Question

find Sn and S infty of the series
$\frac{1}{1 \times 2 \times 3} + \frac{1}{2 \times 3 \times 4} + \frac{1}{3 \times 4 \times 5} + .......$
​

Answer 1

Question :-

Find $S_n$ & $S_\infty$ of the series

$\frac{1}{1*2*3} + \frac{1}{2*3*4} + \frac{1}{3*4*5} + ...........$ ?

Answer:-

⇒ we have given that

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

∴ $T_n = \frac{1}{n(n+1)(n+2)}$

$T_n = \frac{(n+2) - n}{( n ( n +1) ( n+2)) *2}$

$T_n = \frac{1}{2} [ \frac{1}{n(n+1)} - \frac{1}{(n+1)( n+2)} ]$

_-_-_-_-_-_-_-_-_-_-_-_-_-_-

$s_n =$ ∑$T_n =$ ∑$\frac{1}{2} [ \frac{1}{n+ 1)} - \frac{1}{(n+1)(n+2)} ]$

$S_n = \frac{1}{2} [ \frac{1}{1.2} - \frac{1}{2.3} ]$

     $+ \frac{1}{2} [ \frac{1}{2.3} - \frac{1}{3.4} ]$

     $+ \frac{1}{2} [ \frac{1}{3.4} - \frac{1}{4.5} ]$

up to........

      $+ \frac{1}{2} [ \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} ]$

∴ $S_n = \frac{1}{2} [ \frac{1}{2} - \frac{1}{(n+1)(n+2)} ]$

$S_\infty = \frac{1}{2} [ \frac{1}{2} - \frac{1}{(\infty +1 ) ( \infty +2 )} ]$

$S_\infty = \frac{1}{2} * \frac{1}{2} = \frac{1}{6}$

Therfore:-

$S_\ifnty = \frac{1}{6}$

Thank you.....

Answer 2

We need to find general expression for,

$\longrightarrow S_n=\dfrac{1}{1\times2\times3}+\dfrac{1}{2\times3\times4}+\dfrac{1}{3\times4\times5}+\,\dots\,+\dfrac{1}{n(n+1)(n+2)}$

or,

$\displaystyle\longrightarrow S_n=\sum_{k=1}^n\dfrac{1}{k(k+1)(k+2)}$

We have,

• $\displaystyle\boxed{\dfrac{1}{n(n+1)(n+2)\dots(n+k)}=\dfrac{1}{k!}\sum_{r=0}^k(-1)^r\cdot\dfrac{^kC_r}{n+r}}$

Then we get,

$\displaystyle\longrightarrow S_n=\dfrac{1}{2}\sum_{k=1}^n\left[\dfrac{1}{k}-\dfrac{2}{k+1}+\dfrac{1}{k+2}\right]$

$\displaystyle\longrightarrow S_n=\dfrac{1}{2}\sum_{k=1}^n\left[\dfrac{1}{k}-\dfrac{1}{k+1}\right]-\dfrac{1}{2}\sum_{k=1}^n\left[\dfrac{1}{k+1}-\dfrac{1}{k+2}\right]$

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

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

$\displaystyle\longrightarrow\underline{\underline{S_n=\dfrac{n(n+3)}{4(n+1)(n+2)}}}$

For $n\to\infty,$

$\displaystyle\longrightarrow S_\infty=\dfrac{1}{4}\lim_{n\to\infty}\dfrac{n(n+3)}{(n+1)(n+2)}$

$\displaystyle\longrightarrow S_\infty=\dfrac{1}{4}\lim_{n\to\infty}\dfrac{\left(\dfrac{n(n+3)}{n^2}\right)}{\left(\dfrac{(n+1)(n+2)}{n^2}\right)}$

$\displaystyle\longrightarrow S_\infty=\dfrac{1}{4}\lim_{n\to\infty}\dfrac{\left(1+\dfrac{3}{n}\right)}{\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{2}{n}\right)}$

$\displaystyle\longrightarrow S_\infty=\dfrac{1}{4}\cdot\dfrac{1+3\times0}{(1+0)(1+2\times0)}\quad\quad\left[\because\lim_{n\to\infty}\dfrac{1}{n}=0\right]$

$\displaystyle\longrightarrow\underline{\underline{S_\infty=\dfrac{1}{4}}}$