Question

If$\underset{n\to \infty }{lim}(\frac{3}{1\cdot 2\cdot 4}+\frac{4}{2\cdot 3\cdot 5}+\frac{5}{3\cdot 4\cdot 6}+\cdots +\frac{n+2}{n(n+1)(n+3)})$ can be expressed as rational in the lowest form $\frac{m}{n}$ where $m,n\in N,$ then the value of $(m+n)$ is

Answer 1

$T_n=\frac{(n+2)}{n(n+1)(n+3)}$
$\Rightarrow T_n=\frac{1}{2}\left(\frac{2n+4}{n(n+1)(n+3)}\right)$
$\Rightarrow T_n=\frac{1}{2}\left(\frac{(n+3)+(n+1)}{n(n+1)(n+3)}\right)$
$\Rightarrow T_n=\frac{1}{2}[\frac{1}{n(n+1)}+\frac{1}{n(n+3)}]$
$\Rightarrow T_n=\frac{1}{2}(\frac{1}{n}-\frac{1}{n+1})+\frac{1}{6}(\frac{1}{n}-\frac{1}{n+3})$

$\sum T_n=\frac{1}{2}(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots +\frac{1}{n}-\frac{1}{n+1})+\frac{1}{6}(\frac{1}{1}-\frac{1}{4}+\frac{1}{2}-\frac{1}{5}+\frac{1}{3}-\frac{1}{6}+\frac{1}{4}-\frac{1}{7}+\cdots +\frac{1}{n}-\frac{1}{n+3})$
$\Rightarrow \sum T_n=\frac{1}{2}(\frac{1}{1}-\frac{1}{n+1})+\frac{1}{6}(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}-\frac{1}{n+3})$
$\Rightarrow \underset{n\to \infty }{lim}\sum T_n=\frac{1}{2}(\frac{1}{1}-0)+\frac{1}{6}(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}-0)$
$\Rightarrow \underset{n\to \infty }{lim}\sum T_n=\frac{1}{2}+\frac{1}{6}\left(\frac{11}{6}\right)=\frac{18+11}{36}$
$\Rightarrow \underset{n\to \infty }{lim}\sum T_n=\frac{29}{36}=\frac{m}{n}$
$\Rightarrow m=29$ and $n=36$
$\therefore m+n=29+36=65$