Question

find Sn & Sx of the series
$\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + ..........$
​

Answer 1

Question:-

⇒ Find $S_{n}$ & $S_{\infty}$ of the series

$\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4} + ............$

Answer:-

⇒ As it is given that

$\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4} + ............$

∴ $T_{n} = \frac{1}{n(n+1)} = \frac{n + 1 - n}{n(n+1)}$     { last - first]

⇒ $T_n = \frac{1}{n} - \frac{1}{n+1}$  ---------> [ to consecative term ]

therfore:-

 $S_n =$ ∑$T_n = T_1 + T_2 + ......... + Tn$

 $S_n =$ ∑$( \frac{1}{n} - \frac{1}{n+1} )$

$S_n = [ \frac{1}{n} - \frac{1}{1+1} + \frac{1}{2} - \frac{1}{2+1} + \frac{1}{3} - \frac{1}{3+1} +........+ \frac{1}{n} - \frac{1}{n+1} ]$

$[S_n = 1 - \frac{1}{n+1} = \frac{n+1-1}{n+1} = \frac{n}{1+n} ]$

$S_\infty = 1 - \frac{1}{\infty + 1} = 1 - \frac{1}{\infty}$

$S_\infty = 1 - 0$

$S_\infty = 1$

Thank you.....

Answer 2

We need to find general expression for,

$\longrightarrow S_n=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\,\dots\,+\dfrac{1}{n(n+1)}$

or,

$\displaystyle\longrightarrow S_n=\sum_{k=1}^n\dfrac{1}{k(k+1)}$

$\displaystyle\longrightarrow S_n=\sum_{k=1}^n\dfrac{(k+1)-k}{k(k+1)}$

$\displaystyle\longrightarrow S_n=\sum_{k=1}^n\left[\dfrac{1}{k}-\dfrac{1}{k+1}\right]$

$\displaystyle\longrightarrow S_n=\sum_{k=1}^n\dfrac{1}{k}-\sum_{k=1}^n\dfrac{1}{k+1}$

$\displaystyle\longrightarrow S_n=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\,\dots\,+\dfrac{1}{n}\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}+\,\dots\,+\dfrac{1}{n}+\dfrac{1}{n+1}\right)$

$\displaystyle\longrightarrow S_n=1-\dfrac{1}{n+1}$

$\displaystyle\longrightarrow\underline{\underline{S_n=\dfrac{n}{n+1}}}$

For $n\to\infty,$

$\displaystyle\longrightarrow S_\infty=\lim_{n\to\infty}\dfrac{n}{n+1}$

$\displaystyle\longrightarrow S_\infty=\lim_{n\to\infty}\dfrac{1}{\left(\dfrac{n+1}{n}\right)}$

$\displaystyle\longrightarrow S_\infty=\lim_{n\to\infty}\dfrac{1}{\left(1+\dfrac{1}{n}\right)}$

$\displaystyle\longrightarrow S_\infty=\dfrac{1}{1+0}\quad\quad\left[\because\lim_{n\to\infty}\dfrac{1}{n}=0\right]$

$\displaystyle\longrightarrow\underline{\underline{S_\infty=1}}$