<s>[INST] Help with limit calculation Can anyone help me with this limit please:
I have been trying to solve this for 2 hours with no success:
$$\lim_{n\to \infty } \frac {1^3+4^3+7^3+...+(3n-2)^3}{[1+4+7+...+(3n-2)]^2}$$
 [/INST] We don't need to make all the detailed calculations as the highest exponent dominates when $n\to\infty$   
$$1+4+7+...+(3n-2)=\frac n2\left(1+3n-2\right)=\frac{3n^2-n}2$$
$$\implies \left(1+4+7+...+(3n-2)\right)^2=\left(\frac{3n^2-n}2\right)^2=\frac{9n^4}4+O(n^3)$$
$$1^3+4^3+7^3+...+(3n-2)^3=\sum_{1\le r\le n}(3r-2)^3=\sum_{1\le r\le n}(27r^3-54r^2+36r-8)$$
$$=27\sum_{1\le r\le n}r^3-54\sum_{1\le r\le n}r^2+36\sum_{1\le r\le n}r-\sum_{1\le r\le n}8=27\left(\frac{n(n+1)}2\right)^2+O(n^3)=\frac{27n^4}4+O(n^3)$$
So, the limit$(F)$ will be $$\lim_{n\to\infty}\frac{\frac{9n^4}4+O(n^3)}{\frac{27n^4}4+O(n^3)}$$
Dividing the numerator & the denominator by $n^4,$
$$F=\lim_{n\to\infty}\frac{\frac{27}4+O\left(\frac1n\right)}{\frac94+O\left(\frac1n\right)}=\frac{\frac{27}4}{\frac94}$$
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