Question

$$\underset{n\to \mathrm{\infty }}{lim}\left[\frac{1^3+2^3+...n^3}{{(n-1)}^4}\right]$$ 

Answer 1

$$\underset{n\to \mathrm{\infty }}{lim}\left[\frac{1^3+2^3+...+n^3}{{(n-1)}^4}\right]$$
$$=\underset{n\to \mathrm{\infty }}{lim}\left[\frac{{\left[\frac{n(n+1)}{2}\right]}^2}{{(n-1)}^4}\right]$$
$$=\underset{n\to \mathrm{\infty }}{lim}\left[\frac{n^2{(n+1)}^2}{4{(n-1)}^4}\right]$$

Dividing the numerator and the denominator by n⁴: 

$$\underset{n\to \mathrm{\infty }}{lim}\left[\frac{\frac{n^2{(n+1)}^2}{n^4}}{4\frac{{(n-1)}^4}{n^4}}\right]$$
$$=\underset{n\to \mathrm{\infty }}{lim}\left[\frac{\frac{{(n+1)}^2}{n^2}}{4{\left(\frac{n-1}{n}\right)}^4}\right]$$
$$=\underset{n\to \mathrm{\infty }}{lim}\left[\frac{{(1+\frac{1}{n})}^2}{4{(1-\frac{1}{n})}^4}\right]$$
$$\text{ When }n\to \mathrm{\infty },\text{ then }\frac{1}{n}\to 0.$$
$$\frac{{(1+0)}^2}{4{(1-0)}^4}$$
$$=\frac{1}{4}$$