Question

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

Answer 1

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

We know that,
$(1^2+2^2+3^2+...+n^2)=\frac{n(n+1)(2n+1)}{6}$
$\Rightarrow \underset{n\to \infty }{lim}\frac{1}{n^3}(1^2+2^2+3^2+\cdots +n^2)=\underset{n\to \infty }{lim}\frac{n(n+1)(2n+1)}{6n^3}$
$=\underset{n\to \infty }{lim}\frac{(1+\frac{1}{n})(2+\frac{1}{n})}{6}=\frac{1\times 2}{6}=\frac{1}{3}$