\(\dfrac{n^3-1}{n^3+1}=\dfrac{\left(n-1\right)\left(n^2+n+1\right)}{\left(n+1\right)\left(n^2-n+1\right)}=\dfrac{\left(n-1\right)\left[\left(n+1\right)^2-\left(n+1\right)+1\right]}{\left(n+1\right)\left(n^2-n+1\right)}\)
\(\Rightarrow u_n=\dfrac{1.\left(3^2-3+1\right)}{3.\left(2^2-2+1\right)}.\dfrac{2.\left(4^2-4+1\right)}{4.\left(3^2-3+1\right)}...\dfrac{\left(n-1\right)\left[\left(n+1\right)^2-\left(n+1\right)+1\right]}{\left(n+1\right)\left(n^2-n+1\right)}\)
\(=\dfrac{1.2.\left[\left(n+1\right)^2-\left(n+1\right)+1\right]}{\left(2^2-2+1\right)n\left(n+1\right)}=\dfrac{2\left(n^2+n+1\right)}{3n\left(n+1\right)}\)
\(\Rightarrow lim\left(u_n\right)=\dfrac{2}{3}\)
