Question

Find the value of the given limit: ${lim}_{n\to \infty }\frac{1^2+2^2+3^2+\_\_\_\_+n^2}{n^3}$

Answer 1

${lim}_{n\to \infty }\frac{1^2+2^2+3^2+....+n^2}{n^3}{lim}_{n\to \infty }\frac{n(n+1)(2n+1)}{6n^3}={lim}_{n\to \infty }\frac{(n^2+n)(2n+1)}{6n^3}{lim}_{n\to \infty }\frac{2n^3+n^2+2n^2+n}{6n^3}$

$={lim}_{n\to \infty }\frac{2n^3}{6n^3}+\frac{n^2}{6n^3}+\frac{2n^2}{6n^3}+\frac{n}{6n^3}={lim}_{n\to \infty }\frac{1}{3}+0=\frac{1}{3}$