Question

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

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

Answer 1

Answer: (E)

$\displaystyle\lim _{k \rightarrow \infty}\left(\frac{1^{3}+2^{3}+3^{3}+\ldots+k^{3}}{k^{4}}\right)$
$=\displaystyle\lim _{k \rightarrow \infty}\left(\frac{k^{2}(k+1)^{2}}{4} \times \frac{1}{k^{4}}\right)$
$\left\{\because 1^{3}+2^{3}+\ldots+k^{3}=\left[\frac{k(k+1)}{2}\right]^{2}\right\}$
$=\displaystyle\lim _{k \rightarrow \infty}\left(\frac{k^{4}(1+1 / k)^{2}}{4} \times \frac{1}{k^{4}}\right)$
$=\frac{(1+0)^{2}}{4}=\frac{1}{4}$