Worked examples — Series — partial sums, convergence definition

4.3.3 · D3 · Maths › Calculus III — Sequences & Series › Series — partial sums, convergence definition

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The scenario matrix

Har series problem inme se kisi ek cell mein aati hai. Hamare neeche ka kaam har ek ko kam se kam ek baar hit karna hai.

Cell	Ise woh case kya banata hai	Example jo hum use karte hain
A Geometric, $|r|<1$, $r>0$	ratio positive aur shrinking → converges	Ex 1
B Geometric, $|r|<1$, $r<0$	ratio negative aur shrinking → phir bhi converges (alternating)	Ex 2
C Geometric, $|r|\ge 1$	ratio bahut bada → diverges; formula forbidden	Ex 3
D Telescoping	terms ek chain mein cancel ho jaate hain → baaki bache ends par converges	Ex 4
E Terms $\to 0$ par diverges	necessary-not-sufficient trap (harmonic-style)	Ex 5
F Terms $\nrightarrow 0$	$n$-th term test usse turant maar deta hai	Ex 6
G Bounded par koi limit nahin	oscillating partial sums → $\infty$ pe gaye bina diverges	Ex 7
H Real-world word problem	physical situation ko ek geometric series mein translate karo	Ex 8
I Exam twist: shifted index / mixed	ek geometric series ko ek alag start ya split ke neeche chhupi hui pehchano	Ex 9

Matrix ko ek checklist ki tarah padho. Neeche har example par uski cell ka stamp laga hai.

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Ex 1 — Cell A: geometric, positive shrinking ratio

Forecast: har term pehle wali ki half hai. Guess: kya yeh pile kahin ruk jaata hai, aur roughly kahan?

Figure dekho: har amber block pehle wali ki half hai, end-to-end stack ki gayi hai. Woh clearly ek wall ki taraf pile up ho rahi hain.

1. $a$ aur $r$ identify karo. $a{r}^{n-1}$ se match karte hue: first term $a=3$, ratio $r=\frac{1}{2}$. Yeh step kyun? Geometric formula tab hi samajh aata hai jab hum uske do ingredients naam le lein.
2. Passport $|r|<1$ check karo. Yahan $|\frac{1}{2}|=\frac{1}{2}<1$. ✅ Yeh step kyun? Closed form $\frac{a}{1-r}$ exist karta hai sirf isliye kyunki $r^N\to 0$, jiske liye $|r|<1$ chahiye. Isko skip karo aur tum ek forbidden formula use kar sakte ho.
3. Sum apply karo. $S=\frac{a}{1-r}=\frac{3}{1-\frac{1}{2}}=\frac{3}{\frac{1}{2}}=6.$ Yeh step kyun? Yeh $s_N=\frac{a(1-r^N)}{1-r}$ ki limit hai jab $r^N\to 0$.

Verify karo: Partial sums $s_1=3,s_2=4.5,s_3=5.25,s_4=5.625,\dots$ — har ek $6$ ke karib, kabhi us se aage nahin. Gap $6-s_N=6r^N$ zero ki taraf shrink karta hai. ✅

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Ex 2 — Cell B: geometric, negative shrinking ratio (alternating)

Forecast: terms sign flip karti hain aur shrink karti hain. Guess: kya flipping convergence barbaad kar deta hai, ya phir bhi settle ho jaata hai (aur pehle do partial sums ke beech land karta hai)?

Figure dekho: running total target ko overshoot karta hai phir undershoot karta hai, andar ki taraf zig-zag karta hua.

1. $a$ aur $r$ identify karo. $a=8$, $r=-\frac{1}{3}$. Yeh step kyun? Wahi do ingredients; sign $r$ ke andar rehta hai.
2. Passport check karo. $|r|=\frac{1}{3}<1$. ✅ Absolute value matter karta hai, sign nahin. Yeh step kyun? $r^N\to 0$ jab bhi $|r|<1$, chahe $r$ negative ho — sign sirf alternate karta hai jabki size khatam ho jaati hai.
3. Sum apply karo. $S=\frac{8}{1-(-\frac{1}{3})}=\frac{8}{\frac{4}{3}}=6.$ Yeh step kyun? Formula identical hai; bas dhyan rakho $1-r=1+\frac{1}{3}=\frac{4}{3}$.

