alag karna trap reveal karta hai.

Step 3. $n = 1 \sum \infty \frac{1}{n}$ diverge karta hai ( $ln N$ ki tarah badhta hai). Finite number plus $\infty$ equals $\infty$ . Yeh step kyun? Convergence Tests for Series ke hisaab se harmonic tail dominate karta hai; telescoping part isse bacha nahin sakta.

Step 4. Conclusion: series diverge karta hai. $no finite sum$

Verify: partial sums badhte hain: $S_{1} = 1.5$ , $S_{2} \approx 2.17$ , $S_{3} \approx 2.68$ , aur yeh har bound se upar jaate rehte hain (harmonic growth) ✓. Exam chahta tha ki tum assume karo ki yeh telescope karta hai — fix hamesha yeh hai ki pehle simplify karo.

Har example ne kaun sa cell hit kiya?

Green cells (Ex 1,2,3,6,7) ek clean value pe converge karte hain; red cells (Ex 5, 8) diverge karte hain; yellow cell (Ex 4) converge karta hai lekin $b_{1} - L$ pe jahan $L \neq = 0$ . Teen colours cover karo aur tum har scenario dekh chuke ho.

Recall Rapid self-test

$\sum_{n = 1}^{\infty} \frac{1}{( 2 n - 1 ) ( 2 n + 1 )}$ ka sum? ::: $\frac{1}{2}$ Gap-2 telescope mein har end pe kitne terms bachte hain? ::: two $\sum_{n = 3}^{\infty} \frac{1}{n ( n + 1 )}$ ? ::: $\frac{1}{3}$ (lead term hai $b_{3}$ ) $\sum (\frac{n}{n + 1} - \frac{n + 1}{n + 2}) = - \frac{1}{2}$ kyun hai, $\frac{1}{2}$ nahin? ::: kyunki $lim b_{n} = 1 \neq = 0$ , toh sum $= b_{1} - lim b_{n} = \frac{1}{2} - 1$ Kya $\sum \frac{1}{n + n + 1}$ converge karta hai? ::: no — yeh equals $N + 1 - 1 \to \infty$ $\prod_{n = 2}^{\infty} (1 - \frac{1}{n ^{2}})$ ki value? ::: $\frac{1}{2}$

Connections

Telescoping series — parent; yeh page uska full example bank hai.
Partial Fraction Decomposition — cells C1, C2 ke liye engine.
Method of Differences — wahi collapse finite sums mein.
Sequence of Partial Sums — har "Verify" $S_{N}$ dekhta hai.
Limits of Sequences — C4 (tail $\to 1$ ) aur C5 (tail $\to \infty$ ) decide karta hai.
Convergence Tests for Series — C8 ko divergent bolne ke liye zaroori.
Geometric Series — contrast: ek aur exact-sum family.