Question

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

Answer 1

$$\begin{gathered} \underset{n\to \infty }{\lim }\left[\frac{1^3+2^3+...+n^3}{n^4}\right] \\ =\underset{n\to \infty }{\lim }\left[\frac{{\left[\frac{n\left(n+1\right)}{2}\right]}^2}{n^4}\right] \\ =\underset{n\to \infty }{\lim }\left[\frac{n^2{\left(n+1\right)}^2}{4n^4}\right] \\ =\underset{n\to \infty }{\lim }\left[\frac{n^2{\left(n+1\right)}^2}{4n^4}\right] \\ =\underset{n\to \infty }{\lim }\left[\frac{{\left(n+1\right)}^2}{4n^2}\right] \\ =\underset{n\to \infty }{\lim }\left[{\left(\frac{n+1}{n}\right)}^2\times \frac{1}{4}\right] \\ =\underset{n\to \infty }{\lim }\left[{\left(1+\frac{1}{n}\right)}^2\times \frac{1}{4}\right] \\ \mathrm{When}n\to \infty ,\mathrm{then}\frac{1}{n}\to 0. \\ \Rightarrow \frac{1}{4} \end{gathered}$$