Verify karo: $s_1=8$ (over), $s_2=5.333$ (under), $s_3=6.222$ (over), $s_4=5.926$ (under) — ek shrinking zig-zag $6$ ki taraf squeeze karta hua. ✅

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Ex 3 — Cell C: geometric with $|r|\ge 1$ (diverges)

Forecast: ratio $>1$ matlab har term badi hai. Guess: converge ya diverge? Aur kya $\frac{a}{1-r}$ allowed hai?

1. $a$ aur $r$ identify karo. $a=2$, $r=\frac{3}{2}$.
2. Passport check karo. $|\frac{3}{2}|=1.5\ge 1$. ❌ Formula forbidden. Yeh step kyun? Closed form $r^N\to 0$ se paida hua tha. Yahan $r^N=(1.5{)}^N\to \infty$, isliye naam lene ke liye koi finite limit nahin hai.
3. $s_N$ se confirm karo. $s_N=\frac{2(1-1.5^N)}{1-1.5}=\frac{2(1-1.5^N)}{-0.5}=4(1.5^N-1)\to \infty .$ Yeh step kyun? Jab shortcut illegal ho tab hamesha definition ($\lim s_N$) par wapas jao.

Verify karo: $s_1=2,s_2=5,s_3=9.5,s_4=16.25,\dots$ — upar ki taraf accelerate karta hua, koi ceiling nahin. Diverges. ✅

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Ex 4 — Cell D: telescoping

Forecast: parent ke $\frac{1}{n(n+1)}$ jaisa lagta hai par shifted hai. Guess: kya yeh phir bhi telescope karta hai, aur kya yeh $1$ mein ya kisi chhoti cheez mein sum karta hai?

1. Partial fractions. $\frac{1}{(n+1)(n+2)}=\frac{1}{n+1}-\frac{1}{n+2}.$ Yeh step kyun? Har term ko difference ke roop mein likhne se consecutive terms cancel ho jaate hain.
2. $s_N$ likho aur cancel karo. $s_N=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots +\left(\frac{1}{N+1}-\frac{1}{N+2}\right)=\frac{1}{2}-\frac{1}{N+2}.$ Yeh step kyun? Har inner fraction ek baar $+$ ke saath aur ek baar $-$ ke saath aata hai aur khatam ho jaata hai; sirf pehla left piece aur aakhri right piece bachta hai.
3. Limit lo. $s_N=\frac{1}{2}-\frac{1}{N+2}\to \frac{1}{2}.$ Yeh step kyun? Convergence $\lim s_N$ ke roop mein define hoti hai; tail $\frac{1}{N+2}\to 0$.

Verify karo: $s_1=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\approx 0.1667$; $s_2=\frac{1}{2}-\frac{1}{4}=0.25$; $s_3=0.3$; $0.5$ ki taraf chadh raha hai. ✅ (Parent ke $1$ se chhota kyunki humne pehla bada term drop kar diya.)

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Ex 5 — Cell E: terms $\to 0$ par series diverge karti hai

Forecast: terms zero ki taraf shrink hote hain. Guess: converge (crumbs tiny ho jaate hain) ya diverge (harmonic disguise mein)?

1. Terms check karo. $a_n=\frac{1}{2n}\to 0$. Yeh step kyun? $n$-th term test sirf convergence ko rule out kar sakta hai; yahan woh pass karta hai, isliye yeh hume kuch nahin batata — hume gehrai mein jaana hoga.
2. Constant factor out karo. $\sum \frac{1}{2n}=\frac{1}{2}\sum \frac{1}{n}.$ Yeh step kyun? Ek divergent series ko nonzero constant se multiply karna use rescue nahin kar sakta — $\frac{1}{2}\cdot \infty$ phir bhi $\infty$ hai.
3. Known result invoke karo. $\sum \frac{1}{n}$ Harmonic series hai aur diverges. Toh infinity ki half phir bhi infinity hai → diverges. Yeh step kyun? Ek established divergent benchmark reuse karne se grouping argument dobara derive karne ki zarurat nahin padti.

