Mathematics - Sequence & Series Question with Solution | TestHub

$a_n=\frac{(2-1)(3-1)\dots \dots (n-1)}{(2+1)(3+1)\dots (n+1)}\times \frac{\left(2^2+2+1\right)\left(3^2+3+1\right)\dots \left(n^2+n+1\right)}{\left(2^2-2+1\right)\left(3^2-3+1\right)\dots .\left(n^2-n+1\right)}$ Since $(k+1{)}^2-(k+1)+1=k^2+k+1$, we get $a_n=\frac{(1)(2)}{n(n+1)}\frac{\left(n^2+n+1\right)}{3}$

$$\Rightarrow \underset{n\to \infty }{\lim }{a}_n=\frac{2}{3}=0.67$$