Question

The value of the following expression is $\underset{n\to \infty }{lim}\frac{1}{n^3}(1^2+2^2+3^2+\cdots +n^2)$.

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. $\frac{1}{6}$

Answer 1

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

$\underset{n\to \infty }{lim}\frac{1}{n^3}(1^2+2^2+3^2+\cdots +n^2)$
$=\underset{n\to \infty }{lim}\frac{1}{n^3}\left[\frac{n(n+1)(2n+1)}{6}\right]$
$=\underset{n\to \infty }{lim}\frac{(1+\frac{1}{n})(2+\frac{1}{n})}{6}$
$=\frac{1\times 2}{6}=\frac{1}{3}$.