Verify karo: Parent ki tarah group karo, $\frac{1}{2}(1+\frac{1}{2}+(\frac{1}{3}+\frac{1}{4})+\cdots )$ — har bracket $\frac{1}{2}$ se zyada hai, isliye $s_N$ har bound se aage badh jaata hai. Diverges. ✅ Yeh necessary-not-sufficient trap concretely samjha diya.

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Ex 6 — Cell F: terms zero nahin jaate

Forecast: large $n$ ke liye fraction $3$ ke karib settle ho jaata hai. Guess: "roughly $3$ forever" add karne se kya hoga?

1. $\lim a_n$ compute karo. $\underset{n\to \infty }{\lim }\frac{3n+1}{n+2}=\lim \frac{3+\frac{1}{n}}{1+\frac{2}{n}}=\frac{3}{1}=3.$ Yeh step kyun? n-th Term Test for Divergence kehta hai: agar $a_n\nrightarrow 0$, series diverges — isliye pehli cheez check karo ki kya terms vanish karti hain.
2. $0$ se compare karo. $3\ne 0$, isliye $a_n\nrightarrow 0$. ❌ Yeh step kyun? Woh numbers add karna jo hamesha $3$ ke karib rehte hain obviously blow up kar deta hai — partial sums $\approx 3N$ ki tarah grow karte hain.
3. Conclude karo. $n$-th term test se, series diverges turant. Aur koi kaam nahin chahiye.

Verify karo: $a_{100}=\frac{301}{102}\approx 2.951$, $a_{1000}\approx 2.995$ — $3$ ke karib hover karta hua, kabhi $0$ ke karib nahin. Diverges. ✅

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Ex 7 — Cell G: bounded, oscillating, koi limit nahin

Forecast: yeh kabhi infinity ki taraf nahin bhaagta. Guess: kya "bounded rehna" matlab converge karna hai?

1. Partial sums list karo. $s_1=1,s_2=0,s_3=1,s_4=0,\dots$ Yeh step kyun? Convergence $(s_N)$ ke baare mein ek statement hai, isliye directly us sequence par dekho.
2. Kya $(s_N)$ ki koi limit hai? Yeh hamesha $1,0,1,0,\dots$ bounce karta rehta hai — koi single number nahin jiske karib yeh aata ho. Yeh step kyun? Ek limit ek value honi chahiye jiske karib tail arbitrarily close rahe; $1$ aur $0$ $\epsilon =0.4$ alag hain, isliye half-width $0.4$ ka koi band dono ko catch nahin kar sakta.
3. Conclude karo. Bounded par koi limit nahin ⇒ diverges. (Yeh bhi: $a_n=(-1{)}^{n-1}\nrightarrow 0$, isliye term test bhi agree karta hai.)

Verify karo: $s_N$ sirf $\{0,1\}$ values leta hai; limit-candidates ka set ek single point nahin hai, isliye $\lim s_N$ exist nahin karta. Diverges. ✅ Divergence ke liye $\to \infty$ zaroori nahin hai.

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Ex 8 — Cell H: real-world word problem (bouncing ball)

Forecast: yeh girta hai, bounce karta hai upar, girta hai, bounce karta hai upar... Guess: infinite bounces — infinite distance, ya koi finite total?

Figure dekho: neeche ki taraf drop ek baar count hoti hai; har bounce height do baar count hoti hai (upar phir neeche).

