<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] Note that
$$\lim_{n\to\infty}{\frac{1^6+2^6+3^6...+n^6}{(1^2+2^2+3^2...+n^2)(1^3+2^3+3^3...+n^3)}}=\lim_{n\to\infty} \frac{12\sum_1^n k^6}{n^7}=\frac{12}{7}$$
indeed by Stolz-Cesaro
$$\lim_{n\to\infty} \frac{12\sum_1^n k^6}{n^7}=\lim_{n\to\infty} \frac{12(n+1)^6}{(n+1)^7-n^7}=\lim_{n\to\infty} \frac{12(n+1)^6}{7n^6+...+1}=\frac{12}{7}$$
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