If

$$A_n = \frac{1.2.3 + 2.3.4 + 3.4.5 + .... upto n terms}{n(1.2 + 2.3 + 3.4 + .... upto n terms)}$$ 

then $$\lim_{n \rightarrow \infty} A_n$$ is

$$\frac{Numerator}{n}$$ = $$\Sigma\ n\left(n+1\right)\left(n+2\right)\ =\ \frac{\left(\Sigma\ n^3+3\Sigma n^2+2\Sigma\ n\ \right)}{n}$$

$$\frac{n^2\left(n+1\right)^2}{4n}+\frac{3n\left(n+1\right)\left(2n+1\right)}{6n}+\frac{2n\left(n+1\right)}{2n}$$

Denominator 

$$\Sigma\ n\left(n+1\right)\ =\ \Sigma\ n^2+n\ =\ \frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{n\left(n+1\right)}{2}$$

Numerator/Denominator = $$\frac{3}{4}\left[1+\frac{3}{n}\right]$$

When n -> $$\infty\ ,\ \frac{1}{n}->\ 0$$

= 3/4