Question

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

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $0$

Answer 1

The correct option is C $\frac{1}{3}$

$\underset{n\to \infty }{lim}\frac{1^2+2^2+3^2+.......+n^2}{n^3}$
$=\underset{n\to \infty }{lim}\frac{n(n+1)(2n+1)}{6n^3}$
$=\frac{1}{6}\underset{n\to \infty }{lim}\frac{1(1+\frac{1}{n})(2+\frac{1}{n})}{1}$
$=\frac{1}{6}(1+0)(2+0)=\frac{1}{3}$
Hence, option 'C' is correct.