Exercises — Series — partial sums, convergence definition

$a$ aur $r$ kya hain? Template hai $\sum a{r}^{n-1}$. Term by term match karo: powers ke saath multiply hone wala constant $a=5$ hai; jo base $n-1$ ki power tak raise hoti hai woh $r=\frac{2}{3}$ hai.

WHY $r$ ka sign/size matter karta hai? Geometric series tab exactly converge karti hai jab $|r|<1$, kyunki partial sum $s_N=\frac{a(1-r^N)}{1-r}$ hai aur sirf $r^N$ hi $N$ pe depend karta hai. Jab $|r|<1$, to $r^N\to 0$; running total chalना band kar deta hai.

Yahan $|r|=\frac{2}{3}<1$, isliye yeh converges karta hai $S=\frac{a}{1-r}=\frac{5}{1-\frac{2}{3}}=\frac{5}{\frac{1}{3}}=15.$

Telescoping kya banata hai ise? Har term ek difference hai do pieces ka jo same shape $\frac{1}{n}$ ki hain, shifted indices ($n$ aur $n+2$) pe evaluate ki gayi hain. Jab tum consecutive terms add karte ho, to ek term ka subtracted piece do terms baad positive piece ke roop mein wapas aata hai, isliye cancel ho jaata hai.

Kaunse pieces bachte hain? Kyunki gap $2$ hai (na ki $1$), cancellation ke baad do leading positives aur do trailing negatives bache rehte hain: $s_N=\underset{\text{pehle do }+}{\underbrace{1+\frac{1}{2}}}-\underset{\text{aakhiri do }-}{\underbrace{\frac{1}{N+1}-\frac{1}{N+2}}}.$ Beech wala sab cancel ho jaata hai. (Actual value hum L3·Q2 mein calculate karenge.)

$a{r}^{n-1}$ form mein rewrite karo. $\frac{3}{4^n}=3\cdot 4^{-n}=3\cdot 4^{-1}\cdot 4^{-(n-1)}=\frac{3}{4}{\left(\frac{1}{4}\right)}^{n-1}$. To $a=\frac{3}{4}$, $r=\frac{1}{4}$. (Check by $n=1$: $\frac{3}{4^1}=\frac{3}{4}$ ✓.)

Convergence test karo. $|r|=\frac{1}{4}<1$, to yeh converge karta hai. $S=\frac{a}{1-r}=\frac{\frac{3}{4}}{1-\frac{1}{4}}=\frac{\frac{3}{4}}{\frac{3}{4}}=1.$

WHY pehle $n$-th Term Test use karein? Yeh pehle ek sasta sawaal answer karta hai: kya terms $0$ tak shrink bhi ho rahi hain? Agar nahi, to series converge nahi kar sakti (ek necessary condition hai), aur bas.

Term ka limit compute karo. Top aur bottom ko $n^2$ se divide karo: ${\lim }_{n\to \infty }\frac{2n^2+1}{5n^2+n}={\lim }_{n\to \infty }\frac{2+\frac{1}{n^2}}{5+\frac{1}{n}}=\frac{2}{5}\ne 0.$ Kyunki $a_n\to \frac{2}{5}\ne 0$, n-th Term Test for Divergence ke by series diverges.

Sum ke roop mein set up karo. $0.\overline{7}=\frac{7}{10}+\frac{7}{100}+\frac{7}{1000}+\cdots ={\sum }_{n=1}^{\infty }\frac{7}{10}{\left(\frac{1}{10}\right)}^{n-1}$, to $a=\frac{7}{10}$, $r=\frac{1}{10}$.

Sum karo. $|r|=\frac{1}{10}<1$: $S=\frac{\frac{7}{10}}{1-\frac{1}{10}}=\frac{\frac{7}{10}}{\frac{9}{10}}=\frac{7}{9}.$ To $0.\overline{7}=\frac{7}{9}$ — repeating decimals sirf convergent geometric series hote hain chhupe hue.

Ratio identify karo. Yeh geometric hai $a=1$ aur $r=x-2$ ke saath.

Convergence condition. Humein chahiye $|r|<1$, yaani $|x-2|<1$, yaani $-1<x-2<1$, yaani $1<x<3.$ Figure 1 dekho: shaded interval $(1,3)$ exactly woh strip hai jahan ratio $(-1,1)$ ke andar rehta hai. Bahar ratio ke powers $r^{n-1}$ shrink nahi karte aur $s_N$ ka koi limit nahi hota.

Endpoint check (kabhi skip mat karo!):

• $x=3\Rightarrow r=1$: $s_N=N\to \infty$, diverges.
• $x=1\Rightarrow r=-1$: $s_N$ bounce karta hai $1,0,1,0,\dots$, koi limit nahi, diverges.

Interval ke andar sum. $S=\frac{a}{1-r}=\frac{1}{1-(x-2)}=\frac{1}{3-x}$ for $1<x<3$. (Yeh Power series and radius of convergence ka pehla taste hai.)

