Find Sn and S infty of the series

1/1x2x3 + 1/2x3x4 + 1/3x4x5

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1. Question :-

   We need to find general expression for,

   [tex]\longmapsto S_n=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\,\dots\,+\dfrac{1}{n(n+1)}[/tex]

   or,

   [tex]\displaystyle\longmapsto S_n=\sum_{k=1}^n\dfrac{1}{k(k+1)}[/tex]

   [tex]\displaystyle\longmapsto S_n=\sum_{k=1}^n\dfrac{(k+1)-k}{k(k+1)}[/tex]

   [tex]\displaystyle\longmapsto S_n=\sum_{k=1}^n\left[\dfrac{1}{k}-\dfrac{1}{k+1}\right][/tex]

   [tex]\displaystyle\longmapsto S_n=\sum_{k=1}^n\dfrac{1}{k}-\sum_{k=1}^n\dfrac{1}{k+1}[/tex]

   [tex]\displaystyle\longmapsto 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)[/tex]

   [tex]\displaystyle\longmapsto S_n=1-\dfrac{1}{n+1}[/tex]

   [tex]\displaystyle\longmapsto\underline{\boxed{S_n=\dfrac{n}{n+1}\qquad \star}}[/tex]

   For [tex]n\to\infty,[/tex]

   [tex]\displaystyle\longmapsto S_\infty=\lim_{n\to\infty}\dfrac{n}{n+1}[/tex]

   [tex]\displaystyle\longmapsto S_\infty=\lim_{n\to\infty}\dfrac{1}{\left(\dfrac{n+1}{n}\right)}[/tex]

   [tex]\displaystyle\longmapsto S_\infty=\lim_{n\to\infty}\dfrac{1}{\left(1+\dfrac{1}{n}\right)}[/tex]

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

   [tex]\displaystyle\longmapsto\underline{\boxed{S_\infty=1}}[/tex]

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