Question

If $S_n=\frac{n(n+1)(n+2)(n+3)}{12}$ then prove that $\underset{n\to \infty }{Lt}{\sum }_{r=1}^n\frac{1}{t_r}=\frac{3}{4}$

Answer 1

$t_n=S_n-S_{n-1}=\frac{1}{3}n(n+1)(n+2)$
$\therefore \frac{1}{t_n}=3\frac{1}{n(n+1)(n+2)}$
$=3[\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2(n+2)}]Partialfractions$
$=\frac{3}{2}[\frac{1}{n}-\frac{2}{n+1}+\frac{1}{n+2}]$
$=\frac{3}{2}[(\frac{1}{n}-\frac{1}{n+1})-(\frac{1}{n+1}-\frac{1}{n+2})]$
Putting n = 1,2,3,...n and adding
$\therefore {\sum }_{r=1}^n\frac{1}{t_r}==\frac{3}{2}[(1-\frac{1}{n+1})-(\frac{1}{2}-\frac{1}{n+2})]$
$\underset{n\to \infty }{L}t{\sum }_{r=1}^n\frac{1}{t_r}=\frac{3}{2}\left[\right(1-0)-(\frac{1}{2}-0)]=\frac{3}{4}\cdot \frac{1}{2}=\frac{3}{4}$