Help with limit calculation

Question

Can anyone help me with this limit please:

I have been trying to solve this for 2 hours with no success: $$\lim_{n\to \infty } \frac {1^3+4^3+7^3+...+(3n-2)^3}{[1+4+7+...+(3n-2)]^2}$$

Answer 1

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 $$\lim_{n\to\infty}\frac{27 n^4-18 n^3-9 n^2+4n}{9n^4-6n^3+n^2}=\lim_{n\to\infty}\frac{27-18(1/n)-9(1/n^2)+4(1/n^3)}{9-6(1/n)+(1/n^2)}=3.$$

Answer 2

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 $n^4,$

$$F=\lim_{n\to\infty}\frac{\frac{27}4+O\left(\frac1n\right)}{\frac94+O\left(\frac1n\right)}=\frac{\frac{27}4}{\frac94}$$

Answer 3

Approximate both sums by integrals (for example, the numerator by $\int_1^n (3x-2)^3 d x$), and you should be enlightened...