Question

Evaluate:
$\underset{n\to \infty }{lim}\frac{(1^3+2^3+3^3+...+n^3)(1^4+2^4+3^4+...+n^4)}{(1^8+2^8+3^8+...+n^8)}$

Answer 1

$\underset{n\to \infty }{lim}\frac{\sum _{r=1}^n{r}^3\sum _{r=1}^n{r}^4}{\sum _{r=1}^n{r}^8}$

$=\underset{n\to \infty }{lim}\frac{\frac{1}{n}\sum _{r=1}^n{\left(\frac{r}{n}\right)}^3\times \frac{1}{n}\sum _{r=1}^n{\left(\frac{r}{n}\right)}^4}{\frac{1}{n}\sum _{r=1}^n{\left(\frac{r}{n}\right)}^8}$

$=\frac{\underset{0}{\overset{1}{\int }}{x}^{3}dx\times \underset{0}{\overset{1}{\int }}{x}^{4}dx}{\underset{0}{\overset{1}{\int }}{x}^{8}dx}$

$=\frac{[\frac{x^4}{4}{]}_0^1\times [\frac{x^5}{5}{]}_0^1}{[\frac{x^9}{9}{]}_0^1}=\frac{\frac{1}{4}\times \frac{1}{5}}{\frac{1}{9}}=0.45$