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:
$$\lim_{n\to \infty } \frac {1^3+4^3+7^3+...+(3n-2)^3}{[1+4+7+...+(3n-2)]^2}$$

You can simplify the fraction using
$$\sum_{k=1}^{n}k=\frac{n(n+1)}{2},\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6},\sum_{k=1}^{n}k^3=\left(\frac{n(n+1)}{2}\right)^2.$$
The numerator is 
$$\sum_{k=1}^{n}(3k-2)^3=\cdots=\frac 14 (27 n^4-18 n^3-9 n^2+4n).$$
The dinominator is 
$$\left(\sum_{k=1}^{n}(3k-2)\right)^2=\cdots=\frac 14 n^2 (3 n-1)^2=\frac 14(9n^4-6n^3+n^2).$$
Hence, your limit will be
$$\lim_{n\to\infty}\frac{27 n^4-18 n^3-9 n^2+4n}{9n^4-6n^3+n^2}=\lim_{n\to\infty}\frac{27-18(1/n)-9(1/n^2)+4(1/n^3)}{9-6(1/n)+(1/n^2)}=3.$$
