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?

Here is yet another answer, which does not require any of the full, explicit formulae for sums of powers.
Consider the sum
$$
S(n, p) = \sum_{m = 1}^{n} m^p\, .
$$
I can rewrite this as
$$
S(n, p) = n^{p+1}\sum_{m = 1}^{n} \frac{1}{n}\;\left(\frac{m}{n}\right)^p\, .
$$
The sum in the above expression is a Riemann sum which approximates the integral
$$
\sum_{m = 1}^{n}\frac{1}{n}\;\left(\frac{m}{n}\right)^p
\approx \int_0^1 x^p\, dx \;=\; \frac{1}{p+1}\, .
$$
Thus, for large $n$
$$
S(n, p) \approx \frac{n^{p+1}}{p+1}\, .
$$
We can use this expression to replace all three sums in your expression:
$$
\lim_{n\rightarrow\infty} \frac{1^6 + \;...\; + n^6}{\left(1^2 +\;...\;+ n^2\right) \left(1^3 +\;...\;+ n^3\right)}
\;=\;
\lim_{n\rightarrow\infty} \frac{n^7 / 7}{\left(n^3 / 3\right) \left(n^4 / 4\right)}
\;=\; \frac{12}{7}
$$
