Exercises — Telescoping series

Hum kya karte hain: bas substitute karo. $a_n=\frac{1}{n}-\frac{1}{n+1}$. $a_1=1-\frac{1}{2},a_2=\frac{1}{2}-\frac{1}{3},a_3=\frac{1}{3}-\frac{1}{4},a_4=\frac{1}{4}-\frac{1}{5}.$ Kyun mushkil tarike se add nahi karte: master formula already batata hai answer first minus last hai. $N=4$ ke saath: $S_4=b_1-b_5=1-\frac{1}{5}=\boxed{\frac{4}{5}}.$ Yeh kaisa lagta hai: term $a_2$ ka $+\frac{1}{2}$, term $a_1$ ke $-\frac{1}{2}$ ko khatam kar deta hai; $+\frac{1}{3},+\frac{1}{4}$ bhi kha liye jaate hain. Sirf $1$ (front se) aur $-\frac{1}{5}$ (back se) bachte hain.

(a) As written telescoping nahi hai — yeh ek single blob hai, koi difference visible nahi. (Iska koi elementary telescoping split bhi nahi hai; chhod do.) (b) Haan, lekin reversed sign order ke saath: $a_n=b_{n+1}-b_n$ jahan $b_n=\sqrt{n}$. Yeh phir bhi telescoping hai — survivors ka sign bas swap ho jaata hai: $S_N=b_{N+1}-b_1=\sqrt{N+1}-1$. (c) Haan, textbook form: $b_n=\frac{1}{n}$, toh $a_n=b_n-b_{n+1}$. ✓ Kyun (b) phir bhi count karta hai: telescoping ko sirf consecutive cancellation chahiye; $b$ upar jaaye ya neeche yeh sirf sign bookkeeping ka detail hai, koi naya phenomenon nahi.

Step 1 — split karo. Chahiye $\frac{1}{(2n-1)(2n+1)}=\frac{A}{2n-1}+\frac{B}{2n+1}$. Kyun partial fractions? Yeh woh engine hai jo ek fraction ko ek difference mein badalta hai jise hum telescope kar sakte hain. Denominators clear karo: $1=A(2n+1)+B(2n-1)$. $n=\frac{1}{2}$ set karo: $1=A(2)\Rightarrow A=\frac{1}{2}$. $n=-\frac{1}{2}$ set karo: $1=B(-2)\Rightarrow B=-\frac{1}{2}$. $a_n=\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right).$ Step 2 — $b_n$ identify karo. $b_n=\frac{1}{2n-1}$, aur $b_{n+1}=\frac{1}{2n+1}$. Gap-1 telescope. ✓ Step 3 — partial sum. $S_N=\frac{1}{2}(b_1-b_{N+1})=\frac{1}{2}\left(1-\frac{1}{2N+1}\right).$ Step 4 — limit. $\frac{1}{2N+1}\to 0$, toh $\sum =\frac{1}{2}(1-0)=\boxed{\frac{1}{2}}.$

Step 1 — split karo. $\frac{2}{n(n+2)}=\frac{A}{n}+\frac{B}{n+2}$. $2=A(n+2)+Bn$ se: $n=0\Rightarrow A=1$; $n=-2\Rightarrow B=-1$. $a_n=\frac{1}{n}-\frac{1}{n+2}.$ Step 2 — gap of 2 dekho. Yeh $b_n-b_{n+2}$ hai jahan $b_n=\frac{1}{n}$, nahi $b_n-b_{n+1}$. Toh har end par do terms bachte hain. Step 3 — likho aur collapse karo: $S_N=\left(1-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\cdots +\left(\frac{1}{N}-\frac{1}{N+2}\right).$ $+\frac{1}{3}$ baad wale $-\frac{1}{3}$ ko kha leta hai, wagera. Front survivors: $1,\frac{1}{2}$. Back survivors: $-\frac{1}{N+1},-\frac{1}{N+2}$. $S_N=1+\frac{1}{2}-\frac{1}{N+1}-\frac{1}{N+2}.$ Step 4 — limit. Dono tail terms $\to 0$: $\sum =1+\frac{1}{2}=\boxed{\frac{3}{2}}.$

Step 1 — $b_n$ identify karo. Yahan $b_n=\frac{n}{n+1}$, aur term literally $b_n-b_{n+1}$ hai. Koi partial fractions nahi chahiye. Step 2 — partial sum. $S_N=b_1-b_{N+1}=\frac{1}{2}-\frac{N+1}{N+2}.$ Step 3 — crucial limit. Yahan $b_n=\frac{n}{n+1}\to 1$, nahi $0$ (dekho Limits of Sequences). Toh: $\sum =b_1-{\lim }_{n\to \infty }{b}_n=\frac{1}{2}-1=\boxed{-\frac{1}{2}}.$ Kyun answer negative hai: har pair $1$ ki taraf neeche step karta hai, lekin running total $b_1=\frac{1}{2}$ minus ceiling $1$ pe anchored hai. Yahan poori baat yahi "last" survivor hai jo vanish nahi hota — middle cancel ho jaata hai, lekin "aakhri" survivor nahi jaata.

