Given, Lim_n->∞  {1² + 2² + 3² + ...... + n² }/n³

= Lim_n->∞  [{n*(n + 1)*(2n + 1)}/6]/{n(n + 1)/2}²

= Lim_n->∞  [{n*n*n *(1 + 1/n)*(2 + 1/n)}/6]/{n * n *(1 + 1/n)/2}²

= Lim_n->∞  [{n³ *(1 + 1/n)*(2 + 1/n)}/6]/{n² *(1 + 1/n)/2}²

= Lim_n->∞  [{(1 + 1/n)*(2 + 1/n)}/6]/[n⁴ * {(1 + 1/n)/2}² ]

=> Lim_n->∞  [{(1 + 1/n)*(2 + 1/n)}/6]/[n * {(1 + 1/n)/2}² ]

= [{(1 + 1/∞)*(2 + 1/∞)}/6]/[∞*{(1 + 1/∞)/2}²

= [{(1 + 0)*(2 + 0)}/6]/∞                    {since 1/∞ = 0}

= {(1 * 2)/6}/∞

= (2/6)/∞

= (1/3)/∞

= 0

So, Lim_n->∞  {1² + 2² + 3² + ...... + n² }/n³ = 0