There is an alternative solution if you remember the sum of squares:
∑ k = 1 n k 2 = 2 n 3 + 3 n 2 + n 6 {\displaystyle \sum _{k=1}^{n}k^{2}={\frac {2n^{3}+3n^{2}+n}{6}}}
With this we quickly find
lim n → ∞ ∑ k = 1 n ( k n ) 2 1 n = lim n → ∞ 1 n 3 ∑ k = 1 n k 2 = lim n → ∞ 1 n 3 2 n 3 + 3 n 2 + n 6 = lim n → ∞ 2 + 3 1 n + 1 n 2 6 = 1 3 . {\displaystyle {\begin{aligned}\lim _{n\to \infty }\sum _{k=1}^{n}\left({\frac {k}{n}}\right)^{2}{\frac {1}{n}}&=\lim _{n\to \infty }{\frac {1}{n^{3}}}\sum _{k=1}^{n}k^{2}\\&=\lim _{n\to \infty }{\frac {1}{n^{3}}}{\frac {2n^{3}+3n^{2}+n}{6}}\\&=\lim _{n\to \infty }{\frac {2+3{\frac {1}{n}}+{\frac {1}{n^{2}}}}{6}}\\&={\frac {1}{3}}.\end{aligned}}}
