Question

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

Answer 1

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

Dividing the numerator and the denominator by n⁴:

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