Step 1 — log ko difference mein split karo. Logs products/quotients ko sums/differences mein badal dete hain: $\ln \frac{n(n+2)}{(n+1{)}^2}=\ln \frac{n}{n+1}-\ln \frac{n+1}{n+2}.$ Expand karke check karo: $\ln n-\ln (n+1)-\ln (n+1)+\ln (n+2)=\ln n+\ln (n+2)-2\ln (n+1)$ ✓. Step 2 — $b_n$ identify karo. $b_n=\ln \frac{n}{n+1}$, term $=b_n-b_{n+1}$. Gap-1 telescope. Step 3 — partial sum. $S_N=b_1-b_{N+1}=\ln \frac{1}{2}-\ln \frac{N+1}{N+2}.$ Step 4 — limit. Jab $N\to \infty$, $\frac{N+1}{N+2}\to 1$, toh $\ln \frac{N+1}{N+2}\to \ln 1=0$. Isliye $\sum =\ln \frac{1}{2}-0=\boxed{-\ln 2}.$

Step 1 — same split. $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$, toh $b_n=\frac{1}{n}$. Step 2 — starting index ko respect karo. Sum $n=3,4,5,\dots$ run karta hai. Pehla surviving term $b_3$ hai, $b_1$ nahi: $S_N={\sum }_{n=3}^N\left(\frac{1}{n}-\frac{1}{n+1}\right)=b_3-b_{N+1}=\frac{1}{3}-\frac{1}{N+1}.$ Step 3 — limit. $\frac{1}{N+1}\to 0$, toh ${\sum }_{n=3}^{\infty }=\frac{1}{3}-0=\boxed{\frac{1}{3}}.$ Sanity check: $n=1$ se full sum $1$ hai; pehle do terms $\frac{1}{2}+\frac{1}{6}=\frac{2}{3}$ subtract karo toh $1-\frac{2}{3}=\frac{1}{3}$ milta hai. ✓

Step 1 — clever grouping. Notice karo $\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left[\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right].$ Verify karo: common denominator deta hai $\frac{1}{2}\cdot \frac{(n+2)-n}{n(n+1)(n+2)}=\frac{1}{2}\cdot \frac{2}{n(n+1)(n+2)}$ ✓. Yeh shape kyun? Hum chahte hain $b_n-b_{n+1}$. $b_n=\frac{1}{n(n+1)}$ set karne se $b_{n+1}=\frac{1}{(n+1)(n+2)}$ milta hai — perfect gap-1 telescope. Step 2 — partial sum. $S_N=\frac{1}{2}(b_1-b_{N+1})=\frac{1}{2}\left(\frac{1}{1\cdot 2}-\frac{1}{(N+1)(N+2)}\right).$ Step 3 — limit. Tail $\to 0$: $\sum =\frac{1}{2}\cdot \frac{1}{2}=\boxed{\frac{1}{4}}.$

Step 1 — difference engineer karo. $n=(n+1)-1$ use karke: $\frac{n}{(n+1)!}=\frac{(n+1)-1}{(n+1)!}=\frac{n+1}{(n+1)!}-\frac{1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}.$ Yeh trick kyun? Humein telescoping shape chahiye tha; factorials ek step shrink hote hain jab numerator $(n+1)$ se match kare. Toh $b_n=\frac{1}{n!}$. Step 2 — partial sum. $S_N=b_1-b_{N+1}=\frac{1}{1!}-\frac{1}{(N+1)!}=1-\frac{1}{(N+1)!}.$ Step 3 — limit. $\frac{1}{(N+1)!}\to 0$ (factorials bahut bade ho jaate hain), toh $\sum =\boxed{1}.$

Part A — gap-2 theorem. Partial sum likho aur survivors match karo: $S_N={\sum }_{n=1}^N(b_n-b_{n+2})=(b_1+b_2+\cdots +b_N)-(b_3+b_4+\cdots +b_{N+2}).$ $b_3$ se $b_N$ tak sab kuch dono brackets mein appear karta hai aur cancel ho jaata hai. Front survivors: $b_1,b_2$. Back survivors: $-b_{N+1},-b_{N+2}$. $S_N=b_1+b_2-b_{N+1}-b_{N+2}.$ $N\to \infty$ lo: kyunki $b_n\to L$, dono tail terms $\to L$, jo deta hai ${\sum }_{n=1}^{\infty }(b_n-b_{n+2})=b_1+b_2-2L.■$ Part B — $b_n=\frac{1}{n}$ pe apply karo (toh $L=0$). Tab $b_n-b_{n+2}=\frac{1}{n}-\frac{1}{n+2}=\frac{2}{n(n+2)}$, aur theorem deta hai ${\sum }_{n=1}^{\infty }\frac{2}{n(n+2)}=b_1+b_2-0=1+\frac{1}{2}=\frac{3}{2},$ isliye $\sum _{n=1}^{\infty }\frac{1}{n(n+2)}=\frac{3}{4}$. ✓ (Parent note se match karta hai — general theorem us example ko contain karta hai.)