Question

Sum the following series to n terms and to infinity $\frac{4}{\mathrm{1.2.3}}+\frac{5}{\mathrm{2.3.4}}+\frac{6}{\mathrm{3.4.5}}+....$.

Answer 1

$n^{th}$ term $=\frac{n+3}{n(n+1)(n+2)}$

$=\frac{n}{n(n+1)(n+2)}+\frac{3}{n(n+1)(n+2)}$

$=\frac{1}{(n+1)(n+2)}+\frac{3}{n(n+1)(n+2)}$

$=A_n+3B_n$

where $A_n=\frac{1}{(n+1)(n+2)}$ and $B_n=\frac{3}{n(n+1)(n+2)}$

$A_n$ by partial fraction is

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

Summation of $A_n$ will be

$\sum _{n=1}^n[\frac{1}{n+1}-\frac{1}{n+2}]$

$=[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}.....\frac{1}{(n+1)}-\frac{1}{(n+2)}]$

$=\frac{1}{2}-\frac{1}{n+2}$

Similarly for $B_n$

$B_n=\left[\frac{1}{2}(\frac{1}{n}-\frac{2}{n+1}+\frac{1}{n+2})\right]$

$B_n=\frac{1}{2}\{[\frac{1}{n}-\frac{1}{n+1}]-[\frac{1}{n+1}-\frac{1}{n+2}]\}$

Summation of $B_n$ will be

$\frac{1}{2}\{\left[\sum _{n=1}^n\frac{1}{n}-\frac{1}{n+1}\right]-\left[\sum _{n=1}^n\frac{1}{n+1}-\frac{1}{n+2}\right]\}$

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

Combining $A_n+3B_n$

$\Rightarrow S_n={\sum }_{n=1}^n(A_n+3B_n)=[\frac{1}{2}-\frac{1}{n+2}]+\frac{3}{2}[\frac{1}{2}-\frac{1}{n+1}+\frac{1}{n+2}]$

Let $n\to \infty$

$S_{\infty }=[\frac{1}{2}-0]+\frac{3}{2}[\frac{1}{2}-0-0]$

$=\frac{1}{2}+\frac{3}{4}$

$=\frac{5}{4}$