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?

The sum of the $k^{th}$ power of the integers is a polynomial of degree $k+1$ in $n$, with leading term $\dfrac{n^{k+1}}{k+1}$.
Indeed, by the binomial theorem,
$$n^k=\sum_{i=1}^ni^k-\sum_{i=1}^{n-1}i^k\sim\frac{n^{k+1}}{k+1}-\frac{(n-1)^{k+1}}{k+1}=n^k+\text{lower degree terms}.$$
The given limit equals that of
$$\frac{n^7\cdot3\cdot4}{7\cdot n^3\cdot n^4}.$$
