Evaluate limit containing $\sum{n^6}$ Evaluate:
$$\lim_{n\to\infty}{\frac{1^6+2^6+3^6+\ldots+n^6}{(1^2+2^2+3^2+\ldots+n^2)(1^3+2^3+3^3+\ldots+n^3)}}$$
I can solve the denominator as:
$$\frac{n(n+1)(2n+1)}{6}\cdot\frac{n^2(n+1)^2}{4}$$
$$n^7\cdot\frac{(1+\frac{1}{n})(2+\frac{1}{n})}{6}\cdot\frac{(1+\frac{1}{n})}{4}$$
$$=\frac{n^7}{12}$$
How can I reduce the numerator?

Using Cesaro-Stolz (or limit of a Riemann sum) one can easily prove that $$\lim_{n\to\infty} \frac{1^p+2^p+\dots +n^p} {n^{p+1}}=\frac{1}{p+1}, p>-1$$ For the current question divide the numerator and denominator by $n^7=n^3\cdot n^4$ and then the limit is equal to $1/7/((1/3)(1/4))=12/7$.
