Question

Evaluate:

$\underset{n\to \infty }{lim}[\frac{1}{n^3}+\frac{2^2}{n^3}+\frac{3^2}{n^3}+\dots +\frac{n^2}{n^3}]$

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

Answer 1

Answer: (C)

$\frac{1}{3}$

Given, $\underset{n\to \infty }{lim}[\frac{1}{n^3}+\frac{2^2}{n^3}+\frac{3^2}{n^3}+\dots +\frac{n^2}{n^3}]$

$=\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{n}{\left(\frac{r}{n}\right)}^2={\int }_0^1{x}^{2}dx$

$\Rightarrow {\int }_0^1{x}^{2}dx$$={\left(\frac{x^3}{3}\right)}_0^1$

$=(\frac{1}{3}-0)=\frac{1}{3}$

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