Question

The value of $\underset{n\to \infty }{lim}\frac{(1^2+2^2+3^2+\cdots +n^2)(1^3+2^3+3^3+\cdots +n^3)}{1^6+2^6+3^6+\cdots +n^6}$ is

• A. $\frac{7}{12}$
• B. $\frac{5}{7}$
• C. $\frac{6}{13}$
• D. $\frac{9}{17}$

Answer 1

The correct option is A $\frac{7}{12}$

The given limit is:
$\underset{n\to \infty }{lim}\frac{(1^2+2^2+3^2+\cdots +n^2)(1^3+2^3+3^3+\cdots +n^3)}{1^6+2^6+3^6+\cdots +n^6}$
$=\underset{n\to \infty }{lim}\frac{\sum _{r=1}^n{r}^2\times \sum _{r=1}^n{r}^3}{\sum _{r=1}^n{r}^6}$
$=\underset{n\to \infty }{lim}\frac{\frac{1}{n}\sum _{r=1}^n{\left(\frac{r}{n}\right)}^2\times \frac{1}{n}\sum _{r=1}^n{\left(\frac{r}{n}\right)}^3}{\frac{1}{n}\sum _{r=1}^n{\left(\frac{r}{n}\right)}^6}$
$=\frac{\underset{0}{\overset{1}{\int }}{x}^{2}dx\cdot \underset{0}{\overset{1}{\int }}{x}^{3}dx}{\underset{0}{\overset{1}{\int }}{x}^{6}dx}=\frac{{\left[\frac{x^3}{3}\right]}_0^1\cdot {\left[\frac{x^4}{4}\right]}_0^1}{{\left[\frac{x^7}{7}\right]}_0^1}=\frac{\frac{1}{3}\times \frac{1}{4}}{\frac{1}{7}}=\frac{7}{12}$