Question

$\underset{n\to \infty }{lim}\frac{(1^2+2^2+3^2+...+n^2)(1^3+2^3+3^3+...+n^3)}{1^6+2^6+3^6+...+n^6}$.

Answer 1

We know that

$\left(i\right){\sum }_{k=1}^n{k}^2$

$=1^2+2^2+3^2+4^2+...+n^2$

$=\frac{1}{6}n(n+1)(2n+1)$

$=\frac{1}{6}{n}^3(1+\frac{1}{n})(2+\frac{1}{n})$

$\left(ii\right){\sum }_{k=1}^n{k}^3$

$=1^3+2^3+3^3+4^3+...+n^3$

$=\frac{1}{4}{n}^4+\frac{1}{2}{n}^3+\frac{1}{4}{n}^2$

$=n^4(\frac{1}{4}+\frac{1}{2n}+\frac{1}{4n^2})$

$\left(iii\right){\sum }_{k=1}^n{k}^6$

$=1^6+2^6+3^6+4^6+...+n^6$

$=\frac{1}{7}{n}^7+\frac{1}{2}{n}^6+\frac{1}{2}{n}^5-\frac{1}{6}{n}^3+\frac{1}{42}n$

$=n^7(\frac{1}{7}+\frac{1}{2n}+\frac{1}{2n^2}-\frac{1}{6n^4}+\frac{1}{42n^6})$

We have ${lim}_{n\to \infty }\frac{(1^2+2^2+3^2+4^2+...+n^2)(1^3+2^3+3^3+4^3+...+n^3)}{1^6+2^6+3^6+4^6+...+n^6}$

$={lim}_{n\to \infty }\frac{\left(\frac{1}{6}{n}^3(1+\frac{1}{n})(2+\frac{1}{n})\right)\left(n^4(\frac{1}{4}+\frac{1}{2n}+\frac{1}{4n^2})\right)}{n^7(\frac{1}{7}+\frac{1}{2n}+\frac{1}{2n^2}-\frac{1}{6n^4}+\frac{1}{42n^6})}$

$=\frac{\frac{1}{6}(1+0)(2+0)(\frac{1}{4}+0+0)}{(\frac{1}{7}+0+0-0+0)}$ since as $x\to \infty \Rightarrow \frac{1}{x}\to 0$

$=\frac{\frac{1}{6}\times 1\times 2\times \frac{1}{4}}{\frac{1}{7}}$

$=\frac{2}{24}\times 7=\frac{7}{12}$