$s_N$ carefully likhо. Do shifted lists align karo:

Partial sums compute karo. Terms hain $1,-1,1,-1,\dots$ $s_1=1,s_2=0,s_3=1,s_4=0,\dots$ To $s_N=1$ odd $N$ ke liye aur $s_N=0$ even $N$ ke liye.

Figure 2 dekho: running total hamesha $1$ aur $0$ ke beech hop karta rehta hai. Yeh kabhi kisi ek value ke paas nahi jaata, isliye ${\lim }_{N\to \infty }{s}_N$ exist nahi karta.

Conclusion. Divergence ka matlab infinity tak blow up karna nahi hota — do heights ke beech bouncing karna bina settle kiye bhi divergence hai. ($n$-th Term Test se bhi consistent hai: $a_n=(-1{)}^{n-1}\nrightarrow 0$.)

WHY hum ise split kar sakte hain? Agar do series $\sum b_n$ aur $\sum c_n$ dono converge karti hain, to $\sum (b_n+c_n)$ dono limits ke sum tak converge karti hai — kyunki partial sums add hote hain: $s_N=\sum b_n+\sum c_n$, aur sum ka limit, limits ka sum hota hai.

Piece 1 — geometric. ${\sum }_{n=1}^{\infty }\frac{1}{2^n}=\sum \frac{1}{2}(\frac{1}{2}{)}^{n-1}$, $a=r=\frac{1}{2}$, $|r|<1$: converges to $\frac{\frac{1}{2}}{1-\frac{1}{2}}=1$.

Piece 2 — telescoping. ${\sum }_{n=1}^{\infty }\frac{1}{n(n+1)}$: parent note se yeh telescope karta hai $s_N=1-\frac{1}{N+1}\to 1$.

Combine karo. Dono converge karte hain, isliye total converges to $S=1+1=2.$

Har piece analyse karo. $\sum \frac{1}{3^n}$ geometric hai $r=\frac{1}{3}$ ke saath: converges (to $\frac{1}{2}$). Lekin $\sum \frac{1}{n}$ Harmonic series hai: diverges.

WHY poori cheez diverge karti hai. Likho $s_N=A_N+B_N$ jahan $A_N\to \frac{1}{2}$ (bounded) aur $B_N={\sum }_{n=1}^N\frac{1}{n}\to \infty$. Convergent-plus-divergent sum divergent hota hai: agar $s_N$ converge karta, to $B_N=s_N-A_N$ do convergent sequences ka difference hota, hence convergent — jo $\sum \frac{1}{n}$ ke diverge karne ka contradiction hai. To yeh series diverges.

(Moral: ek bura saib poore sum ko kharab kar deta hai. Tum limits tab hi add kar sakte ho jab dono exist karein.)

Step 1 — partial fractions (WHY: ek aisa difference force karne ke liye jo cancel ho). Solve karo $\frac{2}{(2n-1)(2n+1)}=\frac{A}{2n-1}+\frac{B}{2n+1}$. Multiply out karo: $2=A(2n+1)+B(2n-1)$. $n=\frac{1}{2}$ rakhne se $A=1$ milta hai; $n=-\frac{1}{2}$ rakhne se $B=-1$ milta hai. To $\frac{2}{(2n-1)(2n+1)}=\frac{1}{2n-1}-\frac{1}{2n+1}.$

Step 2 — partial sum telescopes. $s_N=\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\cdots +\left(\frac{1}{2N-1}-\frac{1}{2N+1}\right)=1-\frac{1}{2N+1}.$

Step 3 — limit. $s_N=1-\frac{1}{2N+1}\to 1$. Series converges to $1$.

(a) WHY sum $\lim s_N$ ke barabar hota hai. By definition series ki value hai hi uske partial sums ka limit. Top aur bottom ko $N$ se divide karo: ${\lim }_{N\to \infty }\frac{3N}{N+2}={\lim }_{N\to \infty }\frac{3}{1+\frac{2}{N}}=3.$ To yeh converges to $3$.

(b) Terms recover karo. Parent note se, $a_N=s_N-s_{N-1}$ (running total ka jump) hai. $N\ge 2$ ke liye:

$r$ identify karo. $r=\frac{p}{p+1}$ aur first term ($n=1$ pe) $a=\frac{p}{p+1}$ ke saath, yeh geometric hai.

$|r|<1$ kab hota hai? Kyunki $p>0$ hai, $0<\frac{p}{p+1}<1$ hamesha hota hai (numerator denominator se chhota hai, dono positive hain). To yeh har $p>0$ ke liye converge karta hai.

Sum. $S=\frac{a}{1-r}=\frac{\frac{p}{p+1}}{1-\frac{p}{p+1}}=\frac{\frac{p}{p+1}}{\frac{1}{p+1}}=p.$ Ek clean result: poori series exactly $p$ tak sum ho jaati hai.