<s>[INST] 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?
 [/INST] You wish to derive a formula for the sum $\sum_{k=1}^n k^6$ using "high-school methods".
Consider $$\frac{x^7}7-\frac{(x-1)^7}7=x^6-3x^5+5x^4+\cdots\\\frac{x^6}6-\frac{(x-1)^6}6=x^5-\cdots\\\cdots$$
So, defining $$I_n:=\frac{n^7}7+\alpha_1\frac{n^6}6+\alpha_2\frac{n^5}5+\cdots+\alpha_6 n+\alpha_7$$ for some constants $\alpha_i$, you can eliminate all the other factors other than $n^6$. So you end up with $$I_n-I_{n-1}=n^6\\\implies \sum_{k=1}^n k^6=I_n-I_0=\frac{n^7}7+\text{lower order terms}$$
This is all that matters to taking your limit, since all the lower order terms go to zero when divided by the $n^7$ you have on the denominator.
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