1. Distance split karo. Initial fall $=10$. Uske baad, height $h$ tak har bounce $2h$ contribute karta hai (upar + neeche). Yeh step kyun? Ek bounce ka upar aur neeche equal hote hain, isliye unhe $2h$ ke roop mein group karna geometric pattern ko saaf rakhta hai.
2. Bounce series banao. Peak heights: $10(0.6),10(0.6{)}^2,\dots$. Total bounce distance $D_{\text{bounce}}=2{\sum }_{n=1}^{\infty }10(0.6{)}^n=20{\sum }_{n=1}^{\infty }(0.6{)}^n.$ Yeh step kyun? Peaks $r=0.6$ ratio ke saath ek geometric sequence form karte hain; $2$ aur $10$ factor out karne se woh expose ho jaata hai.
3. Geometric part sum karo. $\sum _{n=1}^{\infty }(0.6{)}^n=\frac{0.6}{1-0.6}=\frac{0.6}{0.4}=1.5$ (passport: $|0.6|<1$ ✅). Isliye $D_{\text{bounce}}=20\times 1.5=30$. Yeh step kyun? Yeh $n=1$ se start hote hue $\sum a{r}^{n-1}$ hai jahan $a=0.6,r=0.6$ — ya sirf standard "first term over $1-r$", first term $=0.6$.
4. Initial drop add karo. Total $=10+30=40$ m. Yeh step kyun? Pehli fall bounce nahin thi, isliye use alag se add kiya jaata hai, ek baar count kiya.

Verify karo: Partial totals $10,10+2(6)=22,22+2(3.6)=29.2,29.2+2(2.16)=33.52,\cdots \to 40$ m. Units poori tarah metres mein hain. Infinitely many bounces ke bawajood finite total. ✅

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Ex 9 — Cell I: exam twist (shifted start + ek split)

Forecast: yeh hai geometric, par index games se galat plug-in karna aasaan ho jaata hai. Guess: kya answer wahi hai jaise $n=1$ se shuru hota?

1. Real first term dhundne ke liye pehle kuch terms likho. $5(\frac{1}{4}{)}^3+5(\frac{1}{4}{)}^4+5(\frac{1}{4}{)}^5+\cdots$ Actual first term hai $a=5(\frac{1}{4}{)}^3=\frac{5}{64}$; ratio $r=\frac{1}{4}$. Yeh step kyun? $\frac{a}{1-r}$ mein "$a$" ka matlab hai series ka actual first term, chahe woh kis $n$ par correspond kare.
2. Passport check karo. $|\frac{1}{4}|<1$ ✅. Yeh step kyun? Start index shift karna kabhi ratio nahin badalta, isliye convergence unaffected rehti hai — sirf total badlata hai.
3. True first term se sum karo. $S=\frac{a}{1-r}=\frac{5/64}{1-\frac{1}{4}}=\frac{5/64}{3/4}=\frac{5}{64}\cdot \frac{4}{3}=\frac{5}{48}.$ Yeh step kyun? Jab $a$ aur $r$ sahi hain, standard formula unchanged apply hota hai.

Verify karo (alternative — missing terms subtract karo): $n=0$ se full series hogi ${\sum }_{n=0}^{\infty }5(\frac{1}{4}{)}^n=\frac{5}{1-\frac{1}{4}}=\frac{20}{3}$. $n=0,1,2$ subtract karo: $5+\frac{5}{4}+\frac{5}{16}=\frac{80+20+5}{16}=\frac{105}{16}$. Phir $\frac{20}{3}-\frac{105}{16}=\frac{320-315}{48}=\frac{5}{48}$. ✅ Do routes, same answer.

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Active recall

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Connections

• Parent: partial sums & convergence
• Geometric series — Cells A, B, C, H, I sab yahan rehte hain
• Telescoping series — Cell D
• Harmonic series — Cell E ka benchmark
• n-th Term Test for Divergence — Cells F, G
• Sequences — limits and convergence — har verdict $(s_N)$ sequence ke baare mein ek statement hai
• p-series and the Integral Test, Comparison Test, Ratio & Root Tests — agle tools jab koi closed form exist nahin karta
• Power series and radius of convergence — geometric series with variable $r=x$