Question

$\underset{k\to \infty }{\lim }$ $\left(\frac{1^3+2^3+3^3+...+k^3}{k^4}\right)$ is equal to

• A. $0$
• B. $2$
• C. $\frac{1}{3}$
• D. $\infty$
• E. $\frac{1}{4}$

Answer 1

Answer: (E)

$\underset{k\to \infty }{\lim }\left(\frac{1^3+2^3+3^3+\dots +k^3}{k^4}\right)$
$=\underset{k\to \infty }{\lim }\left(\frac{k^2(k+1{)}^2}{4}\times \frac{1}{k^4}\right)$
$\left\{\because 1^3+2^3+\dots +k^3={\left[\frac{k(k+1)}{2}\right]}^2\right\}$
$=\underset{k\to \infty }{\lim }\left(\frac{k^4(1+1/k{)}^2}{4}\times \frac{1}{k^4}\right)$
$=\frac{(1+0{)}^2}{4}=\frac{1}{4}$