קודם כל מתקיים:
\lim_{n\to\infty}\left(\frac{1^{2}+2^{2}+...+n^{2}}{n^{3}}\right)=\lim_{n\to\infty}\left(\frac{\sum_{k=1}^{n}k^{2}}{n^{3}}\right)
נשתמש בזהות \sum_{k=1}^{n}k^{2}=\frac{n}{6}(n+1)(2n+1) , ולכן:
\begin{align*}
\lim_{n\to\infty}\left(\frac{\sum_{k=1}^{n}k^{2}}{n^{3}}\right)&=\lim_{n\to\infty}\left(\frac{\frac{n}{6}(n+1)(2n+1)}{n^{3}}\right)=\lim_{n\to\infty}\left(\frac{n(n+1)(2n+1)}{6n^{3}}\right)\\&=
\lim_{n\to\infty}\left(\frac{n(2n^{2}+3n+1)}{6n^{3}}\right)=\lim_{n\to\infty}\left(\frac{2n^{2}+3n+1}{6n^{2}}\right)
\end{align*}
נחלק בחזקה הגבוהה:
\lim_{n\to\infty}\left(\frac{\frac{2n^{2}}{n^{2}}+\frac{3n}{n^{2}}+\frac{1}{n^{2}}}{\frac{6n^{2}}{n^{2}}}\right)=\frac{2+0+0}{6}=\frac{1}{3}